Fun with principal ideal domains

A commutative ring $R$ is called a principal ideal domain (PID) if every ideal of $R$ can be generated by a single element. If $R$ is a principal ideal domain, is every subring of $R$ a principal ideal domain? No, definitely not. That is because you can take any integral domain that is not a principal ideal domain, like $\Z[x]$, and take its fraction field. Its fraction field is a PID and the original ring sits inside it as a subring.

Another more interesting example is the ring $\Q[x]$ of polynomials with rational coefficients. It is a PID, yet the subring $\Q[x^2,x^3,x^4,\dots]$ is not. The ideal $(x^2,x^3)$ in this ring is not a principal ideal. By the way, is the ring $\Q[x^2,x^3,\dots]$ Noetherian? Does there exist an ideal in it that needs at least three generators?
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A quick intro to Galois descent for schemes

This is a very quick introduction to Galois descent for schemes defined over fields. It is a very special case of faithfully flat descent and other topos-descent theorems, which I won't go into at all. Typically, if you look up descent in an algebraic geometry text you will quickly run into all sorts of diagrams and descent data. In my opinion, that is a very counterintuitive way to present the basic idea.

What is the descent theorem?

Here is the main topic of this post:

Theorem. Let \(E/F \) be a finite Galois extension of fields with Galois group \(G\). Then the functor
\{\text{quasiproj. \(F\)-schemes}\}&\to \{\text{quasiproj. \(E \) schemes with compatible \(G \) action} \} \\
X&\mapsto X\otimes_F E\end{align*}\] where \(X\otimes_F E \) is given an Galois action via the canonical action on \(E\), is an equivalence of categories.

This is the basic theorem of Galois descent. What does it mean, and how does it work? First, I have to tell you what a compatible Galois action is. Well, if \(X \) is an \(E\)-scheme, then there is a map \(X\to{\rm Spec}(E)\), and there is the usual action of \(G \) on \({\rm Spec}(E)\). Compatible just means that for each \(\sigma\in \), the square

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Dividing a square into triangles of equal area

Take a square and divide it down a diagonal, dividing the square into two triangles. Drawing the opposite diagonal now divides it into four triangles. In these two examples, we divided a square into an even number of triangles, all with equal area. Can we divide a square into an odd number of nonoverlapping triangles, all with equal area? In this question, we do not require that all the triangles be congruent, as in the above examples.

It turns out, you can't. Paul Monksy proved this in a 1970 American Mathematical Monthly paper [1], though John Thomas proved this earlier when the vertices of the triangles are restricted to having rational coordinates.

The proof progresses in several steps. I won't go through every detail, but try and convey the flavour of the proof. The reader is invited to read the proof in its entirety, which is something I just did and I recommend it.
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Finite-dimensional k[x]-modules: projective or not?

Let's suppose $M$ is a nonzero projective $\Z$-module. Can it be finite? Nope. I'm sure there are plenty ways to prove it, but one way is to observe that a projective $\Z$-module is free, and hence if $M$ is nonzero it must have at least one copy of $\Z$. So, $M$ is infinite.

What's the analogue for the ring $k[x]$ where $k$ is a field? If $M$ is a nonzero projective $k[x]$-module, can it be finite? It certainly can't if $k$ is infinite dimensional, since any nonzero $k[x]$-module (whether projective or not) is also a $k$-vector space. What about if $k$ is finite?
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Explicit example showing non-residual finiteness

This is mostly a continuation on the group I gave in the last post, which is given by the presentation
$$G = \langle a,t ~|~ t^{-1}a^2t = a^3\rangle.$$ At the risk of beating a dead horse, I proved that the homomorphism $f:G\to G$ given on generators by $f(t) = t$ and $f(a) = a^2$ is surjective but not injective. Groups for which surjective homomorphisms are isomorphisms are called Hopfian, and so our group $G$ is not Hopfian.

As I've been talking about frequently in the past little while, a group $G$ is called residually finite if for every nontrivial $x\in G$ there exists a homomorphism $\varphi:G\to F$ such that $F$ is finite and $\varphi(x)$ is not the identity of $F$. In the post on residually finite groups, I explained the classic proof that a finitely-generated, residually finite group is Hopfian.

Now the particular group $G$ here that is not Hopfian is finitely-generated, and so of course it can't be residually finite. I was wondering, can we find an explicit nontrivial element $x\in G$ such that for every homomorphism $\varphi:G\to F$ where $F$ is finite, the element $\varphi(x)\in F$ is the identity of $F$?

Yes, that's actually quite easy. It is because in the last post we already proved that the commutator $[t^{-1}at,a]$ is sent to the identity under the endomorphism $f:G\to G$ (recall, which was given by $f(t) = t$ and $f(a) = a^2$). But if we carefully examine the proof of the statement "every finitely-generated residually finite group is Hopfian", we see that the kernel of $f$ is actually contained in every finite-index normal subgroup of $G$. Therefore, in particular, the commutator $[t^{-1}at,a]$, which is nontrivial by Britton's lemma, is mapped to the identity under every homomorphism $G\to F$ where $F$ is a finite group!

Yet another group that is not Hopfian

A few weeks ago I gave an example of a non-Hopfian finitely-presented group. Recall that a group $G$ is said to be Hopfian if every surjective group homomorphism $G\to G$ is actually an isomorphism. All finitely-generated, residually finite groups are Hopfian. So for example, the group of the integers $\Z$ is Hopfian.

Another example of a group that is not Hopfian was given by Gilbert Baumslag and Donald Solitar. Their group is the one-relator group
$$G = \langle a,t ~|~ t^{-1}a^2t = a^3\rangle.$$ …read the rest of this post!

A zero-dimensional ring that is not von Neumann regular

An associative ring $R$ is called von Neumann regular if for each $x\in R$ there exists a $y\in R$ such that $x = xyx$.

Now let $R$ be a commutative ring. Its dimension is the supremum over lengths of chains of prime ideals in $R$. So for example, fields are zero dimensional because the only prime ideal in a field is the zero ideal.

Theorem. Let $R$ be a commutative ring. If $R$ is von Neumann regular, then it is zero dimensional.

The proof follows directly from the definition: suppose $P\subset R$ is a prime ideal of a von Neumann regular ring. If $x\not\in P$ and $y\in R$ is an element such that $x = xyx$, then $x(1 – yx) = 0$. Since $x\not\in P$, we must have $1 = yx$. Therefore, $P$ is maximal.

What about the converse? That's what this counterexample is all about.
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A finitely generated flat module that is not projective

Let's see an example of a finitely-generated flat module that is not projective!

What does this provide a counterexample to?

If $R$ is a ring that is either right Noetherian or a local ring (that is, has a unique maximal right ideal or equivalently, a unique maximal left ideal), then every finitely-generated flat right $R$-module is projective.

So what happens if we drop the Noetherian and local hypotheses?

The Example

Let $R = \prod_{j=1}^\infty F_j$ be an infinite product of fields and let $I = \oplus_{i=1}^\infty F_j$ be the ideal that is the direct sum of all the fields. Then the module $R/I$ is finitely generated. It is also flat, because $R$ is von Neumann regular and in such rings, every module is flat. Why is it not projective?

To see that it is not projective, consider the exact sequence
$$0\to I\to R\to R/I\to 0.$$ If $R/I$ were projective, that would mean that the map $R\to R/I$ splits, which gives a direct sum decomposition $I\oplus R/I\xrightarrow{\sim} R$ where the composition of the map $I\to I\oplus R/I\to R$ is the inclusion $I\to R$. The image of $R/I$ then corresponds to a nonzero ideal in $R$. But any nonzero ideal intersects $I$, so such a splitting is impossible.

This example is part of my new counterexamples project.

Kourkovka Notebook: Open problems in group theory

Every once in a while I spot a true gem on the arXiv. Unsolved Problems in Group Theory: The Kourkovka Notebook is such a gem: it is a huge collection of open problems in group theory. Started in 1965, this 19th volume contains hundreds of problems posed by mathematicians around the world. Additionally, problems solved from past volumes are also included with references.

For example, F.M. Markel proved that if $G$ is a finite supersolvable group with no two conjugacy classes having the same number of elements, then $G$ is actually isomorphic to the symmetric group $S_3$. Pretty cool right? Jiping Zhang extended this theorem by replacing 'supersolvable' by just 'solvable'. Problem 16.3 in the Kourkovka notebook asks the obvious: if $G$ is any finite group where no two conjugacy classes have the same size, is $G\cong S_3$? There are of course many more problems of varying technicality, but there should be something in here for any group theorist.

I've always thought that you can gauge the health of a discipline by the quality of open problems in it. If that's true, then the Kourkovka notebook shows that group theory is thriving very well.

Britton's lemma and a non-Hopfian fp group

In a recent post on residually finite groups, I talked a bit about Hopfian groups. A group $G$ is Hopfian if every surjective group homomorphism $G\to G$ is an isomorphism. This concept connected back to residually finite groups because if a group $G$ is residually finite and finitely generated, then it is Hopfian. A free group on infinitely many generators is an example of a residually finite group that is not Hopfian.

Are there examples of finitely generated groups that are not Hopfian? Such an example would then of course give us an example of a group that is not residually finite.

In this post, we'll see an example of a group that is finitely presented and not Hopfian. Not only that, but I promise the construction is actually not even scary, unlike those finitely presented groups with unsolvable word problem.
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