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

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 2

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 3

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 4

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

$$\rho(1)=\left(\begin{align} 0 & -1 \\ 1 & \ \ 0 \end{align}\right)$$

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 5

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 6

Desmond MacHale wrote an article, "Minimal Counterexamples in Group Theory", Mathematics Magazine, 54 (1981), no. 1, 23–28. 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 7

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 8

Videque

1. Counterexamples in X
2. Counterexamples in Clifford Algebras

Answer 9

Definition: Let A = (S,,',...........) be an ordered pair where S is a nonempty set and (exponent): S X S ---->S is a binary operation on S. Then we call (S,,*',...........) an algebraic structure.

Clearly,this gives a general definition which allows us to define a category AlgStruc i.e. the category of all algebraic structures. Now let's state a simple proposition:

Let 0 in A be a zero element (i.e. * on S for A is ordinary addition and for every x in S, x+0=0+x=x) and let 1 in A be the 1 element ( i.e. ' on S where for every x in S, x'1=1*'x=x). Then the following "subcategory" is empty in AlgStruc: { A ! 0 =1 for every x in S} .

In other words,there is no algebraic structure with at least one binary operation defined on it where the zero element and the 1 element are the same. This is FALSE;consider the trivial semigroup ({0},*) Clearly 0=1 in this case!