Rings as the unions of proper subrings - MathUniPDlucchini/index_files/ring.pdf · 2010-03-21 · IS THERE AN ANALOGUE OF SCORZA’S THEOREM FOR RINGS? EXAMPLES PROOFS EXERCIZE No

RINGS AS THE UNIONS OF PROPER SUBRINGS

Andrea Lucchini

Università di Padova, Italy

Una giornata per SilviaPadova, March 26 - 27, 2010

ANDREA LUCCHINI RINGS AS THE UNIONS OF PROPER SUBRINGS

IS THERE AN ANALOGUE OF SCORZA’S THEOREM FOR RINGS?EXAMPLES

PROOFS

EXERCIZE

No group is the union of two of its proper subgroups.

SCORZA 1926

A group G is a union of three of its pairwise distinct proper subgroupsA, B, C if and only if

A, B, C have index 2 in G;

G/(A ∩ B ∩ C) is isomorphic to the Klein four group.



PROOFS

EXERCIZE

No group is the union of two of its proper subgroups.

SCORZA 1926

A group G is a union of three of its pairwise distinct proper subgroupsA, B, C if and only if

A, B, C have index 2 in G;

G/(A ∩ B ∩ C) is isomorphic to the Klein four group.

QUESTIONS

Are there similar results for rings?

No ring is the union of two of its proper subrings.Is there an analogue of Scorza’s result for rings?



PROOFS

QUESTIONS

Is there an analogue of Scorza’s result for rings?



PROOFS

QUESTIONS


It is sufficient to classify all rings R and all proper subrings S1, S2, S3of R such that

R = S1 ∪ S2 ∪ S3

no non-trivial ideal of R is contained in S1 ∩ S2 ∩ S3.



PROOFS

QUESTIONS



R = S1 ∪ S2 ∪ S3


DEFINITION

A 4-tuple (R, S1, S2, S3) of rings is good if S1, S2, S3 are propersubrings of the ring R so that R = S1 ∪ S2 ∪ S3 and that no non-trivialideal of R is contained in S1 ∩ S2 ∩ S3.



PROOFS

QUESTIONS



R = S1 ∪ S2 ∪ S3


DEFINITION

A 4-tuple (R, S1, S2, S3) of rings is good if S1, S2, S3 are propersubrings of the ring R so that R = S1 ∪ S2 ∪ S3 and that no non-trivialideal of R is contained in S1 ∩ S2 ∩ S3.

THEOREM (ATTILA MAROTI - AL 2009)

All good 4-tuples of rings are completely described by the followingten Examples.



PROOFS

EXAMPLE 1

R =

{(0 00 0

),

(1 00 1

),

(1 00 0

),

(0 00 1

)}≤ M2(Z/2Z).

A commutative ring of order 4 with a multiplicative identity.



PROOFS

EXAMPLE 1

R =

{(0 00 0

),

(1 00 1

),

(1 00 0

),

(0 00 1

)}≤ M2(Z/2Z).

A commutative ring of order 4 with a multiplicative identity.Every non-zero element lies inside a unique subring of order 2.



PROOFS

EXAMPLE 1

R =

{(0 00 0

),

(1 00 1

),

(1 00 0

),

(0 00 1

)}≤ M2(Z/2Z).

A commutative ring of order 4 with a multiplicative identity.Every non-zero element lies inside a unique subring of order 2.R is the direct product of two fields of order 2.



PROOFS

EXAMPLE 2

0 0 0

0 0 00 0 0

,

0 0 01 0 00 0 0

,

0 0 00 0 01 0 0

,

0 0 01 0 01 0 0

≤ M3(Z/2Z)

A commutative ring of order 4.



PROOFS

EXAMPLE 2

0 0 0

0 0 00 0 0

,

0 0 01 0 00 0 0

,

0 0 00 0 01 0 0

,

0 0 01 0 01 0 0

≤ M3(Z/2Z)

A commutative ring of order 4.It has no multiplicative identity since it is a zero ring.



PROOFS

EXAMPLE 2

0 0 0

0 0 00 0 0

,

0 0 01 0 00 0 0

,

0 0 00 0 01 0 0

,

0 0 01 0 01 0 0

≤ M3(Z/2Z)

A commutative ring of order 4.It has no multiplicative identity since it is a zero ring.Every non-zero element lies inside a unique subring of order 2.



PROOFS

EXAMPLES 3 AND 4

R =

{(0 00 0

),

(0 10 0

),

(1 00 0

),

(1 10 0

)}≤ M2(Z/2Z).

Rop =

{(0 00 0

),

(0 01 0

),

(1 00 0

),

(1 01 0

)}≤ M2(Z/2Z).

Two non-commutative rings of order 4.



PROOFS

EXAMPLES 3 AND 4

R =

{(0 00 0

),

(0 10 0

),

(1 00 0

),

(1 10 0

)}≤ M2(Z/2Z).

Rop =

{(0 00 0

),

(0 01 0

),

(1 00 0

),

(1 01 0

)}≤ M2(Z/2Z).

Two non-commutative rings of order 4.They have no multiplicative identity.



PROOFS

EXAMPLES 3 AND 4

R =

{(0 00 0

),

(0 10 0

),

(1 00 0

),

(1 10 0

)}≤ M2(Z/2Z).

Rop =

{(0 00 0

),

(0 01 0

),

(1 00 0

),

(1 01 0

)}≤ M2(Z/2Z).

Two non-commutative rings of order 4.They have no multiplicative identity.Every non-zero element lies inside a unique subring of order 2.



PROOFS

EXAMPLE 5

R =

a 0 0

0 b 00 0 c

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)




PROOFS

EXAMPLE 5

R =

a 0 0

0 b 00 0 c

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)

A commutative ring of order 8 with a multiplicative identity.R is the union of three subrings of order 4 containing themultiplicative identity:

S1 defined by the restriction a + b = 0S2 defined by the restriction a + c = 0S3 defined by the restriction b + c = 0.



PROOFS

EXAMPLE 5

R =

a 0 0

0 b 00 0 c

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)


S1 defined by the restriction a + b = 0S2 defined by the restriction a + c = 0S3 defined by the restriction b + c = 0.

S1 ∩ S2 ∩ S3 =

a 0 0

0 a 00 0 a

∣∣∣a ∈ Z/2Z

is a subring of R, but

it is not an ideal.



PROOFS

EXAMPLE 6

R =

a 0 0

b a 0c 0 a

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)




PROOFS

EXAMPLE 6

R =

a 0 0

b a 0c 0 a

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)


S1 defined by the restriction b = 0S2 defined by the restriction c = 0S3 defined by the restriction b + c = 0.



PROOFS

EXAMPLE 6

R =

a 0 0

b a 0c 0 a

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)



The subset of R obtained by imposing the restriction a = 0 isisomorphic to the ring R of Example 2.



PROOFS

EXAMPLE 6

R =

a 0 0

b a 0c 0 a

∣∣∣a, b, c ∈ Z/2Z

≤ M3(Z/2Z)



The subset of R obtained by imposing the restriction a = 0 isisomorphic to the ring R of Example 2.R can be obtained from Example 2 by adding a multiplicativeidentity 1 and imposing the relation 1 + 1 = 0.



PROOFS

EXAMPLE 7

R :=

{(a c0 b

) ∣∣∣a, b, c ∈ Z/2Z}≤ M2(Z/2Z)

A non-commutative ring of order 8 containing a multiplicativeidentity.



PROOFS

EXAMPLE 7

R :=

{(a c0 b

) ∣∣∣a, b, c ∈ Z/2Z}≤ M2(Z/2Z)

A non-commutative ring of order 8 containing a multiplicativeidentity.R is the union of three subrings of order 4 containing themultiplicative identity:

S1 defined by the restriction c = 0S2 defined by the restriction a + b = 0S3 defined by the restriction a + b + c = 0.



PROOFS

EXAMPLE 7

R :=

{(a c0 b

) ∣∣∣a, b, c ∈ Z/2Z}≤ M2(Z/2Z)



R ∼= Rop.



PROOFS

EXAMPLE 7

R :=

{(a c0 b

) ∣∣∣a, b, c ∈ Z/2Z}≤ M2(Z/2Z)



R ∼= Rop.

R can be obtained from Examples 3 or 4 by adding amultiplicative identity 1 and imposing the relation 1 + 1 = 0.



PROOFS

EXAMPLES 8 AND 9

R =

0 b c d0 b 0 00 0 b 00 0 0 b

∣∣∣b, c, d ∈ Z/2Z

≤ M4(Z/2Z).

Rop =

0 0 0 0b b 0 0c 0 b 0d 0 0 b

∣∣∣b, c, d ∈ Z/2Z

≤ M4(Z/2Z).

Two non-commutative rings of order 8.



PROOFS

EXAMPLES 8 AND 9

R =

0 b c d0 b 0 00 0 b 00 0 0 b

∣∣∣b, c, d ∈ Z/2Z

≤ M4(Z/2Z).

Rop =

0 0 0 0b b 0 0c 0 b 0d 0 0 b

∣∣∣b, c, d ∈ Z/2Z

≤ M4(Z/2Z).

Two non-commutative rings of order 8.They have no multiplicative identity.



PROOFS

EXAMPLES 8 AND 9

R =

0 b c d0 b 0 00 0 b 00 0 0 b

∣∣∣b, c, d ∈ Z/2Z

≤ M4(Z/2Z).

Rop =

0 0 0 0b b 0 0c 0 b 0d 0 0 b

∣∣∣b, c, d ∈ Z/2Z

≤ M4(Z/2Z).

Two non-commutative rings of order 8.They have no multiplicative identity.They are the union of three subrings of order 8:

S1 defined by the restriction c = 0S2 defined by the restriction d = 0S3 defined by the restriction c + d = 0.



PROOFS

EXAMPLE 10

R :=

a 0 0 0b e 0 0c 0 e 0d 0 0 e

∣∣∣a, b, c, d ∈ Z/2Z, a + b = e

≤ M2(Z/2Z)

A non-commutative ring of order 16 containing a multiplicativeidentity.



PROOFS

EXAMPLE 10

R :=

a 0 0 0b e 0 0c 0 e 0d 0 0 e

∣∣∣a, b, c, d ∈ Z/2Z, a + b = e

≤ M2(Z/2Z)

A non-commutative ring of order 16 containing a multiplicativeidentity.R is the union of three subring of order 8 containing themultiplicative identity:




PROOFS

EXAMPLE 10

R :=

a 0 0 0b e 0 0c 0 e 0d 0 0 e

∣∣∣a, b, c, d ∈ Z/2Z, a + b = e

≤ M2(Z/2Z)



R ∼= Rop.



PROOFS

EXAMPLE 10

R :=

a 0 0 0b e 0 0c 0 e 0d 0 0 e

∣∣∣a, b, c, d ∈ Z/2Z, a + b = e

≤ M2(Z/2Z)



R ∼= Rop.

R can be obtained from Examples 8 or 9 by adding amultiplicative identity 1 and imposing the relation 1 + 1 = 0.



PROOFS

EXAMPLE 10

R :=

a 0 0 0b e 0 0c 0 e 0d 0 0 e

∣∣∣a, b, c, d ∈ Z/2Z, a + b = e

≤ M2(Z/2Z)

(R, +) = 〈1, x , y , z〉 ∼= (Z/2Z)4 andz2 = z,x2 = y2 = xy = yx = 0,xz = x , yz = y , zx = 0, zy = 0.

S1 = 〈1, x , z〉, S2 = 〈1, y , z〉, S3 = 〈1, x + y , z〉



PROOFS

In same of the previous examples, R contains and ideal I such thatR/I is isomorphic to one of the other examples.



PROOFS


Examples 5, 7, 10 have Example 1 as an epimorphic image.Example 8 has Example 4 as an epimorphic image.Example 9 has Example 3 as an epimorphic image.



PROOFS



But this is not the case of Example 6.



PROOFS



But this is not the case of Example 6.

THEOREM

A ring R is the union of three of its proper subrings if and only if thereexists a factor ring (of order 4 or 8) of R which is isomorphic to ring ofexamples 1, 2, 3, 4 or 6.



PROOFS

HOW CAN WE PROVE OUR RESULT?

Let (R, S1, S2, S3) be a good 4-tuple of rings and let S = S1∩S2∩S3.Scorza’s Theorem⇒ |R : Si | = 2 for each i ∈ {1, 2, 3} andS1 ∩ S2 = S1 ∩ S3 = S2 ∩ S3 = S.2R ⊆ S1 ∩ S2 ∩ S3 ⇒ 2R = 0.



PROOFS

HOW CAN WE PROVE OUR RESULT?

Let (R, S1, S2, S3) be a good 4-tuple of rings and let S = S1∩S2∩S3.Scorza’s Theorem⇒ |R : Si | = 2 for each i ∈ {1, 2, 3} andS1 ∩ S2 = S1 ∩ S3 = S2 ∩ S3 = S.2R ⊆ S1 ∩ S2 ∩ S3 ⇒ 2R = 0.

S is a subring of R with |R : S| = 4.

[J. Lewin (1967)] Suppose that a ring R contains a subring S offinite index. Then there exists an ideal I of R, contained in S andof finite index in R.R is finite (|R| ≤ 525).



PROOFS

A USEFUL REMARK

Let M be a ring (with or without a multiplicative identity) with2M = 0.

Consider the abelian group M∗ = M ⊕ 〈u〉 with u + u = 0 anddefine a multiplication on M∗ by setting u to be the identity on M∗.

M∗ becomes a ring with a multiplicative identity.



PROOFS

A USEFUL REMARK

Let M be a ring (with or without a multiplicative identity) with2M = 0.

Consider the abelian group M∗ = M ⊕ 〈u〉 with u + u = 0 anddefine a multiplication on M∗ by setting u to be the identity on M∗.

M∗ becomes a ring with a multiplicative identity.

Let (R, S1, S2, S3) be a good 4-tuple . Suppose that R has nomultiplicative identity. Then (R∗, S∗

1 , S∗2 , S∗

3 ) is also a good 4-tuplewhere a unique multiplicative identity was added to the four rings R,S1, S2, and S3.

We may restrict our attention to rings with a multiplicative identity.



PROOFS

PROPOSITION

Let R be a good ring. Then |R| = 4, 8, or 16.

PROOF

There exists x and y such that R = S ⊕ {x , y , x + y , 0},S1 = S ⊕ {x , 0}, S2 = S ⊕ {y , 0}, S3 = S ⊕ {x + y , 0}.



PROOFS

PROPOSITION


PROOF

There exists x and y such that R = S ⊕ {x , y , x + y , 0},S1 = S ⊕ {x , 0}, S2 = S ⊕ {y , 0}, S3 = S ⊕ {x + y , 0}.∀s ∈ S we have: sx ∈ S ⇔ sy ∈ S, xs ∈ S ⇔ ys ∈ S,



PROOFS

PROPOSITION


PROOF


sx ∈ S, sy /∈ S

⇓

sx = s1 ∃ s1 ∈ S, sy = s2 + y ∃ s2 ∈ S

⇓

s(x + y) = s1 + s2 + y /∈ S3



PROOFS

PROPOSITION


PROOF


SR := {s ∈ S | sx ∈ S}, SL := {s ∈ S | xs ∈ S}, T := SL ∩ SR .



PROOFS

PROPOSITION


PROOF


SR := {s ∈ S | sx ∈ S}, SL := {s ∈ S | xs ∈ S}, T := SL ∩ SR .SR and ST are subgroups of S with index at most 2, so T is asubgroup of S with |S : T | ∈ {1, 2, 4}.



PROOFS

PROPOSITION


PROOF


SR := {s ∈ S | sx ∈ S}, SL := {s ∈ S | xs ∈ S}, T := SL ∩ SR .SR and ST are subgroups of S with index at most 2, so T is asubgroup of S with |S : T | ∈ {1, 2, 4}.t ∈ T ⇒ xty , ytx , xtx , yty ∈ S



PROOFS

PROPOSITION


PROOF


SR := {s ∈ S | sx ∈ S}, SL := {s ∈ S | xs ∈ S}, T := SL ∩ SR .SR and ST are subgroups of S with index at most 2, so T is asubgroup of S with |S : T | ∈ {1, 2, 4}.t ∈ T ⇒ xty , ytx , xtx , yty ∈ S{

xt ∈ S ⇒ xty ∈ S2

ty ∈ S ⇒ xty ∈ S1⇒ xty ∈ S1 ∩ S2 = S

{xtx = s1 + bxxty = s2

⇒ xt(x +y) = s1+s2+bx ∈ S3 b = 0



PROOFS

PROPOSITION


PROOF


SR := {s ∈ S | sx ∈ S}, SL := {s ∈ S | xs ∈ S}, T := SL ∩ SR .SR and ST are subgroups of S with index at most 2, so T is asubgroup of S with |S : T | ∈ {1, 2, 4}.t ∈ T ⇒ xty , ytx , xtx , yty ∈ SRTR ⊆ S ⇒ RTR = 0⇒ T = 0⇒ |S| ∈ {1, 2, 4}.



PROOFS

PROPOSITION

Let (R, S1, S2, S3) be a good 4-tuple of rings. If R contains amultiplicative identity 1, then either |R| = 4 or 1 ∈ S1 ∩ S2 ∩ S3.



PROOFS

PROPOSITION

Let (R, S1, S2, S3) be a good 4-tuple of rings. If R contains amultiplicative identity 1, then either |R| = 4 or 1 ∈ S1 ∩ S2 ∩ S3.

PROOF

1 /∈ S

⇓

we may assume 1 /∈ S1, 1 /∈ S2

⇓

S1 and S2 are ideals of R

⇓

S = S1 ∩ S2 is an ideal of R

⇓

S = {0}ANDREA LUCCHINI RINGS AS THE UNIONS OF PROPER SUBRINGS


PROOFS

Thanks



PROOFS

Thanks

to all the participants



PROOFS

Thanks


to the organizers (in particular to Eloisa and Alberto)



PROOFS

Thanks



but especially



PROOFS

Thanks



but especially

thanks to Silvia, for the time she spent with us


Documents

Rings as the unions of proper subrings - MathUniPDlucchini/index_files/ring.pdf · 2010-03-21 · IS THERE AN ANALOGUE OF SCORZA’S THEOREM FOR RINGS? EXAMPLES PROOFS EXERCIZE No