Counterexamples in Algebra

Counterexamples in Algebra August 3, 2015 We use k, F , K to denote the fields, and R to denote the rings. Denote by Z the ring of rational integers, Q the field of rational numbers, R the field of real numbers, and C the field of complex numbers. Denote by A the ring of algebraic integers. 1 Groups A Noncyclic Group of Order 4. Z/2Z × Z/2Z. A Presentation Gives a Trivial Group. hx, y, z|xyx−1 y −1 = y, yzy −1 z −1 = z, zxz −1 x−1 = xi. Two Nonisomorphic Groups with the Same Character Table. D4 and Q8 . A Nonabelian p-Group.    a b 2 |a, b ∈ Z/(p ), a ≡ 1 mod p . Gp = 0 1 This is a nonabelian group of order p3 . Another example of a nonzbelian group of   1 Hp = 0  0 order p3 is   a b  1 c  |a, b, c ∈ Z/(p) .  0 1 In fact, these are the only nonabelian groups of order p3 . On the other hand, every group of order p2 is abelian. Solvable Groups. Every finite group of order < 60, every Abelian group, any p-group. Finite Simple Groups. Cyclic groups Z/pZ, alternating groups An with n ≥ 5, groups of Lie type, sporadic groups. Group Homomorphisms of Additive Group of R. There are linear functions f (x) = ax. There are also nonlinear ones, consider a projection onto one basis element of the vector space R over Q. A Paradoxical Decomposition of a Group. Let F2 be the free group with two generators a, b. Consider S(a), S(a−1 ), S(b), and S(b−1 ) be the set of elements starting with a, a−1 , b, and b−1 respectively. Then we have F2 = hei ∪ S(a) ∪ S(a−1 ) ∪ S(b) ∪ S(b−1 ). We have also F2 = aS(a−1 ) ∪ S(a), and F2 = bS(b−1 ) ∪ S(b). These decompositions are used in the proof of Banach-Tarski Theorem. 2 Rings Commutative Ring with Identity that is Not an Integral Domain. Z × Z, Z/6Z. Commutative Ring without Identity. 2Z, {0, 2} in Z/4Z. Noncommutative Ring without Identity. M2 (2Z). Noncommutative Division Ring with Identity. The real quarternion H. Ring with Cyclic Multiplicative Group. R = Z/nZ with n = 2, 4, pk , 2pk . Any finite fields. Also Z has units {±1} which is isomorphic to Z/2Z and is cyclic. A Subring that is Not an Ideal. Z ⊂ Q. An Order of a Ring is Larger than its Characteristic. Any GF(pn ) for n ≥ 2. A Prime Ideal that is Not a Maximal Ideal. Let R = Z[x]. The ideal P = (x) is a prime ideal since R/P ∼ = Z is an integral domain. Since Z is not a field, P is not a maximal ideal. In PID, every prime ideal is maximal and vice versa. In fact, if R is an integral domain that is not a field, for example Z, then (0) is a prime ideal that is not maximal. A Homomorphic Image Need Not be an Ideal. Z ⊂ Q. An Additive Group Homomorphism that is Not a Ring Homomorphism. The derivative map D : R[x] → R[x]. We have D(f + g) = D(f ) + D(g) but D(f g) = gD(f ) + f D(g). A Multiplicative Group Homomorphism that is Not a Ring Homomorphism. Let f : R → R be f (x) = x2 . The Unique Ring Homomorphism from R to R. The identity. √ A Commutative Ring with Infinitely Many Units. Z[ 2]. A Noncommutative Ring with Infinitely Many Units. M2 (Z). A Non-Dedekind √ Domain. √ √ The ring Z[ −3] is a subring of A ∩ Q( −3) = Z[(1 + −3)/2]. This is not Dedekind since it is not integrally closed. √ √ A Dedekind Domain which is Not a UFD. Z[ −5]. This is a ring of integers in Q( −5). We have √ √ the non-unique factorization 6 = 2 · 3 = (1 + −5)(1 − −5). A UFD which is Not Dedekind. k[x, y]. The Krull-dimension of this ring is 2. A UFD which is Not a PID. Z[x]. Since Z is UFD, Z[x] is a UFD. However, this is not PID because (x, 2) is not principal. A PID which is Not a ED. √ √ The ring of integers in Q( −19). This is Z[(1 + −19)/2]. ∼ × R. A Ring R such Q∞ that R = R Let R = i=1 Z. Then R ∼ = R × R by the following isomorphism: A A A A A f :R→R×R defined by f (x1 , x2 , · · · ) = ((x1 , x3 , · · · ), (x2 , x4 , · · · )) . A Commutative Ring with 4 Elements that is Not Isomorphic to Z/4Z or Z/2Z × Z/2Z. x 0 The matrices over Z/2Z = GF(2). This is isomorphic to GF(2)[x]/(x2 ) by y x   1 0 7→ 1 + (x2 ), 0 1   0 0 7→ x + (x2 ). 1 0 This is not isomorphic to Z/4Z since the characteristic is not 4. This is not isomorphic to Z/2Z × Z/2Z since this ring has two solutions in x2 = 0. Another example is the 4-element subring of Z/16Z, where the multiplication of any pair is zero. A Commutative Ring with Identity that the Converse of CRT Holds. Let R be a commutative ring with identity. The converse of CRT is: If I, J are ideals with I + J 6= R, then R/I ∩ J ∼ 6 R/I × R/J. = Z, F [x] where F is a field. Further, any Dedekind Domain. A Commutative Ring with Identity that the Converse of CRT does Not Hold. Q R= ∞ Z, and I = J = (0). Then I + J 6= R and R/I ∩ J ∼ = R/I × R/J. i=1 A Commutative Ring with Identity that is Noetherian but not Artinian. Z, k[x]. A Commutative Ring with Identity that is neither Noetherian nor Artinian. A the ring of algebraic integers, k[x1 , x2 , · · · ] the ring of polynomials in infinitely many variables. A Local Noetherian Ring. k[[x]] the formal power series ring over a field k. This has a unique maximal ideal (x), and it is Noetherian by Hilbert’s Basis Theorem. Furthermore, this is a DVR. Integral Domains A, B which Contains a Field F but A ⊗F B is Not an Integral Domain. Let A = B = GF(p)(X) and F = GF(p)(X p ). Then A and B are integral domains containing F , but X ⊗ 1 − 1 ⊗ X ∈ A ⊗F B is a nonzero element in A ⊗F B satisfying (X ⊗ 1 − 1 ⊗ X)p = X p ⊗ 1 − 1 ⊗ X p = 0. Hence, A ⊗F B is not an integral domain. A Group Ring which is Not Semisimple. k[x]/(xp − 1) with k = GF(p). This is a group ring kG with a cyclic group G or order p. This is not  semisimple by Maschke’s theorem. This is a local ring with maximal ideal I := ker(kG → − k) = Rad(kG). 3 Fields An Algebraically Closed Field of Finite Characteristic. GF(p). An Infinite Field of Finite Characteristic. GF(p), GF(p)(x) the field of rational functions over GF(p). A Real Transcendental Extension. Q ⊂ Q(π). 1 3 A Real Field which is Not √ Totally Real. Q(2 ). A Totally Real Field. Q( 2). p√ √ A Normal Extension of a Normal Extension may Not√be Normal. Q ⊂ Q( 2) ⊂ Q( 2). √ √ An Algebraic Extension of Infinite Degree. Q( 2, 3, 5, · · · ) over Q, Q over Q, GF(p) over GF(p). A Nontrivial Finite Extension that is Isomorphic to the Ground Field. √ Let F = Q(x) and k = Q( x). Then k is a degree-2 extension of F . However, they are isomorphic. A Finite Extension which Contains Infinitely Many Subextensions. 1 1 Let p be a prime. Let F = GF(p)(x, y) and k = GF(p)(x p , y p ). For any f (y) ∈ GF(p)(y), 1 1 K = F (x p f (y) + y p ) is a nontrivial subextension of k. An Irreducible Polynomial f ∈ Q[x] with Reducible f̄ ∈ Z/pZ[x] for Every p. 2 Let x4 + 1 ∈ Q[x]. If p = 2, then x4 + 1 = (x2 + 1)2 . If p 6= 2, then x4 + 1|x8 − 1|xp −1 − 1. 4 Modules A Noetherian Module which is Not Artinian. Z-module Z. −i An Artinian Module which is Not Noetherian. Z-module M = ∪∞ i=1 (p Z/Z). A Free Module with Infinite Basis. Q-vector space R. An Injective Module which is Not Torsion-Free. Z-module Q/Z A Torsion-Free Module which is Not Flat. Let R = k[x, y] and I = (x, y). Then I is a torsion-free R-module. This is not flat because I ⊗I →I ⊗R is not injective. In fact, 0 6= x ⊗ y − y ⊗ x ∈ Ker (I ⊗ I → I ⊗ R). A Projective Module which is Not Free. Let R = Z/2Z × Z/2Z, and consider Z/2Z × (0) a submodule of R-module R. This is projective since it is a direct summand of free module but it is too small to be free. A Flat Module which is Not Projective. Z-module Q. A Flat Module which is Neither Projective Nor Injective. The Z-module Q ⊕ Z. This is flat because it is a direct sum of flat modules. This is not projective because of Q, not injective because of Z. A Semisimple Module which is Not Simple. CS3 ∼ = C × C × M2 (C). A Module which is Faithful and Flat, but Not Faithfully Flat. Z-module Q.