## Counterexamples in Algebra?

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!

(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)

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My feeling is that this question is far too broad. – Andy Putman Jun 21 2010 at 23:11
I like that the question is general. I think if it's narrowed too much we won't get as many interesting responses. All of the big list type questions that have been successful have been fairly general, so I don't think it hurts as long as we aren't swarmed with questions like this. – jeremy Jun 22 2010 at 0:23
Meta discussion: meta.mathoverflow.net/discussion/459/… – Andrew Stacey Jun 22 2010 at 7:54
Whilst I like lists of counterexamples, I don't think that MO is an appropriate place for one. I've explained why in the meta discussion (NB: please vote for the comment linking to the meta discussion so that it appears "above the fold"). I think that this would work so much better as a wiki page. So I've started one on the nLab: ncatlab.org/nlab/show/counterexamples+in+algebra Obviously, as I'm not an algebraist I didn't understand everything and have probably left out a lot of information in copying it over. I recommend closing this question and redirecting to that nLab page. – Andrew Stacey Jun 22 2010 at 8:33
Andrew, why not keep the question open, in order to generate the examples here that can then be more sensibly organized on your page? It seems likely to me that you will get a lot of good examples with this question that might otherwise be missed. – Joel David Hamkins Jun 22 2010 at 13:16

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 $S=A\times\lbrace0\rbrace$.

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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.

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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.

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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.

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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.

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@Pete: Your example is simple in a writing-down-sense. $\mathbb{Z}[\alpha]$ may well be the shortest expression for such a ring. If one is intressted in proof as well, then your example is much more complicated than $\mathbb{Q}+X\mathbb{R}[X]$ because one has to define and prove several things about algebraic integers first. – Johannes Hahn Aug 11 2010 at 9:47
Every integral domain is a subring of a UFD (for example, its field of fractions). So all you need here is an integral domain which is not a UFD. – Chris Eagle Jun 18 2011 at 8:50

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).

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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! :-)

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## Videque

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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!

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:-? (more stuff to fill the comment) – Andrea Ferretti Jun 22 2010 at 2:08
In what respect is this more interesting than giving the empty set as a counterexample to the statement that all sets are non-empty? – Boyarsky Jun 22 2010 at 3:03
Dear Andrea, I think that most algebraists and algebraic geometers consider that the zero ring IS a ring, corresponding to the (affine) empty scheme. For example the very first example Bourbaki gives for a ring is the zero ring! – Georges Elencwajg Jun 22 2010 at 9:44
Most experts also agree that the zero ring is not an integral domain, and that R is not a prime ideal in the commutative ring R. This means adding to the requirement "xy=0 implies either x=0 or y=0" the requirement "1 is not 0". The cool way to do this in one stroke is to say "the product of a finite collection of elements cannot be zero unless at least one of them is zero". (1 is the product of the empty collection.) To me, saying that the zero ring is a ring but not a domain is much like saying ("correctly") that the empty topological space is a space but not connected. At least on – Tom Goodwillie Jun 25 2010 at 0:20