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!

---

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

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 17

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

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

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.

Answer 30

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 31

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 32

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 33

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

Answer 34

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 35

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 36

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 37

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 38

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 39

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 40

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 41

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 42

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

Answer 43

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 44

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 45

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 46

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 47

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 48

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 49

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 50

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 51

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 52

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 53

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 54

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 55

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 56

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 57

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 58

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 59

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 60

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

Answer 61

OP: [...] counterexamples can illuminate a definition (e.g. a projective module that is not free), [...]

Indeed, let our ring $\ \mathcal R\ $ be the the ring of all continuous functions from the Euclidean sphere $\ \mathbb S\ :=\ \mathbb S^2\ $ (or more generally, $\ \mathbb S\ :=\ \mathbb S^{2\cdot n},\ $ where $\ n\in\mathbb N).\ $ Then module $\ \mathbb T\ $ of all continuous vector fields that are tangent to $\ \mathbb S\ $ is a direct summand of free module $\ \mathcal R^3\ $ hence $\ \mathbb T\ $ is projective but it is not free.

The last property of $\ \mathbb T\ $ that states that $\ \mathbb T\ $ is not free is implied by the Karol Borsuk's theorem about the unruly hair on sphere $\ \mathbb S\ $ that is impossible to brush smoothly.

Answer 62

This one is only about terminology, but while the topic is couterexamples in Algebra so it's tempting to give this one: Lie algebra is not an algebra.

Answer 63

Videque

1. Counterexamples in X
2. Counterexamples in Clifford Algebras