NOTES ON GALOIS THEORY - Department of Mathematics ...people.math.umass.edu/~tevelev/611_2012/galois_notes.pdf · NOTES ON GALOIS THEORY 3 of Fgenerated by Kand by and K( ) denotes

NOTES ON GALOIS THEORY

JENIA TEVELEV

CONTENTS

§1. Algebraic Extensions 2§1.1. Field extensions 2§1.2. Multiplicativity of degree 4§1.3. Algebraic extensions 4§1.4. Adjoining roots 5§1.5. Splitting fields 5§1.6. Algebraic closure 7§1.7. Finite fields 9§1.8. Composite field 9§1.9. Exercises 10§2. Galois Theory 11§2.1. Separable extensions 11§2.2. Normal extensions 13§2.3. Main Theorem of Galois Theory 13§2.4. Fields of invariants 16§2.5. Exercises 17§3. Applications of Galois Theory 17§3.1. Intermediate subfields in separable extensions 17§3.2. Fundamental Theorem of Algebra 18§3.3. Quadratic extensions 18§3.4. Cubic extensions 18§3.5. Galois group of a finite field 19§3.6. Exercises 20§4. Roots of unity 21§4.1. Adjoining roots of unity 21§4.2. Cyclotomic fields 22§4.3. Rational cosines 23§4.4. Abelian extensions of Q 23§4.5. Quadratic reciprocity 25§4.6. Exercises 26§5. Integral extensions of rings 28§5.1. Integral extensions 28§5.2. Examples of the integral closure 30§5.3. Exercises 32§6. Cyclic extensions 32§6.1. Cyclic extensions 32§6.2. Artin’s Lemma 34§6.3. Lagrange resolvents 35

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2 JENIA TEVELEV

§6.4. Exercises 36§7. Norm and Trace 36§7.1. Exercises 37§8. Solvable extensions 37§8.1. Composition series 38§8.2. Solvable groups 39§8.3. Solvable extensions: Galois Theorem. 40§8.4. Solving solvable extensions 42§8.5. Exercises 43§9. Sample midterm 44

§1. ALGEBRAIC EXTENSIONS

§1.1. Field extensions. One rarely studies a single field. A typical situationis to have two fields K ⊂ F . Then F is called a field extension of K. Thiswill be a basic setup for this section. Another notation for this: F/K.

Examples: R ⊂ C, Q ⊂ Q(√

2), etc.Amy field extension F/K can be viewed as a vector space overK (how?)

DEFINITION 1.1.1. The dimension dimK F is called the degree of F over K.Notation: [F : K]. If [F : K] <∞ then F is called a finite extension of K.

For example, [C : R] = 2. The basis of C over R is given by 1 and i.We also have [Q(

√2) : Q] = 2. The basis is given by 1 and

√2. Notice that

this is a non-trivial statement: one needs to know that• 1 and

√2 are linearly independent over Q, i.e. a + b

√2 6= 0 if a or b

is not equal to zero. This is because√

2 is irrational,√

2 6= −a/b.• 1 and

√2 generate Q(

√2) as a vector space over Q. Apriori, an ele-

ment of Q(√

2) is a fraction a+b√

2c+d√

2with a, b, c, s ∈ Q. However, mul-

tiplying the numerator and denominator by c− d√

2 we can rewritethis fraction as a linear combination of 1 and

√2.

We are going to study extensions like this very systematically.

DEFINITION 1.1.2. Consider a field extension K ⊂ F . An element α ∈ Fis called algebraic over K if α is a root of a non-trivial polynomial withcoefficients in K. An element α is called transcendental if it is not algebraic.

• i ∈ C is a root of x2 − 1, so i is algebraic over Q.• π ∈ R is transcendental over Q (Lindemann’s Theorem).• x ∈ C(x) is transcendental over C (obvious).

DEFINITION 1.1.3. Consider a field extension K ⊂ F . Let α ∈ F be alge-braic. A polynomial f(x) ∈ K[x] is called a minimal polynomial of α if f(x)is a monic polynomial of minimal degree such that f(α) = 0.

Let us point out one persistent notational ambiguity. If x is a variablethenK[x] denotes the algebra of polynomials andK(x) denotes the field ofrational functions (i.e. ratios of polynomials) in variable x. But given a fieldextension K ⊂ F and an element α ∈ F , K[α] denotes the minimal subring

NOTES ON GALOIS THEORY 3

of F generated by K and by α and K(α) denotes the minimal subfield of Fgenerated by K and by α. These objects are related as follows:

THEOREM 1.1.4. Consider a field extension K ⊂ F . Let α ∈ F . Consider aunique surjective homomorphism φ : K[x]→ K[α] that sends x to α.

• If α is algebraic then Kerφ is generated by the minimal polynomial f(x),which is irreducible and defined uniquely. Let n = deg f(x). Then[K(α) : K] = n. Moreover, K[α] = K(α). More precisely, elements1, α, . . . , αn−1 form a K-basis of K(α).• If α is transcendental then φ is an isomorphism, which induces an isomor-

phism of fields K(x) ' K(α). In particular, [K(α) : K] =∞.

Proof. Kerφ is an ideal of all polynomials that vanish at α. Kerφ = 0 if andonly if α is transcendental (by definition). In this case

K[x] ' K[α] ⊂ F.Any injective homomorphism of a domain into a field extends to the inhec-tive homomorphism of its field of fractions. So in our case φ extends to theinjective homomorphism K(x)→ F , and its image is obviously K(α).

If α is algebraic then, since K[x] is a PID, Kerφ is generated by a uniquemonic polynomial f(x), hence a minimal polynomial is unique. Since

K[x]/Kerφ ' K[α]

injects in F , which is a domain, K[x]/Kerφ is also a domain, hence f(x) isirreducible and Kerφ = (f) is a maximal ideal. HenceK[x]/Kerφ is a field.Therefore, K[α] is a field. Therefore, K[α] = K(α).

Finally, we notice that 1, α, . . . , αn−1 are linearly independent over K(otherwise we can find a smaller degree polynomial vanishing at α) andspan K[α] over K. Indeed, since f(x) = xn + . . . vanishes at α, we canrewrite αn as a linear combination of smaller powers of α. Then, an easyargument by induction shows that we can rewrite any αm for m > n as asa linear combination of 1, α, . . . , αn−1. �

An often used corollary:

COROLLARY 1.1.5. If α is algebraic overK then [K(α) : K] is equal to the degreeof the minimal polynomial of α.

The number [K(α) : K] is also often called the degree of α over K.

EXAMPLE 1.1.6. [Q(√

2 +√

3) : Q] = 4. This calculation involves a trickwhich will become much more transparent after we discuss Galois theory.The polynomial of degree 4

f(x) =∏

(x±√

2±√

3)

obviously has√

2 +√

3 as a root. To show that it is irreducible, we haveto show that it has no linear and quadratic factors over Q. If it has a linearfactor then

√2 ±√

3 ∈ Q, therefore (√

2 ±√

3)2 ∈ Q, therefore√

6 ∈ Q.Then we use the standard argument: if

√6 = a/b in lowest terms then

a2 = 6b2, therefore 2|a, therefore 4|a2, therefore 2|b, contradiction. If f(x)has a quadratic factor over Q then the sum of two roots of f(x) belongsto Q, which implies that

√2 or

√3 belong to Q, unless the two roots are

4 JENIA TEVELEV

opposite, say√

2+√

3 and−√

2−√

3. But then their product is in Q, whichimplies that

√6 ∈ Q, a contradiction.

EXAMPLE 1.1.7. [Q( 3√

2) : Q] and the minimal polynomial is x3 − 2 (irre-ducible by Eisenstein’s criterion).

§1.2. Multiplicativity of degree.

LEMMA 1.2.1. Consider a tower of field extensions

K ⊂ F ⊂ L.

If [F : K] = n and [L : F ] = m then [L : K] = nm.

Proof. It is easy to prove a bit more: if e1, . . . , em is a basis of L over F andf1, . . . , fn is a basis of F overK then {eifj} is a basis ofL overK (why?) �

This statement has interesting divisibility consequences. For example, ifK/Q is a finite field extension containing

√2, then [K : Q] is even because

[K : Q] = [Q(√

2) : Q][K : Q(√

2)] = 2[K : Q(√

2)].

§1.3. Algebraic extensions.

DEFINITION 1.3.1. An extension K ⊂ F is called algebraic if any element ofF is algebraic over K.

THEOREM 1.3.2. (a) Any finite extension is algebraic.(b) Let α1, . . . , αr ⊂ F be algebraic over K. Then

K(α1, . . . , αr) = K[α1, . . . , αr]

is finite over K. In particular, any element of K(α1, . . . , αr) is algebraic over K.

Proof. If [F : K] < ∞ then 1, α, α2, . . . are linearly dependent over K forany α ∈ F . Therefore, some non-constant polynomial with coefficient in Kvanishes at α, i.e. α is algebraic.

Now suppose that α1, . . . , αr ⊂ F are algebraic over K. Arguing byinduction on r (with the base of induction given by Theorem 1.1.4), let’sassume that K(α1, . . . , αr−1) = K[α1, . . . , αr−1] is finite over K. Since αris algebraic over K, it is also algebraic over K(α1, . . . , αr−1). Using Theo-rem 1.1.4, we see that

K(α1, . . . , αr−1)[αr] = K(α1, . . . , αr−1)(αr) = K(α1, . . . , αr−1, αr)

is finite-dimensional over K(α1, . . . , αr−1). Then [K(α1, . . . , αr) : K] < ∞by Lemma 1.2.1. It follows from part (a) that any element of K(α1, . . . , αr)is algebraic over K. �

One consequence of this theorem is that if α and β are algebraic over Kthen so are α + β, αβ, and α/β. However, it can take some work to writetheir minimal polynomials, as Example 1.1.6 shows.


§1.4. Adjoining roots. In the previous section we studied a fixed extensionK ⊂ F . If α ∈ F is algebraic over K then K(α) is isomorphic to K[x]/(f),where f(x) is a minimal polynomial ofα. An algebraic extensionK ⊂ K(α)generated by one element is sometimes called simple.

Now we will start with a field K and learn how to build its extensionsand compare them. The main building block was discovered by Kronecker:

LEMMA 1.4.1. If f(x) ∈ K[x] is irreducible and monic then K[x]/(f) is a fieldextension of K generated by α := x+ (f). The minimal polynomial of α is f(x).

Proof. Since f is irreducible and K[x] is a PID, K[x]/(f) is a field. It isobviously generated by α. Since f(x) ≡ 0 mod (f), α is a root of f(x).Since f(x) is irreducible over K, f(x) is the minimal polynomial of α. �

We will often want to compare extensions F and F ′ of the same field K.We say that F and F ′ are isomorphic over K if there exists an isomorphismφ : F → F ′ such that φ(a) = a for any a ∈ K. Here is the basic fact:

LEMMA 1.4.2. Let K(α) and K(β) be algebraic extensions of K. TFAE:• α and β have the same minimal polynomial.• There exists and isomorphism of K(α) to K(β) over K that sends α to β.

Proof. If α and β have the same minimal polynomials then the analysisabove shows that both K(α) and K(β) are isomorphic to K[x]/(f), wheref(x) is the common minimal polynomial. Notice that an induced isomor-phism between K(α) and K(β) simply sends α to β. �

EXAMPLE 1.4.3. Fields Q( 3√

2) and Q(ω 3√

2) are isomorphic (here ω is a prim-itive cubic root of unity), because they have the same minimal polynomialx3 − 2. However, they are not equal inside C because Q( 3

√2) is contained

in R but the other field is not.

EXAMPLE 1.4.4. By contrast, it is easy to see that

Q(√

2 +√

3) = Q(√

2−√

3) = Q(√

2,√

3).

So in this case the isomorphism in the Lemma is an automorphism of thefield Q(

√2,√

3).

§1.5. Splitting fields. We would like to adjoin all roots of a polynomial.

DEFINITION 1.5.1. A field F ⊃ K is called a splitting field of f(x) ∈ K[x] if• f splits into linear factors in F [x], and• F is generated by roots of f(x).

In other words, f(x) splits in F but not in any proper subfield of F .

EXAMPLE 1.5.2. Recall that α =√

2 +√

3 is a root of f = x4 − 10x2 + 1.Since α2 ∈ Q(α), we see that

√6 ∈ Q(α). Therefore α

√6 = 2

√3 + 3

√2 ∈

Q(α). Therefore√

2 = (2√

3 + 3√

2)− 2(√

3 +√

2) ∈ Q(α). This shows thatQ(√

3 +√

2) = Q(√

2,√

3), which contains all roots of f(x). So in this caseQ(√

3 +√

2) is a splitting field.

6 JENIA TEVELEV

EXAMPLE 1.5.3. The splitting field of f(x) = x3 − 2 is not equal to Q( 3√

2)because not all of the roots are real. The splitting field is Q( 3

√2, ω). It has

degree 6 over Q by multiplicativity of degree because ω is a root of the qua-dratic polynomial x2 + x + 1. Since ω 6∈ Q( 3

√2), this quadratic polynomial

is irreducible over Q( 3√

2).

LEMMA 1.5.4. Any polynomial f(x) ∈ K[x] has a splitting field. Moreover, anytwo splitting fields L and L′ are isomorphic over K.

Proof. We can of course assume that f is monic. Existence is proved byinduction on deg f : if f(x) does not split then it has an irreducible factorg(x) of degree greater than one. Lemma 1.4.1 gives an extension K ⊂ L =K(α) such that g(x) has a root α ∈ L. Then f(x)/(x − α) ∈ L[x] has asplitting field F ⊃ L by inductive assumption. Then F is a splitting field off(x) ∈ K[x] as well.

Now we have to construct an isomorphism between two splitting fieldsL and L′ over K. It is enough to construct an injective homomorphismφ : L → L′ that preserves K. Indeed, if f(x) =

∏i(x − αi) in L then

f(x) =∏i(x− φ(αi)) in φ(L), so f(x) splits in φ(L), so φ(L) = L′.

We will construct φ step-by-step. Choose a root α ∈ L of f(x). Let g(x)be the minimal polynomial of α. Then g(x) divides f(x), and in particularg(x) splits in L′. Let β be a root of g(x) in L′. Since α and β have thesame minimal polynomial, K(α) and K(β) are isomorphic over K. Fix oneisomorphism,

φ0 : K(α)→ K(β).Now we have a diagram of field maps

L L′

K(α)∪

∧

φ0> K(β)

∪

∧(1)

Notice that L is a splitting field of f(x) over K(α) and L′ is a splittingfield of f(x) over K(β). So ideally, we would like to finish by inductionby continuing to add roots of α. However, notice that the set-up is slightlydifferent: before we were trying to show that L and L′ are isomorphic overK, and now we are trying to construct an isomorphism φ : L → L′ thatextends a given isomorphism φ0 : K(α)→ K(β). So the best thing to do isto generalize our Lemma a little bit to make it more suitable for induction.This is done in the next Lemma. �

LEMMA 1.5.5. Suppose we have a diagram of homomorphisms of fields

L1 L2

K1

∪

∧

ψ> K2

∪

∧(2)

where L1 is a splitting field of f1(x) ∈ K1[x], the polynomial f2(x) ∈ K2[x] splitsin L2, and f2(x) is a polynomial obtained by applying ψ to all coefficients of f1(x).


Then there exists a homomorphism φ : L1 → L2 such that φ|K1 = ψ (i.e. thatmakes a diagram commutative).

Proof. Choose a root α ∈ L1 of f1(x). Let g1(x) be the minimal polynomialof α. Then g1(x) divides f1(x). We have a homomorphism

Ψ : K1[x]→ K2[x]

that extends ψ. Then f2 = Ψ(f1). Let g2 = Ψ(g1). Then g2|f2, and inparticular g2(x) splits in L2. Let β be a root of g2(x) in L2. Let g′2(x) ∈ K2[x]be an irreducible factor of g2(x) with root β. Then

K1(α) ' K1[x]/(g1) and K2(β) ' K2[x]/(g′2).

Notice that Ψ induces a homomorphism K1[x]/(g1) → K2[x]/(g′2). Thisgives an homomorphism

φ0 : K1(α)→ K2(β)

that sends α to β and such that φ0|K1 = ψ. Now we have a commutativediagram of field maps

L1 L2

K1(α)∪

∧

φ0> K2(β)

∪

∧

K1

∪

∧

ψ> K2

∪

∧(3)

Notice that L1 is a splitting field of f1(x) over K1(α) and f2(x) splits in L2.So we are in the same set-up as in the statement of the Lemma, but now[L1 : K1(α)] < [L1 : K1]. So we can finish by induction on [L1 : K1]. �

§1.6. Algebraic closure.

LEMMA 1.6.1. Let K be a field. The following are equivalent:• any polynomial f ∈ K[x] has a root in K.• any polynomial f ∈ K[x] splits in K.• The only algebraic extension of K is K itself.

Proof. Easy. �

If any of these conditions are satisfied thenK is called algebraically closed.

DEFINITION 1.6.2. Let K be a field. A field K containing K is called analgebraic closure of K if

• K is algebraically closed.• K ⊂ K is an algebraic extension.

For example, C is an algebraic closure of R but not of Q (why?)

LEMMA 1.6.3. Let K ⊂ F be a field extension with F algebraically closed. Then

K = {a ∈ F | a is algebraic over F}is an algebraic closure of K.

8 JENIA TEVELEV

Proof. If α, β ∈ K then K(α, β) is finite algebraic over K. In particular, K isa field (obviously algebraic overK). Any polynomial f(x) = xn+a1x

n−1 +. . . + an in K[x] has a root α ∈ F . Then K(a1, . . . , an) is finite over K,K(a1, . . . , an)(α) is finite over K(a1, . . . , an). Therefore, [K(α) : K] < ∞and α is algebraic. �

For example, if K = Q and F = C then K = Q is the field of all algebraicnumbers in C (roots of polynomials with rational coefficients).

LEMMA 1.6.4. Any field K is contained in a field F such that any polynomial inK[x] has a root in F .

Proof. The idea is to use a Kronecker-type construction to adjoin roots ofall polynomials at once. Let K[xf ] be the algebra of polynomials in vari-ables xf , one variable for each irreducible polynomial with coefficientsin K. Consider the ideal

I = 〈f(xf )〉 ⊂ K[xf ]

with one generator for each irreducible polynomial f . Notice that eachpolynomial is a polynomial in its own variable. We claim that I is a properideal. If not, then we can write

1 =s∑i=1

gifi(xfi),

where gi are some polynomials that involve only finitely many variables xf .Let L be a splitting field of the product f1 . . . fs. The formula above remainsvalid in L[xf ]. However, if we plug-in any root of f for xf (for each f ), wewill get 1 = 0, a contradiction. It follows that I is a proper ideal.

Let M be a maximal ideal containing I . Then F := K[xf ]/M is a fieldthat contains K. We claim that any irreducible polynomial f ∈ K[x] has aroot in F . Indeed, f(xf ) ∈M , and therefore xf +M is a root of f in F . �

THEOREM 1.6.5. Any field K has an algebraic closure. It is unique up to anisomorphism over K.

Proof. Applying Lemma 1.6.4 inductively gives an infinite tower of fields

K = F0 ⊂ F1 ⊂ F2 ⊂ . . .such that any polynomial in Fk[x] has a root in Fk+1. Then F = ∪iFi isalgebraically closed as any polynomial in F [x] in fact belongs to some Fk[x],and therefore has a root in F . Applying Lemma 1.6.3 gives an algebraicclosure K.

Let K, K1 be two algebraic closures of K. It suffices to show that thereexists a homomorphism φ : K → K1 over K. Indeed, φ(K) is then anotheralgebraic closure of K contained in K1. Since K1 is algebraic over φ(K), itmust be equal to it.

Finally, we construct φ using Zorn’s lemma. Consider a poset of pairs(F, φ), where K ⊂ F ⊂ K and φ : F → K1 is a homomorphism over K.We say that (F, φ) ≤ (F1, φ1) if F ⊂ F1 and φ is the restriction of φ1 to F .Then any chain has a maximal element (F, φ) (take the union of fields inthe chain and the map φ induced by maps in the chain). This maximal fieldmust be equal to K: if F is properly contained in K then take any α ∈ K\F .


By Lemma 1.5.5, we can extend φ to a homomorphism F (α)→ K1. �

The same argument shows the following slightly more useful statement:

PROPOSITION 1.6.6. Suppose we have a diagram of homomorphisms of fields

L1

K1

∪

∧

ψ> K2

(4)

where L1 is algebraic over K1 and K2 is algebraically closed. Then there existsa homomorphism φ : L1 → K2 such that φ|K1 = ψ (i.e. that makes a diagramcommutative).

§1.7. Finite fields.

THEOREM 1.7.1. For any prime p and positive integer n, there exists a field Fpn

with pn elements. Moreover, any two such fields are isomorphic. We can embedFpm in Fpn if and only if m divides n.

Proof. Let K be a splitting field of the polynomial f(x) = xpn − x ∈ Fp[x].

Since f ′(x) = −1 is coprime to f(x), there are exactly pn roots. Recall thatF : K → K, F (x) = xp is a Frobenius homomorphism. In particular, if αand β are roots of f(x) then ±α ± β and αβ, and α/β are roots as well. Itfollows that K has pn elements and all of them are roots of f(x).

Suppose K is a field with pn elements. The group of units K∗ is Abelianof order pn − 1, and therefore xp

n−1 = 1 for any x ∈ K∗.1 It follows thatK is a splitting field of xp

n − x. But any two splitting fields of the samepolynomial are isomorphic by Lemma 1.5.4.

If Fpm ⊂ Fpn then the latter field is a vector space (of some dimension r)over the former. It follows that

pn = (pm)r = pmr.

It follows that m divides n.Finally, suppose thatm divides n. Then pm−1|pn−1 (easy), and therefore

xpm−1−1|xpn−1−1 (equally easy). It follows that the splitting field of xp

m−xis contained in the splitting field of xp

n − x. �

§1.8. Composite field.

DEFINITION 1.8.1. Let K1K2 be subfields of a field K. Then the compositefield of K1 and K2, denoted by K1K2, is the smallest subfield of K contain-ing K1 and K2.

For example, Q(√

2)Q(√

3) = Q(√

2,√

3) ⊂ C.

1Recall that in fact this analysis implies that K∗ is a cyclic group. Indeed, otherwise wewould have xr = 1 for any x ∈ K∗ and r < pn − 1. However, the polynomial can not havemore roots than its degree.

10 JENIA TEVELEV

PROPOSITION 1.8.2. Let K1, K2 be finite extensions of a field F contained in K.Then

[K1K2 : F ] ≤ [K1 : F ][K2 : F ],with equality iff some basis of K2 over F is linearly independent over K1.

Proof. Let e1, . . . , en (resp. f1, . . . , fm) be a basis of K1 over F (resp. a basisof K2 over F ). Then K1K2 = F (e1, . . . , en, f1, . . . , fm) = K1(f1, . . . , fm).Therefore [K1K2 : K1] ≤ m with equality if and only if {f1, . . . , fm} islinearly independent over K1. The rest of the Proposition follows by mul-tiplicativity of degree. �

§1.9. Exercises.In this set we fix a field extension K ⊂ F unless specified otherwise.

1. (a) Show thatα ∈ F is algebraic overK iffF contains a finite-dimensionalK-vector subspace L (not necessarily a subfield) such that

α · L ⊂ L.(b) If the conditions of (a) are satisfied, letA : L→ L be aK-linear operatorof multiplication by α. Show that the minimal polynomial of this linearoperator in the sense of linear algebra is equal to the minimal polynomialof α in the field-theoretic sense.

2. Consider field extensions E ⊂ F ⊂ K. Suppose that F is algebraicover E and K is algebraic over F . Show that K is algebraic over E.

3. LetS = {α ∈ F |α is algebraic over K}.

(a) Show that S is a field. It is called an algebraic closure of K in F . (b) Showthat an algebraic closure of S in F is equal to S. (c) Show that S is notnecessarily algebraically closed.

4. (a) Show that f(x) = x3+x2+x+3 is irreducible over Q. (b) Considerthe field F = Q(α), where α is a root of f(x). Express 1

2−α+α2 as a Q-linearcombination of 1, α, and α2.

5. Find the degree (over Q) of the splitting field of (a) x4 +x3 +x2 +x+1.(b) x4 − 2.

6. (a) Suppose [F : K] = 2 and charK 6= 2. Show that there exists D ∈ Ksuch that F = K(

√D). (b) Show that (a) fails if [F : K] = charK = 2.

7. For all positive integers n andm, compute the degree [Q(√n,√m) : Q].

8. Let K ⊂ F be an algebraic extension and let R be a subring of F thatcontains K. Show that R is a field.

9. Let f(x) ∈ K[x] be a polynomial of degree 3. Show that if f(x) has aroot in a field extension K ⊂ F of degree 2 then f(x) has a root in K.

10. Let α, β ∈ F be algebraic over K, let f(x) and g(x) be their minimalpolynomials, and suppose that deg f and deg g are coprime. Prove that f(x)is irreducible in K(β)[x].

11. Let K ⊃ Q be a splitting field of a cubic polynomial f(x) ∈ Q[x].Show that if [K : Q] = 3 then f(x) has 3 real roots.

12. Let F = K(α) and suppose that [F : K] is odd. Show that F = K(α2).13. Let f(x) ∈ K[x] be an irreducible polynomial and let g(x) ∈ K[x] be

any non-constant polynomial. Let p(x) be a non-constant polynomial thatdivided f(g(x)). Show that deg f divides deg p.


14. Show that the polynomial x5−t is irreducible over the field C(t) (heret is a variable). Describe a splitting field.

15. Let Fq be a finite field with q elements (q is not necessarily prime).Compute the sum

∑a∈Fq

ak for any integer k.16. (a) Show that the algebraic closure of Fp is equal to the union of its

finite subfields:

Fp =∞⋃n=1

Fpn .

(b) Show that Fp contains proper infinite subfields.17. A complex number α ∈ C is called constructible if [Q(α) : Q] is a

power of 2. Assume without proof that α is constructible if and only if onecan construct α (as a vector on the plane) using the ruler and the compass.(a) Show that cos 20◦ is not constructible. (b) Show that trisection of anangle is not always possible using the ruler and the compass.

§2. GALOIS THEORY

Let K ⊂ F be an algebraic extension. For convenience, in this section wefix an algebraic closure K of K and assume that F ⊂ K.

§2.1. Separable extensions.

DEFINITION 2.1.1. An element α ∈ F is called separable over K if its mini-mal polynomial has no multiple roots.

LEMMA 2.1.2. Let α ∈ F and let f(x) be its minimal polynomial. Then α is notseparable if and only if charK = p and f ′(x) ≡ 0. In this case f(x) = g(xp) forsome polynomial g(x).

Proof. Indeed, f(x) has a multiple root if and only the g.c.d. of f(x) andf ′(x) has positive degree. This greatest common divisor belongs to K[x],and since f(x) is irreducible, it is only possible if f ′(x) ≡ 0. This impliesthat charK = p and f(x) = g(xp) for some polynomial g(x). �

DEFINITION 2.1.3. An algebraic extension F/K is called separable if any α ∈F is separable over K.

THEOREM 2.1.4. Let F/K be an algebraic extension. Suppose that F is generatedover K by elements αi, i ∈ I . Then the following conditions are equivalent:

(1) F/K is separable.(2) αi is separable for any i ∈ I .

If, in addition, F/K is finite then this is equivalent to(3) The number of different embeddings F → K over K is equal to [F : K],

the maximum possible number.

Proof. (1) obviously implies (2). Next we assume that F/K is finite andshow that (2) implies (3). In this case F is generated by finitely many αi’s,so we can assume that I = {1, . . . , r} is a finite set. Then we have the tower

K = F0 ⊂ F1 ⊂ F2 ⊂ . . . ⊂ Fr = F,

where Fk = K(α1, . . . , αk). Each αk is separable over K and hence sep-arable over Fk−1. We have Fk = Fk−1(αk), and therefore the number of

12 JENIA TEVELEV

different embeddings of Fk in K over Fk−1 is equal to [Fk : Fk−1]. But anyhomomorphism F → K can be constructed step-by-step by extending theinclusion K ⊂ K to fields Fk in the tower. It follows that the number ofdifferent embeddings F → K over K is equal to

[F : Fr−1][Fr−1 : Fr−2] . . . [F1 : K] = [F : K].

Moreover, the same reasoning shows that this is the maximum possiblenumber of embeddings.

Now we show that (3) implies (1) (still assuming that F/K is finite).Suppose that α ∈ F is not separable. Then the number of embeddingsK(α) → K is strictly less then [K(α) : K], and considering the towerK ⊂ K(α) ⊂ F gives the contradiction. Indeed, by the above, the num-ber of different embeddings F → K over K(α) is at most [F : K(α)].

Finally, we show that (2) implies (1) in general. Take α ∈ F . Thenα ∈ K(α1, . . . , αk) for a finite subset of generators. Since K(α1, . . . , αk)is finite over K, the finite extension case considered above shows that α isseparable. �

The following theorem is a nice bonus feature of separable extensions.

THEOREM 2.1.5 (Theorem on the Primitive Element). If F/K is a finite sepa-rable extension then F = K(γ) for some γ ∈ F .

REMARK 2.1.6. Eventually we will show that a finite separable extensionF/K contains only finitely many intermediate subfields K ⊂ L ⊂ F .This will follow from the Main Theorem of Galois theory. If we view Fas a finite-dimensional K-vector space, proper subfields of F form a finiteunion of proper vector subspaces. If K is an infinite field, we can alwaysfind a vector γ ∈ F which does not belong to this finite union of sub-spaces. Thus F = K(γ). For example, the field extension Q ⊂ Q(

√2,√

3)contains only three intermediate subfields, namely Q(

√2), Q(

√3), Q(

√6).

So Q(√

2,√

3) is generated by any element not in these subfields, for exam-ple by γ =

√2 +√

3. However, to avoid a logical circle we will prove theTheorem on the Primitive Element first, and then deduce the Main Theo-rem of Galois Theory from it. A proof uses a nice trick, which we will alsouse in the proof of Noether’s normalization theorem later in the course.

Proof. If K is a finite field then F is also finite and we can take γ to be anygenerator of F ∗ (recall that F ∗ is a cyclic group).

Suppose now that K is infinite. We argue by induction on the number ofgenerators of F over K and reduce everything to the following

CLAIM 2.1.7. If F = K(α, β) then we can find γ = β + cα for some c ∈ K suchthat K(α, β) = K(γ).

Since β = γ − cα, if α ∈ K(γ) then β ∈ K(γ) as well. So it is enough toshow that α ∈ K(γ) for some c ∈ K. We are going to choose c later (to givea better motivation for this choice). For now we assume that c is fixed.

Let f(x) (resp. g(x)) be the minimal polynomial of α (resp. β). SinceF/K is separable, their roots α = α1, . . . , αr and β = β1, . . . , βs (in K) arenot multiple. Consider a polynomial

h(x) := g(γ − cx) ∈ K(γ)[x]


Clearly α is a common root of f(x) and h(x). If αi, i ≥ 2 is another commonroot then γ − cαi = βj for some j. It follows that

β + cα = βj + cαi,

and therefore

c =βj − βα− αi

.

There are only finitely many choices for the RHS of this equation. So wecan choose c such that

c 6= βj − βα− αi

, i, j > 1.

In this case α is the only common root of f(x) and h(x). Since f(x) has nomultiple roots, the g.c.d. of f(x) and h(x) is equal to x − α. It follows thatwe can write

x− α = a(x)f(x) + b(x)h(x) for some a(x), b(x) ∈ K(γ)[x].

It follows that x− α ∈ K(γ)[x]. It follows that α ∈ K(γ). �

§2.2. Normal extensions.

DEFINITION 2.2.1. An algebraic extension F/K is called normal if the mini-mal polynomial of any α ∈ F splits in F [x] as the product of linear factors.

THEOREM 2.2.2. Let F/K be algebraic and suppose that F is generated over Kby elements αi, i ∈ I . Let’s also assume that F ⊂ K. Then TFAE:

(1) F/K is normal.(2) Minimal polynomials of αi’s split in F .(3) Any embedding F → K over K has image F .

Proof. It is obvious that (1) implies (2).Let σ : F → K be any homomorphism over K. Then σ(αi) is a root of

the minimal polynomial of αi for any i. It follows that σ(αi) ∈ F for any αi.Therefore σ(F ) ⊂ F . We claim that in fact σ(F ) = F . This is clear if F/K isfinite. But even if it is not finite, any α ∈ F is contained in the splitting fieldF ′ of finitely many of αi’s. The argument above shows that σ(F ′) = F ′,and therefore any α ∈ F is in the image of σ.

Finally, we show that (3) impies (1). Suppose not. Then there existsα ∈ F such that its minimal polynomial does not split in F . Then thereexists an embedding K(α) → K with image not contained in F (just sendα to a root of the minimal polynomial not contained in F ). This embeddingcan be extended to an embedding σ : F → K with σ(F ) 6⊂ K. �

§2.3. Main Theorem of Galois Theory.

DEFINITION 2.3.1. K ⊂ F is a Galois extension if it is separable and normal.

Using our characterizations of normal and separable extensions, this def-inition can be spelled out in three different ways:

• K ⊂ F is an algebraic extension such that the minimal polynomialof any α ∈ F has no multiple roots and splits in F [x] into the prod-uct of linear factors.

14 JENIA TEVELEV

• If K ⊂ F is an algebraic extension generated by elements αi ∈ F ,the requirement is that the minimal polynomial of each αi has nomultiple roots and splits in F [x] into the product of linear factors.• If [F : K] < ∞ and F ⊂ K then the requirement is that there exists

exactly [F : K] homomorphisms σ : F → K over K (that is suchthat σ|K = Id |K), and the image of each of them is F .

DEFINITION 2.3.2. LetK ⊂ F is a field extension. The Galois group Gal(F/K)as the group of all automorphisms σ : F → F such that σ|K = Id |K .

The Main Theorem of Galois Theory gives an amazing strengthening ofsimple observations we made above.

THEOREM 2.3.3. Let K ⊂ F be a finite Galois extension with a Galois groupG = Gal(F/K). Then |G| = [F : K] and there is an inclusion-reversing bijection

{subgroups H ⊂ G} ↔ {towers K ⊂ L ⊂ F}

Namely, a subgroup H corresponds to its fixed subfield

L = FH = {α ∈ F |h(α) = α for any h ∈ H}

and a tower K ⊂ L ⊂ F corresponds to a subgroup

H = Gal(F/L) ⊂ Gal(F/K) = G.

Proof. We embed F into the algebraic closure K of K.Since F/K is separable, the number of homomorphisms F → K over K

is equal to [F : K]. Since F/K is normal, the image of any such homomor-phism is equal to F . Therefore, |G| = [F : K].

Next we show that FG = K. Indeed, FG is clearly a field and we have

K ⊂ FG ⊂ F.

Since F/K Iis a Galois extension, F/FG is a Galois extension as well. Bythe above, we have |Gal(F/FG)| = [F : FG]. But

Gal(F/FG) = Gal(F/K) = G.

Indeed, any automorphism ofF overK belongs toG and therefore fixesFG.And since K ⊂ FG, any automorphism of F over FG fixes K. It followsthat [F : FG] = [F : K] = |G| and so FG = K.

Now take K ⊂ L ⊂ F . We map it to a subgroup H = Gal(F/L) ⊂Gal(F/K) = G. We have proved in the previous step (applied to the exten-sion L ⊂ F ) that FH = L. It follows that the map

{K ⊂ L ⊂ F} → {H ⊂ G}

is one-to-one. It remains to show that this map is onto and that Gal(F/FH) =H for any subgroup H ⊂ G. This follows from Lemma 2.3.4. �

LEMMA 2.3.4. Let F be any field and let G be a finite group of its automorphisms.Then F/FG is a finite Galois extension with Galois group G.

Proof. Let α ∈ F and consider its G-orbit

G · α = {α1, . . . , αr} with α = α1.


Consider the polynomial

f(x) =r∏i=1

(x− αr).

By Vieta formulas, its coefficients are elementary symmetric functions inα1, . . . , αr. These coefficients don’t change when we permute α1, . . . , αr,therefore, they are G-invariant. It follows that f(x) ∈ FG[x].

Since the minimal polynomial of α over FG divides f(x), α is separableover FG. Therefore F/FG is separable.

Since all roots of the minimal polynomial of α are among {α1, . . . , αr},the extension F/FG is normal. Therefore, F/FG is a Galois extension.Clearly, G ⊂ Gal(F/FG). To show that G = Gal(F/FG), it suffices to showthat [F : FG] ≤ |G|. Take any intermediate subfield FG ⊂ L ⊂ F such that[L : FG] < ∞. By the Theorem on the Primitive Element, L = FG(α) forsome α. By the analysis above, the minimal polynomial of α has degree atmost |G|, and therefore [L : FG] = [FG(α) : FG] ≤ |G|. This implies that[F : FG] ≤ |G|. Indeed, suppose α1, . . . , αr ∈ F are linearly independent.Take L = K(α1, . . . , αr). Then [L : K] <∞ and by the above [L : K] ≤ |G|.Therefore, r ≤ |G|. �

COROLLARY 2.3.5. Let H ⊂ G and let L = FH be the corresponding subfield.Then H is a normal subgroup of G if and only if L/K is a normal field extension.In this case Gal(L/K) ' G/H .

Proof. Suppose L/K is a normal extension. Then any automorphism of Fover K preserves L, i.e. we have a “restriction” homomorphism

Gal(F/K)→ Gal(L/K).

Its kernel is obviously Gal(F/L). The restriction homomorphism is ontobecause any automorphism of L/K can be extended to an automorphismof F/K by normality of the latter extension.

In the other direction, suppose L/K is not a normal extension. Thenthere exists g ∈ G such that g(L) 6= L. It is easy to check that gHg−1 is aGalois group of F/g(L). Since g(L) 6= L, it follows by the main theorem ofGalois theory that H 6= gHg−1, i.e. H is not normal. �

REMARK 2.3.6. A simple fact that we will exploit a lot is that the Galoisgroup Gal(F/K) of a finite Galois extension F/K is isomorphic to a sub-group of Sn, where n = [F : K]. For example, by the primitive elementtheorem, F = K(α) for some α ∈ F . Then F is a splitting field of the min-imal polynomial f(x) of α, and Gal(F/K) permutes n roots of f(x). Thisgives an embedding Gal(F/K) ↪→ Sn.

EXAMPLE 2.3.7. Let’s completely analyze the field extension

Q ⊂ Q(√

2,√

3).

We have an intermediate subfield Q(√

2) of degree 2 over Q and it is ele-mentary to check that

√3 is not contained in this subfield. It follows that

Q(√

2,√

3) has degree 4 over Q and is a splitting field of the polynomial

16 JENIA TEVELEV

(x2 − 2)(x2 − 3). In particular, this extension is Galois. Let G be the Galoisgroup. Then

|G| = [Q(√

2,√

3) : Q] = 4andG permutes roots of (x2−2)(x2−3). But not in an arbitrary way: G canonly permute roots of x2−2 (resp. x2−3) among themselves because thesepolynomials are minimal polynomials of

√2 (resp.

√3). So we see that

G ' Z2 × Z2.

It sends√

2 to ±√

2 and√

3 to ±√

3. The group G contains three propersubgroups: H1 fixes

√2, H2 fixes

√3, and H3 can only change the sign of√

2 and√

3 simultaneously. Then H3 fixes√

6 =√

2√

3. So there are exactly3 intermediate subfields: Q(

√2), Q(

√3), and Q(

√6).

Take an element√

2+√

3. Let f(x) be its minimal polynomial. The Galoisgroup G permutes the roots of f(x). So these roots must be ±

√2 ±√

3. Inparticular, f(x) has degree 4, and therefore

√2 +√

3 is a primitive element:

Q(√

2,√

3) = Q(√

2 +√

3).

§2.4. Fields of invariants.

DEFINITION 2.4.1. Whenever a group G acts on a field K (resp. a ring R)by automorphisms, we say that KG (resp. RG) is the field of invariants (resp.the ring of invariants) of G.

EXAMPLE 2.4.2. The symmetric group Sn acts on the field of rational func-tions K = k(x1, . . . , xn) by permuting variables. By Lemma 2.3.4, K/KSn

is a Galois extension with a Galois group Sn. It is clear that KSn containselementary symmetric functions

σ1 =∑i

xi, σ2 =∑i<j

xixj , . . . , σn =∏i

xi.

So KG ⊃ k(σ1, . . . , σn). By the Vieta theorem, k(x1, . . . , xn) is a splittingfield over k(σ1, . . . , σn) of the polynomial

(x− x1) . . . (x− xn) = xn − σ1xn−1 + . . .+ (−1)nσn

without multiple roots. It follows that k(x1, . . . , xn)/k(σ1, . . . , σn) is a Ga-lois extension. Let G be its Galois group. Since G acts faithfully on the setof roots of the polynomial above, we have |G| ≤ n!. On the other hand,

|G| = [k(x1, . . . , xn) : k(σ1, . . . , σn)] ≥ [k(x1, . . . , xn) : k(x1, . . . , xn)Sn ] = n!

Therefore, G = Sn and KG = k(σ1, . . . , σn). In other words, any symmet-ric rational function can be expressed in terms of elementary symmetricfunctions.

Later on we will extend this theorem to polynomial functions:

THEOREM 2.4.3 (Theorem on Symmetric Polynomials). Let

k[x1, . . . , xn]Sn ⊂ [x1, . . . , xn]

be a subring of all polynomials invariant under permutations of variables. Then

k[x1, . . . , xn]Sn = k[σ1, . . . , σn].


§2.5. Exercises.We fix a finite field extension K ⊂ F . We also assume that F ⊂ K.

1. Let α ∈ F and let f(x) be its minimal polynomial over K. Show thatthere exists k ≥ 0 such that all roots of f(x) in K have multiplicity pk andαp

kis separable over K.

2. (a) Show that elements of F separable over K form a field L (called aseparable closure of K in F ). We define

[F : K]s := [L : K].

(b) Show that the separable closure of L in F is equal to L. (b) Prove that thenumber of different homomorphisms F → K over K is equal to [F : K]s.

3. An extension F/K is called purely inseparable if [F : K]s = 1. Showthat F/K is purely inseparable if and only if charK = p and F is generatedover K by elements α1, . . . , αr such that the minimal polynomial of each αihas the form xp

ki − ai for some ai ∈ K and a positive integer ki.4. Let L = Fp(x, y) be the field of rational functions in two variables

and let K = Fp(xp, yp) be its subfield. (a) Show that L/K is an algebraicextension and compute its degree. (b) Show that there exist infinitely manypairwise different intermediate subfields between K and L. (c) Show thatL cannot be expressed as K(α) for some α ∈ L.

5. A field k is called perfect if either char k = 0 or char k = p and theFrobenius homomorphism F : k → k is an isomorphism. Show that if k isperfect then any algebraic extension of k is separable over k and perfect.

6. Let F be a splitting field of the polynomial f ∈ K[x] of degree n. Showthat [F : K] divides n! (do not assume that F is separable over K).

7. Let F ⊂ K be a finite Galois extension ofK and let L ⊂ K be any finiteextension of K. Consider the natural K-linear map L⊗K F → K. (a) Showthat its image is a field, in fact a composite field LF . (b) Show that LF isGalois over L. (c) Show that Gal(LF/L) is isomorphic to Gal(F/L ∩ F ).

8. (a) Find the minimal polynomial over Q of 2√

3 + 3√

3. (b) Compute theGalois group of its splitting field.

9. Let f(x) ∈ Q[x] be an irreducible polynomial of prime degree p. Sup-pose that f(x) has exactly p − 2 real roots. Show that the Galois group ofthe splitting field of f(x) is Sp.

10. For any d ≥ 2, prove existence of an irreducible polynomial in Q[x]of degree d with exactly d− 2 real roots (Hint: take some obvious reduciblepolynomial with exactly d− 2 real roots and perturb it a little bit to make itirreducible).

11. Let G be any finite group. Show that there exist finite extensionsQ ⊂ K ⊂ F such that F/K is a Galois extension with a Galois group G.

12. Let F be a splitting field of the polynomial f(x) ∈ K[x]. Show thatGalF/K acts transitively on roots of f(x) if and only if f(x) is irreducible(do not assume that f(x) is separable).

13. Let F be a splitting field of a biquadratic polynomial x4 + ax2 + b ∈K[x]. Show that Gal(F/K) is isomorphic to a subgroup of D4.

§3. APPLICATIONS OF GALOIS THEORY

§3.1. Intermediate subfields in separable extensions.

18 JENIA TEVELEV

PROPOSITION 3.1.1. Let F/K be a finite separable extension. Then the numberof intermediate subfields K ⊂ L ⊂ F is finite.

Proof. By a theorem on the primitive element, F = K(α). Let L ⊃ F be asplitting field of the minimal polynomial of α. Then L/K is a Galois ex-tension and therefore has finitely many intermediate subfields by the MainTheorem on Galois Theory (MTGT). It follows that the number of interme-diate subfields of F/K is also finite. �

If G = GalL/K and H = Gal(L/F ) = {g ∈ G | g(α) = α} then theseintermediate subfields correspond to intermediate subgroups H ⊆ K ⊆ G.

§3.2. Fundamental Theorem of Algebra.

THEOREM 3.2.1. C is algebraically closed.

Proof. Since we are in char = 0, all field extensions are separable. It sufficesto show (why?) that any finite Galois extension K of R is either R or C.We argue by induction on [K : R]. If [K : R] = 1 then K = R and there isnothing to prove. Suppose that [K : R] > 1.

Let G be the Galois group of K/R. Let H ⊂ G be its 2-Sylow subgroup.Then [KH : R] = [G : H] is odd. Let α ∈ KH . Then the minimal polynomialof α in R[x] has odd degree. But any odd degree polynomial in R[x] has areal root (this is the only place where we use analysis). Therefore KH = R.It follows by MTGT that G = H . So G is a 2-group.

Any p-group has a non-trivial center. Let Γ ⊂ Z(G) be a subgroup oforder 2. Then Γ is normal in G. Therefore, KΓ/R is Galois by MTGT. By in-ductive assumption, KΓ is equal to R or to C.

Finally, K/KΓ is a quadratic extension. By the quadratic formula, anyquadratic polynomial in C[x] splits and any quadratic polynomial in R[x]has a complex root. Therefore, KΓ = R and K = C. �

§3.3. Quadratic extensions. Suppose [F : K] = 2. Since 2 is prime, wehave F = K(α) for any α ∈ F \ K by multiplicativity of degree. Let f ∈K[x] be the minimal polynomial of α. We have f(x) = (x−α)(x− β) in K.

Case 0. F/K is unseparable. This happens if and only if charK = 2 andf(x) = x2 − a for a ∈ K. One example is F2(t2) ⊂ F2(t).

Case 1. F/K is separable, i.e. α and β are different. Since α + β ∈ Kby Vieta’s formulas, F/K is normal and therefore Galois. The Galois groupis Z2 (the only group of order 2). It permutes α and β. Now suppose thatcharK 6= 2. Then an element d = α − β is not Galois-invariant. Therefored 6∈ K by the MTGT. Notice that D = d2 is Z2-invariant, and thereforeD ∈ K. So F = K(d) = K(

√D). D is of course just the discriminant: if

f(x) = x2 + bx+ c then

(α− β)2 = (α+ β)2 − 4αβ = b2 − 4c.

§3.4. Cubic extensions. Let F/K be a cubic extension, i.e. [F : K] = 3.Since 3 is prime, we have F = K(α) for any α ∈ F \K by multiplicativityof degree. Let f ∈ K[x] be the minimal polynomial of α. We have f(x) =(x− x1)(x− x2)(x− x3) in K, here α = x1.


Case 0. F/K is unseparable. This happens if and only if charK = 3 andf(x) = x3 − a for a ∈ K. One example is F3(t3) ⊂ F3(t).

Let’s suppose that F/K is separable, i.e. x1, x2, x3 are different. There aretwo possibilities:

Case A. F/K is normal, i.e. f(x) splits in F . In this case F/K is Galoiswith Galois group Z3 (the only group of order 3). 3 is a prime number, sothere are no intermediate subfields between K and F . One example of thissituation is F3 ⊂ F27 (Fpn/Fp is always normal). Notice that Z3 acts on roots{x1, x2, x3} and S3 contains only one subgroup of order 3, namelyA3. So Z3

permutes roots cyclically.Case B. F/K is not normal, i.e. f(x) does not split in F . In this case F/K

is not Galois. One example is Q ⊂ Q( 3√

2). Let L ⊂ K be the splittingfield of f(x). Then L/K is Galois. Let G = Gal(L/K) be the Galois group.Since G acts on {x1, x2, x3}, we have |G| ≤ 3! = 6. On the other hand,[L : K] ≥ 2[F : K] = 6. So in fact [L : K] = 6 and G = S3.

The symmetric group S3 has four proper subgroups: three subgroupsgenerated by transpositions (ij) for i < j and the alternating group A3.By MTGT, L/K has four intermediate subfields. Since 〈(ij)〉 ' Z2, we have[F 〈(ij)〉 : K] = 6/2 = 3. Since xk ∈ F 〈(ij)〉 for k 6= i, j, we have

F 〈(12)〉 = K(x3), F 〈(13)〉 = K(x2), F 〈(23)〉 = K(x1).

Since |A3| = 3, [FA3 : K] = 2. To describe this field, let’s assume that

charK 6= 2.

Then the element

d = (x1 − x2)(x1 − x3)(x2 − x3) ∈ Fis invariant under A3 but not S3. Therefore d ∈ FA3 \K by the Main Theo-rem of Galois Theory. So we have

FA3 = K(d).

Notice that D = d2 is S3-invariant, and therefore D ∈ K. Since d 6∈ K, D isnot a square in K.D is called the discriminant of f(x). As a symmetric polynomial in roots,

it can be expressed as a polynomial in coefficients of f(x). Notice that d ∈ Kin Case A (because FA3 = K), so in this case D is a square in K. We cansummarize this discussion as follows:

PROPOSITION 3.4.1. Let F/K be a cubic extension and let f(x) be a minimalpolynomial of α ∈ F \K. Let D ∈ K be the discriminant of f(x). Then

• if D = 0 then F/K is unseparable.• if D ∈ (K∗)2 and charK 6= 2 then F/K is Galois.• if D ∈ K∗ \ (K∗)2 then charK 6= 2 and F/K is separable but not Galois.

§3.5. Galois group of a finite field.

THEOREM 3.5.1. Fpn/Fp is a Galois extension with Galois group

Gal(Fpn/Fp) ' Zn.Gal(Fpn/Fp) is generated by the Frobenius map F (x) = xp. Intermediate sub-fields L correspond to divisors k of n.

20 JENIA TEVELEV

Proof. We already proved that Fpn/Fp is a splitting field of f(x) = xpn − x.

Since f ′(x) = −1, this extension is separable and therefore Galois. Since[Fpn : Fp] = n, the Galois group G has order n. The Frobenius map is anelement of G. If F has order d then αp

d= α for any α ∈ Fpn . A polyno-

mial can not have more roots than its degree, therefore d = n and G is acyclic group generated by F . Notice that subgroups H ⊂ G correspondto divisors k|n. Namely, H is generated by F k and has order n/k The laststatement follows from MTGT. �

Notice that we have

(Fpn)H = {α ∈ Fpn |αpk= α} = Fpk .

§3.6. Exercises.1. Let a, b ∈ K and suppose that f(x) = x3 + ax + b has no roots in K.

Let F be a splitting field of f(x). Assume that charK 6= 3. Show that

Gal(F/K) '

{S3 if −4a3 − 27b2 is not a square in K

Z3 if −4a3 − 27b2 is a square in K

2. Let α be a real number such that α4 = 5. (a) Show that Q(iα2) isnormal over Q. (b) Show that Q(α + iα) is normal over Q(iα2). (c) Showthat Q(α+ iα) is not normal over Q.

3. Consider a tower K ⊂ L ⊂ F . Suppose L/K and F/L are finite Galoisextensions. Is it true that F/K is Galois?

4. Compute the Galois group of the polynomial (a) x3−x−1 over Q(√−23);

(b) x3 − 2tx+ t over C(t) (the field of rational functions in one variable).5. Let L/K be a Galois extension with the Galois group S6. (a) Find the

number of intermediate subfields F betweenK and L such that [L : F ] = 9.(b) Let M be the intersection of all fields F in part (a). Find [M : K].

6. Let F/K be a finite Galois extension and let L be an intermediate sub-field between F and K. Let H be the subgroup of Gal(F/K) mapping L toitself. Prove that H is the normalizer of Gal(F/L) in Gal(F/K).

7. Let p1, . . . , pr ∈ Z be distinct primes and let

K = Q(√p1, . . . ,

√pr).

(a) For any non-empty subset S ⊂ {1, . . . , r}, let aS =∏i∈S

pi. Show that all

2r − 1 intermediate subfields of the form Q ⊂ Q(√aS) ⊂ K are different.

(b) Compute the Galois group Gal(K/Q).8. (continuation of the previous problem). Describe explicitly all inter-

mediate subfields L such that either [L : Q] = 2 or [K : L] = 2.9. Let K ⊂ L ⊂ K and suppose that L/K is separable. Show that there

exists the unique minimal (by inclusion) Galois extension F/K such thatL ⊂ F ⊂ K. Show that if L/K is finite then F/K is finite. F is called theGalois closure of L in K.

10. Let F/K be a finite Galois extension with a Galois group G. Let H ⊂G be a subgroup and let L = FH . Let N =

⋂g∈G

gHg−1. Characterize the

field FN in terms of the tower K ⊂ L ⊂ F .


11. Let F/K be a finite Galois extension with a Galois group G. Let H ⊂G be a subgroup and let L = FH . Show that the number of fields of theform g(L) for g ∈ G is equal to |G|

|NG(H)| .12. Let Fpn be a finite field with pn elements and let F : Fpn → Fpn be the

Frobenius map, F (x) = xp. Show that F is diagonalizable (as an Fp-linearoperator) if and only if n divides p− 1.

13. Let F/K be a splitting field of a polynomial f(x) = (x − ai) . . . (x −ar) ∈ K[x] without multiple roots. Let

∆ =∏

1≤i<j≤n(ai − aj)2

be the discriminant of f(x). (a) Show that ∆ ∈ K. (b) LetG ⊂ Sn be the Ga-lois group of F/K acting on roots of f(x). Suppose charK 6= 2. Show thatG ⊂ An if and only if ∆ is a square in K. (c) Let F = C(x1, . . . , xn) be thefield of rational functions in n variables. SupposeAn acts on F by even per-mutations of variables. Show that FAn is generated over C by elementarysymmetric functions σ1, . . . , σn in variables x1, . . . , xn and by

√∆.

14. Let G be a subgroup of the group of automorphisms of C(z) (rationalfunctions in one variable) generated by automorphisms z 7→ 1 − z andz 7→ 1/z. Show that G has 6 elements and that the field of invariants C(z)G

is generated by one function. Find this function.15. Compute the Galois group of the polynomial x4 − 4x2 − 1 over Q.16. Suppose f(x) ∈ Q[x] is an irreducible polynomial such that one of its

complex roots has absolute value 1. Show that f(x) has even degree and ispalindromic: if f(x) = a0 + a1x+ . . .+ anx

n then a0 = an, a1 = an−1, etc.17. Suppose charK 6= 2 and let f(x) ∈ K[x] be an irreducible polynomial

of degree 5 such that its discriminant is a square in K∗. Find all possibleGalois groups for its splitting field.

§4. ROOTS OF UNITY

§4.1. Adjoining roots of unity.

DEFINITION 4.1.1. Let φ(n) = |Z∗n| be the Euler function, i.e. the number ofelements in Zn coprime to n.

PROPOSITION 4.1.2. Suppose n is coprime to charK. Let F/K be the splittingfield of xn − 1 (in which case we say that F is obtained from K by adjoining n-th roots of unity). Then [F : K] divides φ(n) and Gal(F/K) is isomorphic to asubgroup of (Z/nZ)∗.

Proof. Considerµn = {α ∈ F |αn = 1}.

Notice that µn is cyclic (as any finite subgroup in the multiplicative groupof a field) and has order n, Indeed, since n is coprime to charK,

(xn − 1, nxn−1) = 1.

It follows that xn − 1 has no multiple roots. Let G = Gal(F/K). Since Fis the splitting field of xn − 1, G acts faithfully on µn. Since any element of

22 JENIA TEVELEV

G is an automorphism of F , the action of G on µn preserves multiplication.So G is isomorphic to a subgroup of

Aut(µn) ' Aut(Zn) ' Z∗n,

which has φ(n) elements. In particular, [F : K] = |G| divides φ(n). �

§4.2. Cyclotomic fields. Now we explore the case K = Q. Let ζn ∈ C be aprimitive n-th root of unity, for example we can take ζn = e2πi/n. Since anyn-th root of 1 is a power of ζn, the splitting field of xn − 1 is equal to Q(ζn).

DEFINITION 4.2.1. Q(ζn) is called the cyclotomic field (Etymology: cyclotomyis the process of dividing the circle into equal parts, from cycl- + -tomy).

THEOREM 4.2.2. Let ζ = ζn. The cyclotomic field Q(ζ) has degree φ(n) over Q.The Galois group of Q(ζ)/Q is isomorphic to Z∗n. The minimal polynomial of ζ is

Φn(x) =∏

0<k<n(k,n)=1

(x− ζk)

(the cyclotomic polynomial). We have xn − 1 =∏d|n

Φd.

Proof. Let f(x) be the minimal polynomial of ζ. We know by Prop. 4.1.2that deg f = [Q(ζ) : Q] divides φ(n) and that Gal(Q(ζ)/Q) ⊂ Z∗n.

CLAIM 4.2.3. f(ζk) = 0 whenever (k, n) = 1.

This implies that f(x) has at least φ(n) roots, and therefore that

[Q(ζ) : Q] = deg f(x) = φ(n), f(x) = Φn(x), and Gal(Q(ζ)/Q) = Z∗n.

Proof of the Claim. If p is a prime divisor of k then ζk = (ζk/p)p and ζk/p

is also a primitive n-th root of unity. Arguing by induction on k it sufficesto show that f(ζp) = 0 if p is prime (and does not divide n). Arguing bycontradiction, suppose that f(ζp) 6= 0. We have a factorization

xn − 1 = f(x)g(x).

Since f(ζp) = 0, g(ζp) = 0. It follows that ζ is a root of g(xp). Therefore,

g(xp) = f(x)h(x). (1)

By Gauss lemma, polynomials f(x), g(x), and g(x) all have integer coef-ficients. So we can reduce (1) modulo p:

g(xp) ≡ f(x)h(x) mod p

≡ g(x)p mod p (Frobenius!) (2)

Let f(x) and g(x) be polynomials in Zp[x] obtained by reducing f(x) andg(x) modulo p. By (2) f(x) divides g(x)p, and therefore f(x) and g(x) arenot coprime. Therefore, xn−1 = f(x)g(x) has a multiple root in some finitefield Fpl . But since (p, n) = 1, (nxn−1, xn − 1) = 1, and therefore xn − 1 hasno multiple roots in Fpl . This is a contradiction. �


§4.3. Rational cosines. The following Lemma was left without proof inour discussion of the Hilbert 3d problem.

LEMMA 4.3.1. If cos 2πmn ∈ Q then it is equal to 1, 1

2 , 0, −12 , or−1. In particular,

arccos136∈ Qπ.

Proof. Suppose cos 2πmn ∈ Q. We can assume that (m,n) = 1. Then

ζ = cos2πmn

+ i sin2πmn∈ C.

is a primitive n-th root of 1. Recall that [Q(ζ) : Q] = φ(n). On other hand,since cos 2πm

n ∈ Q, we have

Q(ζ) ⊂ Q(i sin

2πmn

)= Q

(√cos2

2πmn− 1

)= Q(

√r),

where r is a rational number. So Q(ξ) is at most a quadratic extension of Q,and therefore,

φ(n) = [Q(ξ) : Q] = 1 or 2.Take a prime decomposition

n = pk11 . . . pkss .

By the Chinese theorem, we have an isomorphism of rings

Z/nZ = Z/pk11 ⊕ . . .⊕ Z/pkss .

This isomorphism induces an isomorphism of groups of invertible elements

(Z/nZ)∗ = (Z/pk11 )∗ × . . .× (Z/pkss )∗.

This gives a formulaφ(n) = φ(pk11 ) . . . φ(pks

s ).It is clear that φ(pk) = pk − pk−1 because a number is coprime to pk if andonly if it is coprime to p, and Z/pkZ contains exactly pk−1 elements that aredivisible by p. So we have

φ(n) = (pk11 − pk1−11 ) . . . (pks

s − pks−1s ).

If φ(n) ≤ 2 then each pi ≤ 3 and each ki ≤ 2. Going through the list ofpossibilities, we see that the only solutions are

n = 1, 2, 3, 4, 6.

This gives the Claim. �

§4.4. Abelian extensions of Q. The role of cyclotomic fields can be appre-ciated for the following theorem

THEOREM 4.4.1 (Kronecker–Weber). Any Galois extensionK/Q with an AbelianGalois group is contained in some cyclotomic field Q(ζn).

This remarkable theorem is very difficult, and attempts to generalize itto Abelian extensions of fields of algebraic numbers led to the developmentof Class Field Theory (and to modern Langlands program). Let’s just provethe easiest case, first observed by Gauss:

24 JENIA TEVELEV

THEOREM 4.4.2. Any quadratic extension K/Q is contained in some Q(ζn).

We will need the following definition:

DEFINITION 4.4.3. Since F∗p is cyclic and has even order, we have

F∗p/(F∗p)2 ' {±1}.

The induced homomorphism

F∗p � {±1}, ν 7→(ν

p

)is called the quadratic (or Legendre) symbol. It is equal to 1 if ν is a squarein Fp and −1 otherwise. Since the quadratic symbol is a homomorphism, itis multiplicative: (

ν

p

)(ν ′

p

)=(νν ′

p

).

Proof of Theorem 4.4.2. Any quadratic extension of Q has the form Q(√n),

where n is a square-free integer (why?). Let n = p1 . . . pr be the primedecomposition. Notice that ζl ∈ Q(ζm) if l|m. It follows that if

√pi ∈ Q(ζli)

for any i then√n ∈ Q(ζl1...lr). So it suffices to prove the Theorem for Q(

√p)

when p is prime.The case p = 2 is easy:

√2 = 2 cosπ/4 = ζ8 + ζ−1

8 .

Suppose now that p is odd. Let ζ = ζp and consider the Gauss sum

S =∑ν∈F∗p

(ν

p

)ζν ∈ Q(ζ).

CLAIM 4.4.4 (Gauss).

S2 =(−1p

)p

Given the claim,√p ∈ Q(ζp) if (−1

p ) = 1 and i√p ∈ Q(ζp) if (−1

p ) = −1. Inthe latter case

√p ∈ Q(ζ4p) (because i ∈ Q(ζ4)). The theorem is proved. �

Proof of the Claim. This is one long ingenious calculation

S2 =∑ν,µ∈F∗p

(ν

p

)(µ

p

)ζν+µ =

∑ν,µ∈F∗p

(νµ

p

)ζν+µ =

(a neat trick is to replace ν with νµ for any fixed µ ∈ F∗p)

=∑ν,µ∈F∗p

(νµ2

p

)ζνµ+µ =

(multiplicativity of the quadratic symbol)

=∑ν,µ∈F∗p

(ν

p

)ζµ(ν+1) =


(separate the cases ν = −1 and ν 6= −1)

=∑µ∈F∗p

(−1p

)ζ0 +

∑ν 6=−1

(ν

p

) ∑µ∈F∗p

ζµ(ν+1) =

(−1p

)(p− 1)−

∑ν 6=−1

(ν

p

)+∑ν 6=−1

(ν

p

) ∑µ∈Fp

ζµ(ν+1) =

(it is easy to see (why?) that∑µ∈Fp

(ζν+1)µ = 0)

=(−1p

)(p− 1)−

∑ν 6=−1

(ν

p

)=

(it is easy to see (why?) that∑ν∈Fp

(νp

)= 0)

= p

(−1p

)−∑ν∈Fp

(ν

p

)= p

(−1p

).

This proves the Claim. �

§4.5. Quadratic reciprocity. The most famous fact about quadratic sym-bols is the reciprocity law

THEOREM 4.5.1 (Quadratic Reciprocity, or Gauss’ Theorema Aureum).(p

q

)(q

p

)= (−1)

p−12· q−1

2

if p and q are odd primes.

Along with multiplicativity of the quadratic symbol and the formula(2p

)= (−1)

p2−18

(proved in the exercises), the reciprocity law can be used to quickly decidewhen a given number is a square modulo p.

Proof of the Quadratic Reciprocity. It is easy to check (see the exercises) that(p

q

)≡ p

q−12 mod q.

By (4.4.4), we have

S2 = p

(−1p

)= (−1)

p−12 p,

where S is the Gauss sum. So we have

Sq−1 = (−1)p−12

q−12 p

q−12 ≡ (−1)

p−12

q−12

(p

q

)mod q,

where from now on we work in the ring Z[ζp]. So “a ≡ b mod q” is just anotation for “a− b ∈ (q)”. On the other hand,

Sq ≡∑ν∈F∗p

(ν

p

)qζνq mod q (Frobenius!)

26 JENIA TEVELEV

≡∑ν∈F∗p

(ν

p

)ζνq ≡

∑ν∈F∗p

(νq

p

)(q

p

)ζνq ≡

(q

p

)S mod q.

We can combine two formulas for Sq to get the quadratic reciprocity law,but we have to be careful because we are doing calculations in Z[ζp] ratherthan in Z. So far we have proved that

(−1)p−12

q−12

(p

q

)S ≡

(q

p

)S mod q

This implies

(−1)p−12

q−12 S2 −

(q

p

)(p

q

)S2 ≡ 0 mod q,

i.e.

(−1)p−12

q−12 S2 −

(q

p

)(p

q

)S2 = qz,

where z ∈ Z[ζp]. Since S2 = ±p, z is obviously a rational number.

CLAIM 4.5.2. Z[ζp] ∩Q = Z.

One can give an ad hoc proof, but a conceptual reason will become clearlater after we discuss integral extensions of rings. Given the Claim, wehave

(−1)p−12

q−12 S2 −

(q

p

)(p

q

)S2 ≡ 0 mod q in Z.

Since S2 = ±p and (p, q) = 1, we can cancel S2. This shows that

(−1)p−12

q−12 −

(p

q

)(q

p

)is divisible by q in Z. Since its absolute value is at most 2, it is equal to 0. �

§4.6. Exercises.1. Let n andm be coprime integers. Show that Φn(x) (the n-th cyclotomic

polynomial) is irreducible over Q(ζm).2. Compute Φ5(x), Φ8(x), Φ12(x).3. Find all angles θ ∈ Qπ such that cos(θ) can be written as a + b

√D,

where a, b ∈ Q and D ∈ Z is square-free. Find a and b for each θ.4. Let

αr =

√2 +

√2 +

√2 + . . .+

√2 (r nested radicals).

(a) Show that the minimal polynomial fr(x) ∈ Q[x] of αr can be computedrecursively as follows: fr(x) = fr−1(x2−2), where f1(x) = x2−2. (b) De-scribe all roots of fr(x) in terms of cosines of various angles. (c) Show thatQ(αr)/Q is a Galois extension and compute its Galois group.

5. Let q be an odd prime and let a be an integer coprime to q. Then(a

q

)≡ a

q−12 mod q.


6. Let p be a on odd prime. Let α be a primitive 8-th root of unity in Fpand let y = α + α−1. (a) Show that yp = (−1)

p2−18 y. (b) Show that y2 = 2.

(c) Show that(

2p

)= (−1)

p2−18 .

7. Compute (20134567)

8. (a) Let K be an algebraic closure of K. Show that there exists theunique maximal (by inclusion) subfield K ⊂ Kab ⊂ K such that Kab/K isGalois and the Galois group Gal(Kab/K) is Abelian. (b) Deduce from theKronecker–Weber Theorem that

Qab =⋃n≥1

Q(ζn).

9. Let f(x) ∈ Z[x], deg(f) > 0. Show that the reduction of f(x) modulo phas a root in Fp for infinitely many primes p. [Hint: if f(x) = x, your argu-ment should become the Euclid’s argument for the infinitude of primes.]

10. Let Φn(x) be the n-th cyclotomic polynomial, a a non-zero integer, pa prime. Assume that p does not divide n. Prove that Φn(a) ≡ 0 mod pif and only if a has order n in (Z/pZ)∗.

11. Fix an integer n > 1. Use the previous two exercises to show2 thatthere exist infinitely many primes p such that p ≡ 1 mod n.

12. Let G be a finite Abelian group. (a) Show that there exists a posi-tive square-free integer n and a subgroup Γ ⊂ Z∗n such that G ' Z∗n/Γ.(b) Show that there exists a Galois extension K/Q with a Galois group G.(One of the most famous open problems, the inverse problem of GaloisTheory, is to prove the same statement for any finite group G).

13. (a) Show that∣∣∣∣∣∣∣∣∣∣∣

1 1 . . . 1α1 α2 . . . αnα2

1 α22 . . . α2

n...

.... . .

...αn−1

1 αn−12 . . . αn−1

n

∣∣∣∣∣∣∣∣∣∣∣=∏i>j

(αi − αj).

(b) For any k ≥ 0, let pk =n∑i=1

αki . Show that∣∣∣∣∣∣∣∣∣∣∣

p0 p1 . . . pn−1

p1 p2 . . . pnp2 p3 . . . pn+1...

.... . .

...pn−1 pn . . . p2n−2

∣∣∣∣∣∣∣∣∣∣∣=∏i>j

(αi − αj)2.

14. (continuation of the previous problem). (a) Let p be an odd primenumber. Show that the discriminant of the cyclotomic polynomial Φp(x) isequal to (−1)

p−12 pp−2. (b) Use (a) to give a different proof of the Kronecker–

Weber theorem for quadratic extensions.

2This proof is due to J. Sylvester. The result is a special case of the Dirichlet’s theorem onprimes in arithmetic progressions: for any coprime integers a and n, there exist infinitelymany primes p such that p ≡ a mod n. The proof uses complex analysis.

28 JENIA TEVELEV

§5. INTEGRAL EXTENSIONS OF RINGS

§5.1. Integral extensions. We would like to extend the notion of “algebraicextension” to rings. All rings in this section are commutative, with 1.

DEFINITION 5.1.1. Let R ⊂ S be rings.• An element α ∈ S is called integral over R if there exists a monic

polynomial f ∈ R[x] such that f(α) = 0.• The ring S is called integral over R if every α ∈ S is integral over R.• The integral closure of R in S is the set of all elements α ∈ S integral

over R.• The ring R is called integrally closed in S if R is equal to its integral

closure in S.• Let R be an integral domain. Its integral closure in the field of frac-

tions K is called a normalization or an integral closure of R.• An integral domainR is called normal or integrally closed ifR is equal

to its integral closure in the field of fractions K of R.

Here is the main technical result:

THEOREM 5.1.2. Let R ⊂ S be rings. Let α ∈ S. Then TFAE:(1) α is integral over R.(2) R[α] ⊂ S is a finitely generated R-module.(3) α ∈ T for some subring T ⊂ S that is a finitely generated R-submodule.

Proof. (1) ⇒ (2). There exists a monic polynomial f(x) ∈ R[x] of degree nsuch that f(α) = 0. It follows that αn ∈ R + Rα + . . . + Rαn−1. Arguingby induction on k, we see that αk ∈ R + Rα + . . .+ Rαn−1 for all k ≥ n. Itfollows that

R[α] = R+Rα+ . . .+Rαn−1

is finitely generated as an R-module.(2)⇒ (3). Take T = R[α].(3) ⇒ (1). If R is Noetherian then T , being a finitely generated R-

module, satisfies ACC for submodules. So if we denote Mk ⊂ T to bean R-submodule generated by 1, α, . . . , αk then Mn−1 = Mn for some n,i.e. αn ∈Mn−1. This shows that α is integral over R.

In general, one can argue as follows. Let e1, . . . , em be generators of theR-module T . For every i we can write

αei =∑j

aijej for some aij ∈ R.

This impliesAej = 0 for any j,

where A is the matrix [αδij − aij ]. Multiplying this identity by the adjointmatrix of A on the left gives

(detA)ej = 0 for any j.

Since 1 ∈ T , we can write 1 as an R-linear combination of e1, . . . , em. Thisimplies detA = 0. Therefore α is a root of the monic polynomial f(x) =det [xδij − aij ], i.e. the characteristic polynomial of [aij ]. �


LEMMA 5.1.3. Consider rings R ⊂ S ⊂ T . If S is a finitely generated R-moduleand T is a finitely generated S-module then T is a finitely generated R-module.

Proof. If {ei} generate S as an R-module and {fj} generate T as an S-module then and {eifj} generate T as an R-module. �

LEMMA 5.1.4. Consider rings R ⊂ S and suppose α1, . . . , as ∈ S are integralover R. Then R[α1, . . . , αs] is a finitely generated R-module.

Proof. Let

R0 = R, Ri = R[α1, . . . , αi] for i = 1, . . . , s.

Since αi is integral over R, it is aposteriori integral over Ri−1 as well forevery i = 1, . . . , s. It follows that Ri is a finitely-generated Ri−1-modulefor every i = 1, . . . , s. Applying Lemma 5.1.3 inductively shows that Rs =R[α1, . . . , αs] is a finitely generated R-module. �

COROLLARY 5.1.5. Suppose R ⊂ T ⊂ S, T is integral over R, and α ∈ S isintegral over T . Then α is integral over R.

Proof. There exists a monic polynomial f(x) = xn+a1xn−1 + . . .+an ∈ T [x]

such that f(α) = 0. Therefore α is integral over R[a1, . . . , an]. There-fore R[α, a1, . . . , an] is a finitely generated R[a1, . . . , an]-module. On theother hand, since a1, . . . , an are integral over R, Lemma 5.1.4 shows thatR[a1, . . . , an] is a finitely-generated R-module. Therefore R[α, a1, . . . , an] isa finitely generated R-module. So α is integral over R. �

COROLLARY 5.1.6. The integral closure ofR in S is a ring. This ring is integrallyclosed in S.

Proof. Let α, β ∈ S be integral over R. Then R[α, β] is a finitely generatedR-module by Lemma 5.1.4. Since α + β, αβ ∈ R[α, β], we can apply Theo-rem 5.1.2 (3) to conclude that α + β, αβ are integral over R. Therefore, theintegral closure T of R in S is a ring.

It remains to show that T is integrally closed. Suppose α ∈ S is integralover T . By Corollary 5.1.5, α is integral over R. Therefore α ∈ T . �

An important example:

THEOREM 5.1.7. Any UFD R is integrally closed in its field of fractions K.

Proof. Take α = p/q ∈ K, where p, q ∈ R and we can assume that theyare coprime. Suppose α is integral over R. Then f(α) = 0 for some monicpolynomial f(x) = xn + a1x

n−1 + . . .+ an ∈ R[x]. Therefore

pn + a1pn−1q + a2p

n−2q2 + . . .+ anqn = 0

and so q|pn. Since p and q are coprime, it follows that q is a unit in R,i.e. α ∈ R. So R is integrally closed in K. �

REMARK 5.1.8. Applying this argument to R = Z gives the standard proofof the “rational roots” theorem: a rational root of a monic integral polyno-mial is in fact an integer.

COROLLARY 5.1.9. Z[ζp] ∩Q = Z.

30 JENIA TEVELEV

Proof. Suppose α ∈ Z[ζp] ∩ Q. Since ζp is a root of a monic polynomialxp − 1 ∈ Z[x], it is integral over Z. Therefore, Z[ζp] is integral over Z.Therefore α is integral over Z. But Z is a UFD, so it is integrally closed in Q.Since α ∈ Q is integral over Z, we see that in fact α ∈ Z. �

COROLLARY 5.1.10 (Theorem on Symmetric Polynomials).

k[x1, . . . , xn]Sn = k[σ1, . . . , σn].

Proof. We already know that

k(x1, . . . , xn)Sn = k(σ1, . . . , σn).

So it suffices to show that

k[x1, . . . , xn] ∩ k(σ1, . . . , σn) = k[σ1, . . . , σn].

Suppose α ∈ k[x1, . . . , xn] ∩ k(σ1, . . . , σn). Since every xi is a root of amonic polynomial

∏(X − xi) ∈ k[σ1, . . . , σn][X], every xi is integral over

k[σ1, . . . , σn]. Therefore, k[x1, . . . , xn] is integral over k[σ1, . . . , σn]. There-fore α is integral over k[σ1, . . . , σn]. But k[σ1, . . . , σn] is a polynomial ring inn variables3. So it is a UFD by the Gauss lemma. Therefore it is integrallyclosed in k(σ1, . . . , σn). Since α ∈ k(σ1, . . . , σn) is integral over k[σ1, . . . , σn],we see that in fact α ∈ k[σ1, . . . , σn]. �

REMARK 5.1.11. In practice, there exists a simple algorithm to write anysymmetric polynomial f ∈ k[x1, . . . , xn]Sn as a polynomial in σ1, . . . , σn.This algorithm also proves that k[x1, . . . , xn]Sn = k[σ1, . . . , σn] by “lexico-graphic induction”. Order the variables x1 > x2 > . . . > xn and then orderall monomials in k[x1, . . . , xn] lexicographically. This is a total order. Thisalso gives a partial order on k[x1, . . . , xn]: we can compare two polynomialsby comparing their leading monomials. If f ∈ k[x1, . . . , xn]Sn has a leadingmonomial axk11 . . . xkn

n then it is easy to see that k1 ≥ k2 ≥ . . . ≥ kn. More-over, f−aσk1−k21 σk2−k32 . . . σkn

n is also Sn-invariant and its leading monomialis smaller than the leading monomial of f .

§5.2. Examples of the integral closure.

EXAMPLE 5.2.1. Let R = C[x, y]/(y2 − x3). We have a homomorphism

ψ : C[x, y]→ C[t], x 7→ t2, y 7→ t3.

The image is a subring C[t2, t3] (polynomials in t without a linear term).Let I be the kernel. Clearly (y2 − x3) ⊂ I . We claim that in fact they areequal. Let f(x, y) ∈ I . Modulo (y2 − x3), we can write

f(x, y) = a(x) + b(x)y.

So we have a(t2) + b(t2)t3 = 0. Coefficients in a (resp. b) contribute only toeven (resp. odd) degree monomials in t. So a(x) = b(x) = 0 and thereforef(x, y) = 0. So (y2 − x3) = I . By the first isomorphism theorem we have

R ' C[t2, t3],

3This is actually not obvious. We will see how to prove things like this carefully whenwe discuss transcendence degree later.


in particular R is a domain. Notice that t = t3/t2 belongs to the field offractions K of R. In particular, K = C(t). Since t 6∈ R but t is a root ofthe polynomial T 2 − t2 ∈ R[T ], R is not integrally closed. We claim that itsintegral closure is C[t] ⊂ K. Indeed, C[t] is integral over R. Also, C[t] is aUFD and therefore integrally closed. So C[t] is an integral closure of R.

DEFINITION 5.2.2. Let K be an algebraic extension of Q. The integral clo-sure of Z in K is called the ring of algebraic integers in K. Notation: OK .

EXAMPLE 5.2.3. Recall that Z[√

2] is a PID and therefore a UFD. ThereforeZ[√

2] is algebraically closed in its field of fractions, which is Q(√

2). Also,√2 is obviously integral over Z: it is a root of a monic polynomial x2 − 2.

So Z[√

2] is an integral closure of Z in Q(√

2).

EXAMPLE 5.2.4. We claim that Z[√

5] is not integrally closed in Q(√

5), andtherefore an integral closure of Z in Q(

√5) is strictly larger than Z[

√5].

Indeed, the golden ratio 1+√

52 belongs to Q(

√5) and is integral over Z: it is

a root of a monic polynomial x2 − x− 1.

THEOREM 5.2.5. Let K = Q(√D), where D is a square-free integer. Then

OK = Z[ω],

where

ω =

{√D if D = 2, 3 mod 4

1+√D

2 if D = 1 mod 4

Proof. Since√D is a root of x2−D = 0 and 1+

√D

2 is a root of x2−x+ 1−D4 = 0

(if D ≡ 1 mod 4), we see that Z[ω] is integral over Z. It remains to showthat if α = a+ b

√D ∈ Q(

√D) is integral over Z then α ∈ Z[ω]. Let f(x) be

the minimal polynomial ofα over Q. Sinceα is a root of a monic polynomialin Z[x], it follows, by Gauss lemma, that f(x) ∈ Z[x]. If b = 0 then a ∈ Zbecause Z is integrally closed. So let’s assume that b 6= 0. Then f(x) hasdegree 2. Write f(x) = x2 − px + q. The Galois group GalK/Q sends α toa− b

√D. Therefore this is also a root of f(x), and we have

p = 2a, q = a2 − b2D.If a ∈ Z then b2D ∈ Z, and, since D is square-free, b ∈ Z as well. In this caseα ∈ Z[

√D]. Since a = p/2 another possibility is that a = a′/2, where a′ is

an odd integer. Since (a′)2 − 4b2D ∈ 4Z, we see that 4b2D ∈ (a′)2 + 4Z andtherefore that 4b2D ≡ 1 mod 4. It follows that b = b′/2, where b′ is an oddinteger. Therefore α ∈ Z[ω]. �

REMARK 5.2.6. Notice that D can be negative. For example, since −1 ≡ 3mod 4, we see that Z[i] is an integral closure of Z in Q(i). This also followsfrom the fact that Z[i] is a UFD. Another interesting example if Z[

√−5]. By

the theorem above, this ring is integrally closed. However, it is not a UFD.Indeed, 6 has two different factorizations into irreducible elements:

(1 +√−5)(1−

√−5) = 2 · 3.

We should mention here a famous theorem of Heegner–Stark:

32 JENIA TEVELEV

THEOREM 5.2.7 (Heegner–Stark). Let K = Q(√d), where d is a negative inte-

ger. Then OK is a UFD if and only if

d ∈ {−1,−2,−3,−7,−11,−19,−43,−67,−163}.

This was first conjectured by Gauss. On the other hand, it is not knownhow to classify UFDs of this form with d > 0.

§5.3. Exercises.1. Let R be a domain with the field of fractions K. Let F/K be an alge-

braic extension and let S be the integral closure of R in F . For any α ∈ F ,show that there exists r ∈ R such that rα ∈ S.

2. Suppose n,m ≥ 2 are coprime positive integers. Show that

C[x, y]/(xn − ym)

is a domain and find its normalization.3. Let A be an integrally closed domain with the field of fractions K. Let

F/K be an algebraic extension of fields. Let α ∈ F . Show that α is integralover A if and only if its minimal polynomial has coefficients in A.

4. Prove that the Gauss Lemma holds not only in a UFD but in any in-tegrally closed domain R in the following form: suppose f(x) ∈ R[x] is amonic polynomial and f(x) = g(x)h(x), where g(x), h(x) ∈ K[x] are monicpolynomials (here K is the field of fractions of R). Then g(x), h(x) ∈ R[x].

5. Let A ⊂ B be domains and suppose that B is integral over A. LetI ⊂ B be an ideal. Show that B/I is integral over A/A ∩ I .

6. (a) Let A ⊂ B be rings and suppose that B is integral over A. Showthat A is a field if and only if B is a field. (b) Let A ⊂ B be rings andsuppose that B is integral over A. Let p ⊂ B be a prime ideal. Show that pis a maximal ideal of B if and only if p ∩A is a maximal ideal of A.

7. Let A be an integrally closed domain with the field of fractions K. LetF/K be a Galois extension with the Galois group G. Let B be the integralclosure of A in F . Show that G preserves B and that BG = A.

8. Let k be a field and let A be a finitely generated k-algebra. Let B ⊂ Abe a k-subalgebra such that A is integral over B. Show that A is a finitelygeneratedB-module andB is a finitely generated k-algebra (Hint: considera subalgebra of B generated by coefficients of monic equations satisfied bygenerators of A).

§6. CYCLIC EXTENSIONS

§6.1. Cyclic extensions. Any quadratic extension F/K (if charK 6= 2) canbe obtained by simply adding a quadratic root (of the discriminant) F =K(√D). It turns out that a very similar description is available for any

Galois extension with a cyclic Galois group:

THEOREM 6.1.1. Fix n ≥ 2. Suppose that charK does not divide n and that Kcontains all n-th roots of 1.

• Let α ∈ K be a root of xn − a for some a ∈ K. Then K(α)/K is Galois,and the Galois group is cyclic of order d, where d|n and αd ∈ K.• If F/K is a Galois extension with a cyclic Galois group of order n thenF = K(α) for some α ∈ F such that αn ∈ K.


Proof. Let ζ ∈ K be a primitive n-th root of 1.One direction is easy: Let α be a root of xn− a for some a ∈ K. Then ζkα

is also a root for any 1 ≤ k ≤ n − 1. It follows that xn − a splits in K(α).Since all the roots are distinct, we see that K(α)/K is Galois. Let G be theGalois group. For any g ∈ G, we have gα = ζkα for some unique k ∈ Z/nZ.This gives an injective homomorphism

G→ Z/nZ, g 7→ k.

Therefore, G is cyclic of order d|n. Let σ be a generator. Then σ(α) = να,where νd = 1. We have

σ(αd) = [σ(α)]d = νdαd = αd.

It follows that αd ∈ K.Now a less obvious implication: Let F/K be a Galois extension with a

cyclic Galois group G of order n. Let σ be a generator of the Galois group.It suffices to prove the following:

CLAIM 6.1.2. There exists α ∈ F ∗ such that σ(α) = ζα.

Indeed, given the Claim, the G-orbit of α is

{α, σ(α) = ζα, σ2(α) = ζ2α, . . . , σn−1(α) = ζn−1α}.

Since G acts transitively on the set of roots of the minimal polynomial of α,we see that this minimal polynomial is equal to

f(x) = (x− α)(x− ζα) . . . (x− ζn−1α).

In particular, [K(α) : K] = n, and therefore K(α) = F . Finally,

σ(αn) = ζnαn = αn,

and therefore αn = a ∈ K (it also follows that f(x) = xn − a).It remains to prove the Claim. We have a K-linear operator σ : F → F

such that σn = Id and we are trying to find an eigenvector for σ with aneigenvalue ζ. We give two proofs.

Proof A. Since σn = Id, the minimal polynomial of σ (viewed as a K-linear operator) divides Tn − 1. By our assumptions, the latter polyno-mial is separable and splits over K. Therefore, σ is diagonalizable over K.We claim that all eigenvalues are different. Equivalently, all n-th roots of 1appear as eigenvalues. Equivalently, any two eigenvectors with the sameeigenvector are linearly dependent. Indeed, suppose x, y ∈ F \ {0} areeigenvectors with the same eigenvalue ν. Then

σ(x/y) = σ(x)/σ(y) = (νx)/(νy) = x/y.

It follows that x/y is G-invariant, and therefore belongs to K by MTGT.Therefore, x and y are linearly dependent and we are done.

For practical purposes, it would be nice to find a way to produce eigen-vectors explicitly. Lagrange has discovered a nice trick for doing this.

Proof B. Consider the following K-linear operator on F :

A = Id +ζ−1σ + . . .+ ζ−(n−1)σn−1.

34 JENIA TEVELEV

By Lemma 6.2.1 below, this operator is not identically 0. Let β ∈ F be anyelement such that α := A(β) 6= 0. Then

α = β + ζ−1σ(β) + . . .+ ζ−(n−2)σn−2(β) + ζ−(n−1)σn−1(β) (3)

and

σ(α) = σ(β) + ζ−1σ2(β) + . . .+ ζ−(n−2)σn−1(β) + ζ−(n−1)σn(β) =

= ζβ + σ(β) + ζ−1σ2(β) + . . .+ ζ−(n−2)σn−1(β) = ζα.

We are done! �

§6.2. Artin’s Lemma.

LEMMA 6.2.1 (Artin).

(1) Take a field extension F/K and let σ1, . . . , σr be different automorphismsof F over K. Let α1, . . . , αr ∈ F and consider the following K-linearoperator A : F → F :

A = α1σ1 + . . .+ αrσr.

If A = 0 then α1 = . . . = αr = 0.(2) In fact, more is true: let F be a field, let Γ be a group, and let σi : Γ→ F ∗,

for i ∈ I , be different homomorphisms. Then they are linearly indepen-dent over F : if α1σ1 + . . . + αrσr = 0 as a function Γ → F for someα1, . . . , αr ∈ F then in fact α1 = . . . = αr = 0.

Proof. The first part follows from the second part by taking Γ = F ∗ (anyautomorphism F → F obviously induces a multiplicative homomorphismF ∗ → F ∗). To prove the second part, suppose we have a relation

α1σ1 + . . .+ αrσr = 0. (4)

We can assume that r is the minimal possible. Then r ≥ 2 and αi 6= 0 forany i. Since σ1, σ2 are different, there exists z ∈ Γ such that σ1(z) 6= σ2(z).Then we have

α1σ1(xz) + . . .+ αrσr(xz) = 0

for any x ∈ Γ, and therefore

α1σ1(z)σ1 + . . .+ αrσr(z)σr = 0

is another linear relation on our homomorphisms. We subtract it from (4)multiplied by σ1(z), which gives

α2(σ1(z)− σ2(z))σ2 + α3(σ1(z)− σ3(z))σ3 + . . . = 0.

Since σ1(z) 6= σ2(z), this is a non-trivial relation. But it has fewer than rterms, a contradiction. �


§6.3. Lagrange resolvents. Suppose that charK does not divide n and thatK contains all n-th roots of 1. Let F/K be a Galois extension with a cyclicGalois group of order n. Let σ be a generator. As we have seen before,F = K(α) for some α ∈ F such that αn ∈ K. Moreover, α can be computedas the following expression, called the Lagrange resolvent:

Eζ(β) = β + ζ−1σ(β) + . . .+ ζ−(n−2)σn−2(β) + ζ−(n−1)σn−1(β).

Let’s push this a little bit further. As a function of β, Eζ(β) is an K-linearfunction on F . We can define Eζk(β) for any 0 ≤ k < n in the same way.For example, E1(β) = β+σ(β) + . . .+σn−1(β). Artin’s Lemma tells us thateach of these resolvents is not equal to zero (for some β). This gives

COROLLARY 6.3.1. The action of σ on F (viewed as a K-vector space) is diago-nalizable with eigenvalues 1, ζ, . . . , ζn−1 and eigenvectors given by Lagrange re-solvents.

In fact, it is possible to show that we can use Lagrange resolvents withthe same β. To see this, let’s introduce a basis of F as a K-vector space andthe corresponding coordinates x1, . . . , xn. The function

E1(β)Eζ(β) . . . Eζn−1(β)

then can be viewed as an L-valued polynomial of degree n in n variables.Let’s assume for simplicity that K is an infinite field. Then this polyno-mial function does not vanish for some K-values of xi’s, i.e. for some β.It follows that we can find β ∈ F such that

E1(β), Eζ(β), . . . , Eζn−1(β)

are non-zero eigenvectors for σ with eigenvalues 1, ζ, . . . , ζn−1. It also fol-lows that vectors

β, σ(β), . . . , σn−1(β)are linear independent. In fact, their linear independence is equivalent tolinear independence of Lagrange resolvents because the transition matrixbetween the two systems of vectors is the Vandermonde matrix

1 1 . . . 11 ζ . . . ζn−1

1 ζ2 . . . ζ2(n−1)

......

. . ....

1 ζn−1 . . . ζ

with non-zero determinant.

We proved the special case of the following quite general

THEOREM 6.3.2 (Normal Basis Theorem). Let F/K be a finite Galois extensionof degree n with the Galois group G = {e = σ0, σ1, . . . , σn−1}. Then there existsβ ∈ F such that elements

β = σ0(β), σ1(β), . . . , σn−1(β)

form a basis of F over K.

The proof is not hard and can be found in Lang’s “Algebra”.

36 JENIA TEVELEV

§6.4. Exercises.1. Let K = C[z−1, z]] be the field of Laurent series (series in z, polyno-

mials in z−1). Let Km = C[z−1m , z

1m ]] ⊃ K. (a) Show that Km/K is Galois

with a Galois group Z/mZ. (b) Show that any Galois extension F/K witha Galois group Z/mZ is isomorphic to Km. (c) Show that

Kab =⋃m≥1

Km,

the field of so called Puiseux series4 (see the previous worksheet for the def-inition of Kab.

2. Let F/K be a Galois extension with a cyclic Galois group G of order p,where charK = p. Let σ be a generator of G. (a) Show that there existsα ∈ F such that σ(α) = α+ 1. (b) Show that F = K(α), where α is a root ofxp − x− a for some a ∈ K.

3. Suppose that charK = p and let a ∈ K. Show that the polynomialxp− x− a either splits in K or is irreducible. Show that in the latter case itsGalois group is cyclic of order p.

§7. NORM AND TRACE

DEFINITION 7.0.1. Let F/K be a separable extension of degree n (not nec-essarily Galois) and let σ1, . . . , σn : F → K be the set of all embeddingsover K. Let α ∈ F . We define its trace

TrF/K(α) = σ1(α) + . . .+ σn(α)

and normNF/K(α) = σ1(α) . . . σn(α).

EXAMPLE 7.0.2. We have NC/R(a+ ib) = (a+ ib)(a− ib) = a2 + b2.

One has to be careful: the norm and the trace depend on the extensionF/K and not just on α ∈ F . But this dependence is easy to understand:

LEMMA 7.0.3. Let F/K and L/F be separable extensions and let α ∈ F . Then

TrL/K(α) = [L : F ] TrF/K(α) and NL/K(α) = NF/K(α)[L:F ].

Proof. Any embedding L → K over K is an extension of some embeddingσ : F → K. By separability, there are [L : F ] possible extensions. �

As a consequence of Artin’s Lemma 6.2.1, we see that

COROLLARY 7.0.4. The trace TrF/K(α) 6= 0 for some α ∈ F .

There are two simple ways to compute the trace and the norm:

LEMMA 7.0.5. Let α ∈ K be a separable element with the minimal polynomialf(x) = xk + a1x

k−1 + . . .+ ak. Then

TrK(α)/K(α) = −a1 and NK(α)/K(α) = (−1)kak.

The trace gives an additive homomorphism TrF/K : F → K. The norm gives amultiplicative homomorphism NF/K : F ∗ → K∗.

4Isaac Newton proved that the field of Puiseux series in in fact algebraically closed.


Proof. Notice that embeddings K(α) → K just send α to all possible rootsof f(x). So the lemma follows from Vieta formulas. �

LEMMA 7.0.6. Let α ∈ F and let A be a K-linear operator F → F of left multi-plication by α. Then TrF/K(α) = Tr(A) and NF/K(α) = det(A).

Proof. Let e1, . . . , er be a basis of F over K(α). Then as a K-vector space, Fis a direct sum of vector subspaces

F = K(α)e1 ⊕ . . .⊕K(α)er.

Choosing a basis of F compatible with this decomposition, we see that thematrix of A in this basis is block-diagonal with r = [F : K(α)] blocks,where each block is a matrix of the left multiplication by α in K(α). Soit suffices to prove the lemma for the extension K(α)/K. In this case wechoose a basis 1, α, . . . , αk−1 of K(α), where k = [K(α) : K]. Let f(x) =xk + a1x

k−1 + . . .+ ak be the minimal polynomial of α. The matrix of A inthis basis is

0 0 . . . 0 −ak1 0 . . . 0 −ak−1

0 1 . . . 0...

.... . .

......

0 0 . . . 1 −a1

So we are done by Lemma 7.0.5. �

§7.1. Exercises. 1. Letα =√

2 +√

7. Compute the rational numberNQ(α)/Q(α).2. Let F/K be a Galois extension with a cyclic Galois group G. Let σ be

a generator of G. Show that

Ker[TrF/K ] = Im[IdF −σ].

In other words, if β ∈ F then TrF/K(β) = 0 iff β = α−σ(α) for some α ∈ F .3. Let F/K be a Galois extension with a cyclic Galois group G. Let σ be

a generator of G. Let β ∈ F . (a) There exists θ ∈ F such that α 6= 0, where

α = θ+βσ(θ)+βσ(β)σ2(θ)+βσ(β)σ2(β)σ3(θ)+. . .+βσ(β) . . . σn−2(β)σn−1(θ).

(b) Show that NF/K(β) = 1 if and only if β = α/σ(α) for some α ∈ F ∗.4. Let K = Q(ζ), where ζ is a primitive n-th root of unity. Show that if

n = pr for some prime p then NK/Q(1− ζ) = p.

§8. SOLVABLE EXTENSIONS

Galois discovered Galois theory while proving that a general quinticequation x5 + a1x

4 + . . . + a5 = 0 is not solvable in radicals. Moreover,he showed that only equations with solvable Galois groups are solvable inradicals (using Lagrange resolvents). Hence the name: a solvable group.Before proving this classical result, let’s make a digression into group the-ory to discuss composition series.

38 JENIA TEVELEV

§8.1. Composition series. These results are very general and the argumentsrely only on the first and the second isomorphism theorem. So they holdnot just for groups but also for R-modules, etc.

DEFINITION 8.1.1. A group G is called simple if it is not trivial, and has nonormal subgroups other than {e} and G itself.

EXAMPLE 8.1.2. Z/pZ for prime p, An for n ≥ 5, PGLn(Fp) if n and p arelarge enough, e.g. Klein’s 168-group PGL3(F2).

DEFINITION 8.1.3. Let G be a group. A sequence of subgroups

G = G1 ⊃ G2 ⊃ G3 ⊃ . . . ⊃ Gr = {e} (5)

is called a filtration of G. A refinement of (5) is a filtration obtained by insert-ing a finite number of subgroups between terms of (5). A filtration is callednormal if each Gi+1 is normal in Gi. A normal filtration is called compositionseries if each quotient Gi/Gi+1 (called composition factor) is a simple group.

LEMMA 8.1.4. Consider any normal filtration (5). After removing repeating termsGi = Gi+1, we can assume that all subquotients are non-trivial. If G is finite thenwe can refine this filtration to composition series.

Proof. If one of the quotients Gi/Gi+1 is not simple then let HEGi/Gi+1 bea proper normal subgroup. We can refine the filtration as follows:

Gi ⊃ p−1(H) ⊃ Gi+1,

where p : Gi → Gi/Gi+1 is the quotient map. The refinement proceduremust stop after at most log2 |G| − 1 steps. �

DEFINITION 8.1.5. Two normal filtrations of the same group, say (5) and

G = H1 ⊃ H2 ⊃ H3 ⊃ . . . ⊃ Hs = {e} (6)

are called equivalent if (after removing repeating terms) r = s and the se-quence of consequent quotients

G1/G2, G2/G3, . . . , Gr−1/Gr = Gr−1

can be rearranged so that they are respectively isomorphic to

H1/H2, H2/H3, . . . , Hs−1/Hs = Hs−1.

We have a quite miraculous theorem of Jordan and Holder:

THEOREM 8.1.6. Any two composition series of a group G are equivalent.

Since composition series is a normal filtration which can not be refined,the Jordan–Holder theorem is a corollary of the following even more gen-eral theorem of Schreier:

THEOREM 8.1.7. Any two normal filtrations of G have equivalent refinements.

Proof. Given normal filtrations (5) and (6), and for each i = 1, . . . , r− 1 andj = 1, . . . , s, we define

Gij = Gi+1(Hj ∩Gi).Notice that since Gi+1 is normal in Gi, Gij is a subgroup of Gi and we canrefine our first filtration by inserting blocks

Gi = Gi1 ⊃ Gi2 ⊃ . . . ⊃ Gis = Gi+1.


Since Hj+1 is normal in Hj , Gi,j+1 is normal in Gij . So this refinement is anormal filtration. Similarly, we can define

Hji = Hj+1(Gi ∩Hj)

for j = 1, . . . , s − 1 and i = 1, . . . , r. This gives a refinement of the secondfiltration. Both refined filtrations have (r−1)(s−1)+1 elements, namelyGijand {e} in the first case andHji and {e} in the second case, where the rangefor indices is i = 1, . . . , r − 1, j = 1, . . . , s − 1. We claim that these normalfiltrations are equivalent, and more precisely, the quotients of Gij and Hji

are isomorphic (of course many of these quotients will be simultaneouslytrivial). This is the content of the next lemma. �

LEMMA 8.1.8 (Zassenhaus Lemma). Let U, V be subgroups of a group G. Letu, v be normal subgroups of U , V , respectively. Then u(U ∩ v) is normal inu(U∩V ), (u∩V )v is normal in (U∩V )v, and the quotient groups are isomorphic.

Proof. We are going to prove that(u(U ∩ V )

)/(u(U ∩ v)

)' (U ∩ V )/

((u ∩ V )(U ∩ v)

). (5)

The Lemma will follow from symmetry. Since U ∩ V normalizes u, we seethat u(U ∩ V ) is a subgroup. Set

H = U ∩ V and N = u(U ∩ v).

Then H normalizes N and by the second isomorphism theorem for groupswe have

HN/N ' H/H ∩N.A small calculation shows that

H ∩N = U ∩ V ∩(u(U ∩ v)

)= (u ∩ V )(U ∩ v)

andHN = (U ∩ V )u(U ∩ v) = (U ∩ V )u = u(U ∩ V ).

This gives (5). �

§8.2. Solvable groups.

DEFINITION 8.2.1. A normal filtration (5) of a group G is called Abelian(resp. cyclic) if each quotient Gi/Gi+1 is Abelian (resp. cyclic). A groupG is called solvable if it has an Abelian filtration.

As a corollary of Lemma 8.1.4, we have

COROLLARY 8.2.2. An Abelian filtration of a finite solvable group can be refinedto cyclic composition series, i.e. such that subsequent quotients are cyclic groupsZ/pZ (for various primes p).

REMARK 8.2.3. Notice that the Jordan–Holder theorem for finite solvablegroups immediately follows from the unique factorization of integers (theproduct of all primes appearing in composition factors Z/pZ is obviouslyequal to |G|). In general, the Jordan–Holder theorem can be viewed as aform of unique factorization for groups.

Here are some basic examples of solvable groups:

40 JENIA TEVELEV

LEMMA 8.2.4. A finite p-group G (i.e. a group of size pn for some n) is solvable.Any Abelian group is solvable. A dihedral group is solvable.

Proof. Induction on |G|. A basic fact about finite p-groups is that their cen-ter Z(G) is not-trivial (because if x 6∈ Z(G) then the number of elementsconjugate to x is divisible by p). We have a normal filtration

G ⊃ Z(G) ⊃ {e}.The quotient group G/Z(G) is a p-group of smaller size, so by induction ithas an Abelian filtration. Pulling it back to G gives an Abelian refinementof the inclusion G ⊃ Z(G). �

LEMMA 8.2.5. A subgroup H of a solvable group G is solvable.

Proof. An Abelian filtration (5) induces a filtration

H = H1 = H ∩G1 ⊃ H2 = H ∩G2 ⊃ . . . ⊃ Hr = H ∩Gr.Since Gi+1 is normal in Gi, Hi+1 = Gi+1 ∩H is normal in Hi = Gi ∩H . Theinclusion Hi ⊂ Gi induces the inclusion Hi/Hi+1 ⊂ Gi/Gi+1. ThereforeHi/Hi+1 is Abelian. �

§8.3. Solvable extensions: Galois Theorem. In this section we will as-sume for simplicity that

charK = 0.Alternatively, one can assume that all extensions we consider are separableand their degrees are not divisible by characteristic.

DEFINITION 8.3.1. Let F/K be a finite field extension.• The extension F/K is called solvable if there exists a Galois extensionL/K containing F with a solvable Galois group.• The extension F/K is solvable in radicals if there exists a tower

K = L0 ⊂ L1 ⊂ . . . ⊂ Lrsuch that F ⊂ Lr and such that Li = Li−1( ni

√ai) for some ai ∈ Li−1.

THEOREM 8.3.2. F/K is solvable if and only if it is solvable in radicals.

Proof. All fields appearing in the proof will be subfields of the fixed alge-braic closure K. Let F/K be a solvable extension. Let L/K be the Galoisextension containing F with a solvable Galois group G of size n.

Let K(ζn) be the splitting field of xn − 1. Consider the diagram of fields

L(ζn)

L K(ζn)

K

By the next Lemma (proved in the homework), L(ζn)/K(ζn) is a Ga-lois extension and its Galois group H is isomorphic to Gal(L/L ∩ K(ζn)).The latter group is a subgroup of G. So H is solvable.


LEMMA 8.3.3. Let L ⊂ K be a finite Galois extension of K and let F ⊂ K be anyfinite extension of K. Consider the diagram of field extensions

LF

L F

K

Then the composite field LF is Galois over F and the Galois group Gal(LF/F ) isisomorphic to Gal(L/L ∩ F ).

A cyclic tower of subgroups

H = H1 ⊃ H2 ⊃ . . . ⊃ Hr = {e}gives rise to a tower of subfields

K(ζn) = J1 ⊂ J2 ⊂ . . . ⊂ Jr = L(ζn),

whereJi = L(ζn)Hi .

By the Main Theorem of Galois theory, L(ζn)/Ji is Galois with a Galoisgroup Hi. Since Hi+1 is normal in Hi, Ji+1/Ji is a Galois extension withGalois group Hi/Hi+1, which is cyclic.

Since Ji+1/Ji is a cyclic extension of degree d|n (by Lagrange Theorem),and Ji contains n-th roots of unity, we can apply Theorem 6.1.1. We seethat on each step Ji+1 = Ji(α), where some power of α belongs to Ji−1. Itfollows that F/K is solvable in radicals.

Conversely, suppose F/K is solvable in radicals, i.e. F is contained in afield L that admits a tower

K ⊂ L1 ⊂ . . . ⊂ Lr = L

such that on each step Li = Li−1(α), where αk ∈ Li−1 for some k. Let n bethe l.c.m. of k’s that appear. Consider the tower of fields

K ⊂ K(ζn) ⊂ L1(ζn) ⊂ . . . ⊂ Lr(ζn) = M,

where each consecutive embedding is Galois with an Abelian Galois groupon the first step (by Theorem 4.1.2) and a cyclic Galois group for the remain-ing steps (by Theorem 6.1.1). However, we are not quite done yet becauseM/K is not necessarily Galois.

Let g1, . . . , gk : M → K be the list of all embeddings over K, where g1 isthe identity. Each of the embeddings gi(M) ⊂ K has the same property asabove: in the corresponding tower

K ⊂ gi(K(ζn)) ⊂ gi(L1(ζn)) ⊂ . . . ⊂ gi(M), (6)

each consecutive embedding is Galois with an Abelian Galois group. No-tice that the composite field M = g1(M) . . . gk(M) is Galois over K andadmits a tower of field extensions

K ⊂ g1(M) ⊂ g1(M)g2(M) ⊂ . . . ⊂ g1(M) . . . gk(M) =M

42 JENIA TEVELEV

Consider the i-th step of this tower

N ⊂ Ngi(M),

where N = g1(M) . . . gi−1(M). We can refine this inclusion of fields bytaking a composite of (6) with N . By Lemma 8.3.3, each consecutive em-bedding in this mega-tower is Galois with an Abelian Galois group. By theMain Theorem of Galois Theory, this tower of subfields ofM correspondsto an Abelian filtration of Gal(M/K). Therefore this group is solvable. �

§8.4. Solving solvable extensions.

DEFINITION 8.4.1. A polynomial f ∈ K[x] is called• solvable if its Galois group is solvable.• solvable in radicals if for any root β of f(x) in K there exists a tower

K = L0 ⊂ L1 ⊂ . . . ⊂ Lrsuch that β ∈ Lr and such that Li = Li−1( ni

√ai) for some ai ∈ Li−1.

Theorem 8.3.2 implies that these two notions are equivalent. For exam-ple, a sufficiently general polynomial f(x) ∈ Q[x] of degree n has Galoisgroup Sn, and therefore is not solvable for n > 4. On the other hand, anyequation of degree at most 4 is solvable in radicals because its Galois groupis a subgroup of S4, and the latter group is solvable. The proof of the Galoistheorem on solvable extensions is constructive, so one can actually “solve”solvable extensions. Let’s consider an equation of degree 3:

x3 + a1x2 + a2x+ a3 = 0 ∈ K[x].

Since we are going to apply the Galois theorem, let’s assume right awaythat K contains a primitive cubic root of unity ω and that charK 6= 2, 3.To simplify calculations, lets make a change of variables x 7→ x−a1/3. Thiskills a1. So we can assume that

a1 = 0.

Let F be the splitting field. In this field

f(x) = (x− x1)(x− x2)(x− x3).

Let G = Gal(F/K) ⊂ S3. A cyclic filtration {e} ⊂ A3 ⊂ S3 gives a cyclicfiltration

{e} ⊂ A3 ∩G ⊂ Gand the corresponding tower

F ⊃ FA3∩G ⊃ K.

The extension F/FA3∩G has Galois group A3 ∩G ⊂ Z/3Z which acts bycyclically permuting the roots x1 → x2 → x3 → x1. Let’s write down allLagrange resolvents:

E1 = x1 + x2 + x3

Eω = x1 + ω2x2 + ωx3

Eω2 = x1 + ωx2 + ω2x3


It suffices to derive formulas for the Lagrange resolvents, since then wecan compute the roots x1, x2, x3 by solving a system of linear equations. ByVieta formulas, we have

E1 = −a1 = 0

and so it suffices to compute

E3ω, E

3ω2 ∈ FA3∩G.

The extension FA3∩G/K is cyclic with a Galois group contained in S3/A3 'Z/2Z: this quotient group is generated by a transposition σ that exchangesx2 ↔ x3. By our general recipe, instead of computing E3

ω and E3ω2 directly,

we want to compute their Lagrange resolvents

E3ω ± σ(E3

ω) and E3ω2 ± σ(E3

ω2),

which have the property that their squares belong to K. Then we computeE3ω and E3

ω2 by solving a system of linear equation.Here the calculation is simplified by the fact that

σ(E3ω) = E3

ω2 .

So the Lagrange resolvents for the second step are simply

E3ω ± E3

ω2 .

It remains to compute these resolvents and then to solve the system of twolinear equations in two variables to find E3

ω and E3ω2 . Note that E3

ω + E3ω2

is invariant under σ, i.e. it is in fact a symmetric polynomial in x1, x2, x3,i.e. it should be possible to express it in terms of coefficients of f(x). A littlecalculation shows that

E3ω + E3

ω2 = −27a3.

A square of E3ω − E3

ω2 should also be invariant, in fact we have

(E3ω − E3

ω2)2 = 27(4a32 + 27a2

3) = −27∆2,

where D is the discriminant. This gives formulas for roots of the cubicequation discovered by the Italian mathematician Niccolo Tartaglia.

§8.5. Exercises. 1. Show that a finite field extension F/K is solvable if andonly if GalL/K is solvable, where L is the Galois closure of F in K.

2. Let p be a prime number and let ζp ∈ C be a primitive p-th root ofunity. Show that Gal(Q(ζp,

p√

2)/Q) is a semidirect product of Z/pZ and F∗p.3. Suppose D4 acts on F = C(x1, . . . , x4) by permutations of variables

(here we identify variables with vertices of the square). Show that FD4 isgenerated over C by 4 functions and find them.

4. Let M be a module over a ring R. A sequence of submodules

M = M1 ⊃M2 ⊃ . . . ⊃Mr = 0

is called a filtration of M (of length r). A module M is called simple if itdoes not contain any submodules other than 0 and itself. A filtration iscalled simple if each Mi/Mi+1 is simple. A module M is said to be of finitelength if it admits a simple finite filtration. Two filtrations of M are called

44 JENIA TEVELEV

equivalent if they have the same length and the same collection of subquo-tients {M1/M2,M2/M3, . . . ,Mr−1/Mr} (up to isomorphism and renumber-ing). Prove that if M has finite length then any two simple filtrations of Mare equivalent and any filtration of M can be refined to a simple filtration.

5. (a) Let f(x) ∈ K[x] be an irreducible separable polynomial with roots

α = α1, α2, . . . , αn ∈ K.Suppose that there exist rational functions θ1(x), . . . , θn(x) ∈ K(x) suchthat αi = θi(α) for any i. Suppose also that

θi(θj(α)) = θj(θi(α))

for any i, j. Show that the Galois group of the splitting field of f(x) isAbelian5. (b) Give an example of the situation as in part (a) with K = Qand such that the Galois group of f(x) is not cyclic. Give a specific polyno-mial f(x), and compute its roots and functions θi.

§9. SAMPLE MIDTERM

1. Let L/K be a finite Galois extension with Galois group G. A mapf : G→ L∗ is called a Galois 1-cocycle if it has the following property:

f(στ) = f(σ)σ(f(τ)) for any σ, τ ∈ G.Show that a map f : G→ L∗ is a Galois 1-cocycle if and only if there existsα ∈ L∗ such that

σ(α) = f(σ)α for any σ ∈ G.

2. Let p be a prime number. Find the Galois group of f(x) = x4+p ∈ Q[x].

3. Let L/K be a Galois extension with Galois group SL2(F7). Find thenumber of subfields K ⊂ F ⊂ L such that [F : K] = 48.

4. Let p be a prime number. Let K be a field. Let a ∈ K. Suppose thatone of the following two conditions holds:

(a) charK = p.(b) charK 6= p and K contains all p-th roots of 1.

Show that f(x) = xp − a ∈ K[x] is either irreducible or has a root in K.

5. Let L/K be a finite Galois extension with Galois group G. Show thatthe following properties are equivalent:

(a) L is a splitting field of an irreducible polynomial of degree n.(b) G contains a subgroup H of index n such that ∩

g∈GgHg−1 = {e}.

6. Let K be a field. Show that if Gal K/K is Abelian then any finiteseparable extension of K is Galois.

5This was proved by Abel himself.

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