Counterexamples in algebra?

Question

This is certainly related to "What are your favorite instructional counterexamples?", but I thought I would ask a more focused question. We've all seen Counterexamples in analysis and Counterexamples in topology, so I think it's time for: Counterexamples in algebra.

Now, algebra is quite broad, and I'm new at this, so if I need to narrow this then I will- just let me know. At the moment I'm looking for counterexamples in all areas of algebra: finite groups, representation theory, homological algebra, Galois theory, Lie groups and Lie algebras, etc. This might be too much, so a moderator can change that.

These counterexamples can illuminate a definition (e.g. a projective module that is not free), illustrate the importance of a condition in a theorem (e.g. non-locally compact group that does not admit a Haar measure), or provide a useful counterexample for a variety of possible conjectures (I don't have an algebraic example, but something analogous to the Cantor set in analysis). I look forward to your responses!

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You can also add your counter-examples to this nLab page: http://ncatlab.org/nlab/show/counterexamples+in+algebra

(the link to that page is currently "below the fold" in the comment list so I (Andrew Stacey) have added it to the main question)

Answer 1

In the category of rings, epimorphisms do not have to be surjective: $\mathbb{Z}\hookrightarrow \mathbb{Q}$.

Answer 2

I like Lance Small's example of a right but not left Notherian ring: matrices of the form $\begin{pmatrix}a & b\\ 0 & c\end{pmatrix}$ where $a\in\mathbb{Z}$ and $b,c\in\mathbb{Q}$.

Answer 3

A non-abelian group, all of whose subgroups are normal: the quaternion group, $$Q=\langle\thinspace a,b\thinspace|\thinspace a^4=1,a^2=b^2,ab=ba^3\thinspace\rangle$$

Answer 4

The ring $A = \prod_{n=1}^{\infty} \mathbb{F}_2$ has some interesting/disturbing properties.

For example, the affine scheme $X := {\rm{Spec}}(A)$ has non-open connected components (since it has infinitely many open points), all local rings on $X$ are noetherian (in fact they're all $\mathbb{F}_2$ since $a^2 = a$ for all elements $a$) even though $A$ is not noetherian, and if $I$ is an ideal that isn't finitely generated then ${\rm{Spec}}(A/I) \hookrightarrow X$ is formally unramified (since closed immersion), finite type, and flat but not étale (since not finitely presented) and not open, in contrast with the noetherian case.

Answer 5

1) (Nagata) There are noetherian domains of infinte Krull dimension: Localize $k[x_1,x_2,...]$ at the prime ideals $(x_1),(x_2,x_3),(x_4,x_5,x_6),...$.

2) (Malcev) Every commutative cancellative monoid embeds into a group. This is false in the non-commutative case. A very instructive counterexample is given by $\langle a,b,c,d,x,y,u,v : ax=by, cx=dy, au=bv \rangle$.

3) The Theorem of Cantor-Bernstein for sets does not carry over to algebraic structures. For example, the fields $K=\overline{\mathbb{Q}(x_1,x_2,...)}$ (or $K=\mathbb{C}$) and $K(t)$ embed into each other, but they are not isomorphic.

Answer 6

An exact sequence that does not split: $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$, where the first map is multiplication by 2.

Answer 7

A number ring which is a principal ideal domain (and, hence, a unique factorization domain) but is not Euclidean: the ring of integers of ${\bf Q}(\sqrt{-19})$. See Th Motzkin, The Euclidean algorithm, Bull Amer Math Soc 55 (1949) 1142-1146, available at http://projecteuclid.org/euclid.bams/1183514381

Answer 8

The group $\mathbb{Z}^4$ is not the fundamental group of any 3-manifold, proved by Stallings in this 1962 paper. It follows that there is no algorithm for recognizing 3-manifold groups.

Answer 9

• Does $R[x] \cong S[x]$ imply $R \cong S$? ( Taken from this link. )

• Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy-(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y-(1-z^2))}$$ Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$. In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3)

• http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf

Excellent Counterexamples.

Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.

The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) =\{e\}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse? That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)= \{e\}$?

The answer is yes there are groups with non-trivial center which are isomorphic to $\mathscr{S}(G)$. The answer is given at this link

Next one:

• Does there exists a finite group $G$ and a normal subgroup $H$ of $G$ such that $|Aut(H)|>|Aut(G)|$

Arturo Magidin posed this question some time ago at MATH.SE

• Question. Can we have a finite group $G$, normal subgroups $H$ and $K$ that are isomorphic as groups, $G/H$ isomorphic to $G/K$, but no $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H) = K$?

• Answer was provided by Vipul Naik. Link is given here.

Question was posed by Zev Chonoles at $\textbf{MATH.SE}$

• I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example?

• Answer from this link: Take $G = \mathbb{Z}_4 \times \mathbb{Z}_2$, $H$ generated by $(0,1)$, $K$ generated by $(2,0)$. Then $H \cong K \cong \mathbb{Z}_2$ but $G/H \cong \mathbb{}Z_4$ while $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

Answer 10

A number field where the ring of integers is Euclidean but not norm-Euclidean: ${\bf Q}(\sqrt{69})$. See David A Clark, A quadratic field which is Euclidean but not norm-Euclidean, Manuscripta Mathematica 83 (1994) 327-330.

Answer 11

In group theory, Lagrange's Theorem states that the order of a subgroup divides the order of the group, however the converse is false. The usual counterexample given is the alternating group $A_4$ of order 12 which has no subgroup of order 6.

Answer 12

From Milnor's book "Algebraic K-Theory":

A (nonzero!) associative ring for which a free module of rank 2 is isomorphic to a free module of rank 1: The ring of endomorphisms of an infinite-dimensional vector space.

Answer 13

Two finite non-isomorphic groups with the same order profile: let $C_n$ be the cyclic group of $n$ elements, let $Q=\langle\thinspace a,b\thinspace|\thinspace a^4=1,a^2=b^2,ab=ba^3\thinspace\rangle$ be the quaternion group, then $C_4\times C_4$ and $C_2\times Q$ are not isomorphic (the first is abelian, the second is not) but both have 1 element of order 1, 3 elements of order 2, and 12 elements of order 4.

By contrast, if two finite abelian groups have the same order profile, then they are isomorphic.

Answer 14

The ring $R = k[x,y]/(x^2, xy)$ is a simple example of a local commutative noetherian ring that is not Cohen-Macaulay. It is sometimes referred to as the "Emmy Ring."

This ring is very useful for showing how unintuitive non-CM rings can be. For instance, letting $I = (x)$, then $\operatorname{depth} R/I = 1 > 0 = \operatorname{depth} R$; in particular the (innocuous looking) inequality

$ \operatorname{depth} R/I + \operatorname{grade} I \leq \operatorname{depth} R $

need not hold. Here $\operatorname{grade} I$ is the length of the longest regular sequence in $I$.

Answer 15

Tarski's monsters: infinite groups in which every proper non-trivial subgroup is of prime order $p$. They are $2$-generated simple groups.

They were constructed by Olshanskii and as far as I remember they were also constructed independently by Rips, maybe even before Olshanskii, but he did not bother publishing it. Can anyone confirm this?

Answer 16

Grigorchuk 1984 example of a finitely generated group with intermediate growth (there are no such linear group).

Answer 17

A finite group in which a product of two commutators need not be a commutator: This is Exercise 3.27 in Rotman, The Theory of Groups, a construction attributed to Carmichael. Let $G$ be the subgroup of $S_{16}$ generated by the eight permutations $(ac)(bd)$, $(eg)(fh)$, $(ik)(jl)$, $(mo)(np)$, $(ac)(eg)(ik)$, $(ab)(cd)(mo)$, $(ef)(gh)(mn)(op)$, and $(ij)(kl)$. Then the commutator subgroup of $G$ is generated by the first four of these elements, and has order 16. It contains $\alpha=(ik)(jl)(mo)(np)$, but $\alpha$ is not a commutator.

Rotman remarks elsewhere that the smallest group in which there is a product of commutators which is not a commutator is a group of order 96.

Answer 18

Two famous cases that come to mind are:

1. Nagata's counterexample to Hilbert's fourteenth problem.

2. Counterexamples by various people to (the original version of) the Burnside problem.

Answer 19

There are finitely presented groups whose word problem is undecidable in computability theory.

Answer 20

An infinite group with exactly two conjugacy classes. See G. Higman, B. H. Neumann, and H. Neumann, Embedding theorems for groups, J. London Math. Soc. 24 (1949), 247-254.

Answer 21

Higman's group $G=\left< a_1,\ldots, a_4 | \forall i\in\mathbb{Z}/4\mathbb{Z}: a_i=[a_{i+1},a_i] \right>$, which has no subgroups of finite index. See: G. Higman, A finitely generated infinite simple group, J. London Math. Soc. 26 (1951), 61-64.

Answer 22

I think the Prüfer group $G=\mathbb{Z}_{p^\infty}$ deserves its place here.

For example,

• it has two nonisomorphic subgroups $H_1, H_2$ such that $G/H_1 \simeq G/H_2$,
• moreover, it has a proper subgroup $0 \neq H \subseteq G$ such that $G \simeq G/H$,
• it is an infinite group whose proper subgroups are all finite,
• moreover, it is a non-cyclic group whose all proper subgroups are cyclic,
• it is Artinian and not Noetherian (as a $\mathbb{Z}$-module), ...

Answer 23

An infinitely generated and non-Noetherian subring of a polynomial ring:

$$R=K[x,xy,xy^2,\ldots, xy^n,\ldots] \subset S=K[x,y].$$

Explanation The ring $R$ is graded and monomial: it is spanned by the monomials $x^ay^b$ that are contained in it, whose exponents are the lattice points in the cone $C=\{(a,b)=(0,0)$ or $a>0, b\geq 0\}.$ The minimal generators of the homogeneous ideal $R_{+}$ of positive degree elements correspond to the minimal generators $(1,n), n\geq 0$ of the lattice cone $C\cap\mathbb{Z}^2.$ Thus $R_{+}$ (respectively, $R$) is infinitely-generated with ideal (respectively, $K$-algebra) minimal generators $x,xy,xy^2,\ldots, xy^n,\ldots.$

Answer 24

Harry Hutchins "Examples of commutative rings" may be of interest. It is based on his 1978 Chicago Ph.D. thesis under Kaplansky, and not surprisingly it serves as a useful complement to Kaplansky's excellent textbook Commutative Rings (most references to proofs refer to Kaplansky). There is also a 3 page list of errata, updates,... dated July 1983, which is distributed with the book.

Hutchins, Harry C. 83a:13001 13-02
Examples of commutative rings. (English)
Polygonal Publ. House, Washington, N. J., 1981. vii+167 pp. $13.75. ISBN 0-936428-05-8

The book is divided into two parts: a brief sketch of commutative ring theory which includes pertinent definitions along with main results without proof (but with ample references), and Part II, the 180 examples. The examples do cover a very large range of topics. Although most of them appear elsewhere, they are enhanced by a fairly complete listing of their properties. Example 67, for instance, is M. Hochster's counterexample to the polynomial cancellation problem, and it lists a number of properties of the two rings that were not given in the original paper Proc. Amer. Math. Soc. 34 (1972), no. 1, 81 - 82; MR 45 #3394. Some of the examples appear more than once, since many rings exhibit more than one interesting property. (R=Kx, y, z is used in Examples 6 and 22.) The examples are grouped into areas, but a drawback is that these have not been labeled and separated off. In addition, the Index is for Part I and definitions only, and this means that searching for a specific example with certain properties can be time consuming. The book can be used as a supplement to one of the standard texts in commutative ring theory, and it does appear to complement the monograph by I. Kaplansky Commutative rings, Allyn and Bacon, Boston, Mass., 1970; MR 40 #7234; second edition, Univ. Chicago Press, Chicago, Ill., 1974; MR 49 #10674. --Reviewed by Jon L. Johnson

Answer 25

If $f$ and $g$ are relatively prime in ${\mathbf Q}[X]$ then the mapping ${\mathbf Q}[X]/(fg) \rightarrow {\mathbf Q}[X]/(f) \times {\mathbf Q}[X]/(g)$ given by $h \bmod fg \mapsto (h \bmod f, h \bmod g)$ is a ring isomorphism. This is a special case of the Chinese remainder theorem.

If we replace ${\mathbf Q}[X]$ with its subring ${\mathbf Z}[X]$, which is a UFD, then for relatively prime $f$ and $g$ in ${\mathbf Z}[X]$ the mapping ${\mathbf Z}[X]/(fg) \rightarrow {\mathbf Z}[X]/(f) \times {\mathbf Z}[X]/(g)$ given by $h \bmod fg \mapsto (h \bmod f, h \bmod g)$ is a ring homomorphism, but it is not necessarily an isomorphism since it need not be surjective (though it is injective). For example, if $f = X-1$ and $g = 1+X+\cdots + X^{n-1}$ where $n > 1$ then the natural mapping $${\mathbf Z}[X](X^n-1) \rightarrow {\mathbf Z}[X]/(X-1) \times {\mathbf Z}[X]/(1+X+\cdots + X^{n-1})$$ does not have $(0,1)$ in its image. The reason is that if $f(X) \in {\mathbf Z}[X]$ is mapped to $(0,1)$ then $f(X) = (X-1)g(X)$ for some $g(X) \in {\mathbf Z}[X]$ and then $1 = (\zeta_p-1)g(\zeta_p)$ for any prime $p$ dividing $n$, which says $\zeta_p-1$ is a unit in ${\mathbf Z}[\zeta_p]$, and that's false.

Answer 26

Thompson's group T is a finitely presented infinite simple group.

Answer 27

A polynomial, solvable in radicals, whose splitting field is not a radical extension (of $\bf Q$). Let $f(x)$ be any cyclic cubic, that is, any cubic with rational coefficients, irreducible over the rationals, with Galois group cyclic of order 3. Then $f(x)=0$ is solvable in radicals (every cubic is), so the splitting field $K$ of $f$ over $\bf Q$ is contained in a radical extension of $\bf Q$, but $K$ is not itself a radical extension of $\bf Q$. The degree of $K$ over $\bf Q$ is 3, so for $K$ to be radical over $\bf Q$ it would have to be an extension of $\bf Q$ by the cube root of some element of $\bf Q$, but such extensions are not normal.

Answer 28

Two non-zero commutative rings with unity, one a subring of the other, but with different unities. Let $R={\bf Z}/10{\bf Z}$, $S=2R$, then $R$ and $S$ are commutative rings with unity, $S$ is a subring of $R$, but the identity element of $S$ isn't the identity element of $R$. If we view $R$ as $\lbrace0,1,\dots,9\rbrace$ with operations modulo 10, so $S=\lbrace0,2,4,6,8\rbrace$, then the multiplicative identity in $S$ is 6.

This works more generally if $\gcd(m,n)=1$, $R={\bf Z}/mn{\bf Z}$, and $S=mR$. It works even more generally if $A$ and $B$ are non-zero commutative rings with unity, $R=A\times B$, and $S=A\times\lbrace0\rbrace$.

Answer 29

Sweedler's Hopf algebra. It is the Hopf algebra generated by two elements $x, g$ with relations $g^2 = 1$, $x^2 = 0$, and $gxg = - x$. The coproduct is given by $$ \Delta(g) = g \otimes g, \quad \Delta(x) = x \otimes 1 + g \otimes x,$$ the counit by $$ \varepsilon(g) = 1, \quad \varepsilon(x) = 0,$$ and the antipode by $$ S(g) = g, \quad S(x) = - gx.$$ It is noncommutative and noncocommutative, is quasitriangular and coquasitriangular, but is not a quantum double.

Answer 30

This quasigroup is not isomorphic to any loop (i.e. quasigroup with identity):

* | a   b   c
-------------
a | a   c   b
b | c   b   a
c | b   a   c

See e.g. Latin squares: Equivalents and equivalence.