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

While a finite abelian group is determined by its character table, this is not true for (finite) nonabelian groups. E.g., the dihedral and quaternion groups of order 8 (or more generally two nonabelian groups of order p³ for a prime p) are nonisomorphic but have the same character table.

Answer 2

An infinitely-generated Noetherian ring: $\mathbb{Q},$ the field of rational numbers.

Answer 3

A subring of a UFD need not be a UFD.

An example by M. Zafrullah: Let R be the set of real numbers and Q be the set of rational numbers. Then the polynomial ring R[X] is a UFD (since it is a PID), but its subring Q + XR[X] is not a UFD.

Answer 4

My favorite counter example in Galois theory is the field extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$.

Here are some cases to which it provides a counter example:

1. It is a non-Galois extension (so providing counter example to "every extension is Galois").

2. The Galois group of its Galois closure is non-abelian.

3. Although the intersection of it with $\mathbb{Q}(\zeta_3 \sqrt[3]{2})$ is $\mathbb{Q}$, these fields are not linearly disjoint.

[Here $\zeta_3=e^{\frac{2\pi i}{3}}$.]

Answer 5

A principal ideal with two non-associate generators (i.e. generators that are not unit multiples of each other).

In the ring $k[x,y,z]/(x-xyz)$, we have $([x])=([xy])$, but there is no unit $u$ such that $[xy]=u[x]$. See http://blog.jpolak.org/?p=534

Answer 6

Please forgive me if someone has already posted this...

Let $X > Y > Z$ be a tower of groups with $Y$ and $Z$ being normal subgroups of $X$ and $Y$, respectively. $Z$ need not be a normal subgroup of $X$.

An example: $D_4 >$ Klein's $4$-group $> Z/2Z$.

Answer 7

An example showing that, in the standard definition of a ring (without $1$), it won't do to replace the left and right distributive laws with the single law$$(u+v)\cdot(x+y)=u\cdot x+u\cdot y+v\cdot x+v\cdot y$$as was done on p. 18 of my old copy (July 1957 printing) of Kelley's General Topology.

(Spoiler alert: if you click on the gray box, an example of what can go wrong is revealed.)

Take an additive group of order $3$, choose a nonzero element $c$, and define $x\cdot y=c$.

Answer 8

I like Amnon Yekutieli's example of a module whose completion is not complete.

Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an $A$-module. Algebraically you can define the completion $\hat{M}$ of $M$ as the inverse limit of the modules $M / I^k M$ (with the canonical quotient maps $M / I^{k+1} M \to M / I^k M$). There is a canonical module morphism $M \to \hat{M}$ and you can call $M$ ($I$-adically) complete if this is an isomorphism.

I used to think that the completion of an arbitrary module is complete! Rest assured that this is true if $A$ is Noetherian. But it does fail for the simplest example of a non-Noetherian ring: take $A = k[x_1, x_2, \ldots]$ the ring of polynomials in countably many variables, and $M = A$. For the ideal $I = \langle x_1, x_2, \ldots \rangle$ generated by the all variables, the completion $\hat{M}$ is, as one would expect from the finite dimensional case, the ring of power series in countably many variables (these power series should have only finitely many monomials of any given degree, so something like $\sum_i x_i$ does not count). However this module is not $I$-adically complete: indeed, look at the sequence of polynomials $\sum_{i=1}^n x_i^i$. If it did converge to a power series, by comparing coefficients, it is clear that the limit would have to be $\sum_{i=1}^\infty x_i^i$. (Since all power series in $I^k \hat{M}$ have only monomials of degree at least $k$, elements of the completion of $\hat{M}$ have a well-defined coefficient for any monomial.) But it does not in fact converge to that since it is easy to check that the tails, $\sum_{i=j}^\infty x_i^i$ do not lie in any $I^k \hat{M}$, i.e, you can't have an equality of the form $\sum_{i=j}^\infty x_i^i = m_1 g_1 + \cdots + m_l g_l$, where the $m_i$ are finitely many monomials: every term on the RHS mentions one of the finitely many variables present in the $m_i$, but there is no such "finite cover by variables" for the LHS.

I learned this example from Amnon Yekutieli's paper On Flatness and Completion for Infinitely Generated Modules over Noetherian Rings.

Answer 9

Counter-example to the idea that algebraic duals cannot become simpler.

Consider the free $\mathbb{Z}$-module on a countable set $V=\mathbb{Z}^{(\omega)}$. The dual is $V^{\ast}\cong \mathbb{Z}^{\omega}$, a countable direct product of $\mathbb{Z}$'s, which is not free and also much bigger than $V$. Strangely, the double dual is $V^{\ast\ast}\cong V$!

Answer 10

I am also making this list for my record. The examples here are marvelous. I will put something that were not in this list. These examples are from Field Theory.

An Algebraic Extension of Infinite Degree.

$\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5, \cdots)$ over $\mathbb{Q}$.

A Nontrivial Finite Extension that is Isomorphic to the Ground Field.

Let $F=\mathbb{Q}(x)$ and $k=\mathbb{Q}(\sqrt x)$. Then $k$ is a degree-2 extension of $F$. However, they are isomorphic.

A Finite Extension which Contains Infinitely Many Subextensions.

Let $p$ be a prime. Let $F=\mathbb{GF}(p)(x,y)$ and $k=\mathbb{GF} (p) (x^{\frac 1 p},y^{\frac 1 p})$. For any $f(y)\in \mathbb{GF}(p)(y)$, $$ K=F(x^{\frac 1 p} f(y) + y^{\frac 1 p})$$ is a nontrivial subextension of $k$.

Answer 11

This is probably more of an example than a counterexample. Consider the following binary operation table defined on a three element set with zero:

    0 1 2
0   0 0 0
1   0 0 1
2   0 2 2

V. Murskii showed that the equational theory of this algebra has no logically equivalent (in equational logic) finite theory. Lyndon earlier showed that every two element algebra with one binary operation did have a finite basis, and Perkins found a six element semigroup with no finite basis. I don't know the status of algebras with a single ternary operation.

Gerhard "Ask Me About System Design" Paseman, 2010.06.21

Answer 12

The quaternion group of order $8$ has a real irreducible character of degree $2,$ but the associated representation can not be realized over the real field.

Answer 13

I have been maintaining this small blog for a few months https://counterexamplesinalgebra.wordpress.com/

Answer 14

A finitely generated module with a non-finitely generated submodule: Consider the polynomial ring $k[x_1, x_2, ...]$ as a module over itself. The submodule generated by $\{ x_1, x_2, ...\}$ is not finitely generated.

Answer 15

Examples of modules not having a basis.

It is well known that a vector space always has a basis. A module may not have a basis. Here are some examples:

• The module $\mathbb{Z}/n\mathbb{Z}$ of the integers modulo $n$. This module has torsion.
• The module $\mathbb{Q}$ of the rational numbers over the integers. This module is torsion-free.
• The module $F[X]$ over the ring $F^\prime[X]$ of polynomials that have coefficient of $X$ equal to $0$. This module is finitely generated and torsion-free.

For more details, you can have a look here.

Answer 16

The (several) results of A.H. Schofield answering Artin's question (in the negative) by constructing skew-field extensions $K \subset L$ such that the right and left degrees are different deserve a mention.

Answer 17

A finitely generated soluble group isomorphic to a proper quotient group

Let $\mathbb{Q}_2$ be the ring of rational numbers of the form $m2^n$ with $m, n \in \mathbb{Z}$ and $N = U(3, \mathbb{Q}_2)$ the group of unitriangular matrixes of dimension $3$ over $\mathbb{Q}_2$. Let $t$ be the diagonal matrix with diagonal entries: $1, 2, 1$, put $H = \langle t, N \rangle$ and $w=\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$. Then the group $G=H/\langle w^2 \rangle$ is finitely generated soluble and isomorphic to a proper quotient subgroup.

For more details you can see here.

Answer 18

Matrices in $\text{Mat}_2(\mathbb{Z})$ not conjugate to their transpose by $\text{GL}_2(\mathbb{Z})$.

A matrix and its transpose are similar over any field (cf. here), thus a matrix $M\in\text{Mat}_2(\mathbb{Z})$ is conjugate to $M^\top$ by a matrix in $\text{GL}_2(\mathbb{Q})$.

But there are matrices $M\in\text{Mat}_2(\mathbb{Z})$ that are not conjugate to $M^\top$ by any matrix in $\text{GL}_2(\mathbb{Z})$. E.g. $$ M=\begin{pmatrix}2&5\\-9&-22\end{pmatrix} $$ See here for the details. Note that even $M\in\text{SL}_2(\mathbb{Z})$.

This shows that linear algebra over rings, even nicest ones like $\mathbb{Z}$ a Euclidean domain, becomes a lot more challenging than over fields.

Answer 19

Radical of a primary ideal is prime but not every ideal whose radical is prime is primary. Here is a cute counterexample: Let $I=(x^2,xy)\in F[x,y]$ where $F$ is a field. The radical $\sqrt{I}$ of $I$ is $(x)$ which is prime but $I$ is not primary; $xy\in I$, $x\not\in I$ but no power of $y$ belongs to $I$.

This is from page 154 of Commutative Algebra Vol. 1 by Zariski and Samuel. Now that I check, this is the 1975 printing which I bought on 1979. How time flies when you are having fun! :-)

Answer 20

This is obscure, but Mogiljanskaja gave an example of two (and even an infinite sequence of) non-isomorphic semigroups such that their power semigroups are isomorphic. This resolved a question of Schein and Tamura. I don't remember the example now, but I should have the papers in my family house. If this is of any interest to anybody I could try to write it up since the papers are not readily available -- my copies came from Russia by snail mail and it cost a bit. There are two papers. The first shows an example of a sequence of semigroups and is quite simple. The second one adds one semigroup to the sequence and that's a bit more intricate, but still elementary. The semigroups can also easily be made commutative, which Mogiljanskaja remarks on.

Answer 21

Desmond MacHale wrote an article, "Minimum Counterexamples in Group Theory", Mathematics Magazine, 54 (1981), no. 1, 23–28; jstor. I've found this paper useful in an introductory algebra class and I like the philosophy of the paper, "Is X true? No, probably not. So what is a smallest counterexample?" A variation on the group theory (and Irish!) tune of MacHales appears here. A followup article is "Constructing a minimal counterexample in group theory" by Arnold Feldman, also in Mathematics Magazine (1985).

Answer 22

Regarding Schur's lemma:

For a finite group $G$ and $V$ a finite-dimensional irreducible representation of $G$ over a field $K$, there exist endomorphisms of this representation that are not scalar multiples of the identity. For example, take $G=\mathbb{Z}_4$, $K=\mathbb{R}$, and $\rho:\mathbb{Z_4}\rightarrow GL(\mathbb{R}^2)$ given by

$$\rho(1)=\begin{pmatrix} 0 & -1 \\ 1 & \ \ 0 \end{pmatrix}$$

Then since $\rho(1)$ has no real eigenvalues the representation is irreducible. But on the other hand, $\mathbb{Z}_4$ is abelian and $\rho(1): \mathbb{R}^2\rightarrow\mathbb{R}^2$ is an endomorphism of this representation.

This is why it is important $K$ be algebraically closed.

Answer 23

If $x$ and $y$ are elements of an associative ring such that $xy\ne1=yx$ then there is a mutually inverse pair of invertible matrices one of which is lower triangular but not upper triangular and the other is upper triangular but not lower triangular. $$\begin{pmatrix} y & 0 \\ 1-xy & x \\ \end{pmatrix}^{-1} = \begin{pmatrix} x & 1-xy \\ 0 & y \\ \end{pmatrix} $$ To construct such a ring, consider the monoid $M$ of functions $\mathbb N\to \mathbb N$. Let $x$ be the function $n\mapsto n+1$ and let $y$ be the function $n\mapsto n-1$ if $n>0$ and $0\mapsto 0$. View $x$ and $y$ as members of the monoid algebra $\mathbb ZM$. This example also relates to proofs of general forms of Schanuel's lemma.

Answer 24

If $K/\Bbb Q$ is a number field and $\mathcal{O}_K$ its ring of integers, $\mathcal{O}_K$ need not have a power basis as a $\Bbb Z$-module; i.e., $\mathcal{O}_K\neq\Bbb Z[\alpha]$ for any $\alpha\in\mathcal{O}_K$! An example is given by $K = \Bbb Q(\alpha)$, where $\alpha$ is a root of $T^3 - T^2 - 2T - 8$ (a $\Bbb Z$-basis is instead given by $\{1,\alpha,(\alpha^2 + \alpha)/2\}$).

See Keith Conrad's notes for more detail on this example.

Answer 25

The Krull topology on an absolute Galois group is not the profinite topology.

For instance, let $G_\mathbb{Q}=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ for some fixed algebraic closure $\overline{\mathbb{Q}}/\mathbb{Q}$. It is easy to see that $G_\mathbb{Q}$ has uncountably many (normal) subgroups of index $2$, because they are $1-1$ with one-dimensional quotients of a $\mathbb{F}_2$-vector space of infinite dimension. Being of finite index, they are all open in the profinite topology, hence closed since a profinite group is compact when endowed with the profinite topology. Since there are only countably many quadratic extensions of $\mathbb{Q}$, one of the above subgroups cannot be closed in the Krull topology, by the fundamental theorem of infinite Galois theory.

Answer 26

There exist non-isomorphic finite groups $G_1$ and $G_2$ such that $\mathbb{Q}[G_1]$ is isomorphic to $\mathbb{Q}[G_2]$. Just take the two distinct non-abelian groups of order $p^3$ where $p$ is an odd prime.

In fact, an example due to Everett Dade cleverly builds on this example to construct non-isomorphic finite groups $H_1$ and $H_2$ such that $k[H_1]$ is isomorphic to $k[H_2]$ for every field $k$.

Answer 27

A very basic one: Over the field of two elements, the symmetric matrix $\left(\begin{matrix}1&1\\1&1\end{matrix}\right)$ is nilpotent and thus not diagonalizable.

Answer 28

Here's one from universal algebra: the class of mono-unary algebras $\mathfrak A=(A,f)$ such that $f$ is a permutation with a unique fixed point does not have the unique factorization property for direct decomposition.

Example. For $n\equiv1\pmod3$ let $\mathfrak A_n=(A,f)$ where $|A|=n$ and $f$ is a permutation of order $3$ with a unique fixed point; then $\mathfrak A_{100}\cong\mathfrak A_{10}\times\mathfrak A_{10}\cong\mathfrak A_4\times\mathfrak A_{25}$, and $\mathfrak A_4,\mathfrak A_{10},\mathfrak A_{25}$ are indecomposable.

Another example, smaller if less pretty. Let $$\mathfrak A=\langle[5],\ (1)(2\ 3)(4\ 5)\rangle,$$ $$\mathfrak B=\langle[9],\ (1)(2\ 3\ 4\ 5)(6\ 7\ 8\ 9)\rangle,$$ $$\mathfrak C=\langle[5],\ (1)(2\ 3\ 4\ 5)\rangle,$$ $$\mathfrak D=\langle[9],\ (1)(2\ 3)(4\ 5)(6\ 7\ 8\ 9)\rangle;$$ then $\mathfrak A$, $\mathfrak B$, $\mathfrak C$, $\mathfrak D$ are indecomposable, and $\mathfrak A\times\mathfrak B\cong\mathfrak C\times\mathfrak D$.

Answer 29

Assertion: If $S$ is an associative ring with identity and $M, N$ are unital left $S$-modules, then $\hom_S(M,N)$ is a unital left $S$-module.

Th is is false in general. Consider the matrix ring $S:=M_2({\mathbb Z})$, and let $M={\mathbb Z}^2=N$ be equipped with natural $S$-module structure. We have $\hom_S(M,N)\hookrightarrow S$, as additive groups; moreover, if $\phi\in \hom_S(M,N)$, then $\phi(A{\bf v})=A(\phi({\bf v}))$ for every $A\in M_2({\mathbb Z})$ and ${\bf v}\in M$. Thus, the scalar matrices, and only those, are the elements in $\hom_S(M,N)$, since the only matrices that commute with all four of the elementary matrices are precisely the scalar matrices. By the isomorphism ${\mathbb Z}\cong {\mathbb Z} I_2$, we have $X:=\hom_S(M,N)\cong {\mathbb Z}$, and this makes $\hom_{\mathbb Z}(X,X)\cong {\mathbb Z}$.

Now, to make $X=\hom_S(M,N)$ into a unital left $S$-module, we need to give a unital ring homomorphism $S\rightarrow \hom_{\mathbb Z}(X,X)\cong {\mathbb Z}$. But there is no such. To see this, we notice that the elementary matrices in $S$ are nilpotent elements, and therefore, must be mapped to the unique nilpotent element $0\in {\mathbb Z}$. Since a ring homomorphism is additive, the only homomorphism $S\rightarrow {\mathbb Z}$ must be trivial.

Answer 30

You might find several answers in Harry Hutchins's book on Examples of Commutative Rings.