Posted by Jason Polak on 16. January 2017 · Write a comment · Categories: commutative-algebra · Tags:

An abelian group $A$ is a left $E = {\rm End}(A)$-module via $f*a = f(a)$. If $B$ is a direct summand of $A$ as an abelian group, then ${\rm Hom}(B,A)$ is also a left $E$-module and is in fact a direct summand of $E$ as an $E$-module, so it is $E$-projective. In particular, if $B = \Z$, then ${\rm Hom}(B,A)\cong A$ as $E$-modules. Thus $A$ is a projective $E$-module whenever $A$ has $\Z$ as a direct-summand.

These observations allow us to construct projective modules that often aren’t free over interesting rings. Take the abelian group $A = \Z\oplus \Z$ for instance. Its endomorphism ring $E$ is the ring $M_2(\Z)$ of $2\times 2$ matrices with coefficients in $\Z$. As we have remarked, $\Z\oplus \Z$ must be projective as an $M_2(\Z)$-module.

Is $\Z\oplus\Z$ free as an $M_2(\Z)$-module? On the surface, it seems not to be, but of course we need proof. And here it is: for each element of $\Z\oplus \Z$, there exists an element of $M_2(\Z)$ annihilating it. Such a thing can’t happen for free modules.

One might wonder, is every $M_2(\Z)$-module projective? Or in other words, is $M_2(\Z)$ semisimple? Let’s hope not! But $M_2(\Z)$ is thankfully not semisimple: $\Z/2\oplus\Z/2$ is a $M_2(\Z)$-module that is not projective: any nonzero element of $M_2(\Z)$ spans a submodule of infinite order, and therefore so must any nonzero element of a nonzero projective.

Posted by Jason Polak on 16. January 2017 · Write a comment · Categories: exposition · Tags:

A neutrino is a ultra low mass chargless subatomic particle that is produced in a variety of nuclear reactions such as beta decay. Neutrinos are incredibly abundant, but because of their size and lack of charge, they pass through almost anything and are extremely difficult to detect. Ray Jayawardhana’s book Neutrino Hunters is a fascinating glimpse into the great strides made by physicists to actually detect and understand these mysterious particles.

Neutrino Hunters progresses historically from Wolfgang Pauli’s hypothesising the existence of neutrino to its experimental confirmation and analysis as a central player in the workings of the universe. Several colourful and intriguing portraits of physicists appear along with the incredible experiments that were devised to understand the neutrino. Particularly fascinating were the early neutrino detectors, filled with hundreds of liters of dry-cleaning fluid that had to be placed deep underground to avoid interference. As time goes on, the detectors become more complex and even more ingenious, though I’ll leave out specifics so as not to spoil the book. Suffice it to say, some of the feats accomplished with complicated apparatus are truly astounding.

After the story of the then-state of the art is told, the author explains some future experiments and unresolved speculations, such as the possibility of a fourth neutrino flavour and neutrino communication.

The author manages to keep an excellent balance between precise scientific explanation for the nonspecialist and lively historical recounting. As someone who usually is terribly bored with history, I was never bored reading this book. This is not surprising, as the author is a physicist himself and does not fall into the habit (usually possessed by journalists without scientific training) of including vast amounts of irrelevant detail.

Overall, Jayawardhana is precise without ever being overly technical, and Neutrino Hunters should be easily readable by anyone with a basic knowledge of the atom and a yearning to glimpse into the subatomic world and the origins of the universe. Highly recommended.

Posted by Jason Polak on 11. January 2017 · Write a comment · Categories: commutative-algebra

Wolmer Vasconcelos [1] gave the following classification theorem about commutative local rings of global dimension two:

Theorem. Let $A$ be a commutative local ring of global dimension two with maximal ideal $M$. If $M$ is principal or not finitely generated, then $A$ is a valuation domain. Otherwise $M$ is generated by a regular sequence of two elements, and $A$ will be Noetherian if and only if it is completely integrally closed.

In this post we shall prove a small part of this theorem: that if $A$ is a commutative local ring of global dimension two and $M$ is a principal ideal, then $A$ is a valuation domain (i.e. for all $a,b\in A$ either $a | b$ or $b | a). As always, we use the term local ring to mean a commutative ring with a unique maximal ideal.
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Posted by Jason Polak on 07. January 2017 · Write a comment · Categories: exposition · Tags:

Math blogging is a fun part of being a mathematician. For me, it’s an aid to reading literature and an outlet for my writing prediliction. Blogging is cool because you write whatever you want without having to worry about the sometimes arbitrary and muddled standards of publication. But how do you do it? In this post I’ll give six tips on math blogging that should help.

1. Don’t worry about failure

Blogging is such a carefree medium that there’s no reason to worry about failed or unfinished posts. I have a folder of such posts just in case I want to resurrect any of them, and it contains around 80 posts–or just under half of the number of all the posts published. These range from very preliminary to finished and polished, totalling about 48000 words, or about twice the number of words in Hampton Fancher’s script for the movie Blade Runner. And you know what? It was fun to write those too, but in the end I decided they were not the right material for Aleph Zero Categorical.

2. Don’t worry about sophistication

Surprisingly, it’s actually fine to write about finitely generated abelian groups even though you’re working on interuniversal Teichmuller theory. It’s also fine to have a sophisticated blog. Mathematics actually needs much more exposition, so there’s really no need to restrict yourself to certain topics because you think other mathematicians will look down on it.

3. Keep your blog focused

Pick a topic and stick with it. For my blog, it’s math and related fields, like computer science and applications, though mostly I just write about algebra. There have been occasions where I’ve been tempted to post reviews of books I’ve read in other fields like biology but I’ve resisted because that was outside the scope I set out, and I doubt it would make sense to my audience. Writing outside the scope is a slipperly slope: first it’ll be one or two posts on chemistry, and then pretty soon you’ll be writing on bizarre topics like politics and South American mushrooms.

4. Update regularly

I require myself to produce one post per month. Only once or twice did that fail in the past five or so years, and on average I’m way above that. Not only will a regular update requirement keep you blogging, but it will keep your readers around. Most math blogs miss months, so I figure I’m safe.

5. Heed the format

A blog post is not supposed to be long, and people don’t visit blogs to read proofs of the four colour theorem. I’ve definitely written posts that were too long, and in the end those were not so popular. Keep posts to a main idea, keep it under a thousand words, and your blog will be far more readable for it.

6. In the end, it doesn’t have to be math

Weird advice for a post on math blogging right? But in the end if you don’t enjoy math blogging, you might still enjoy food blogging or posting pictures of rocks.

A ring of left global dimension zero is a ring $R$ for which every left $R$-module is projective. These are also known as semisimple rings of the Wedderburn-Artin theory fame, which says that these rings are precisely the finite direct products of full matrix rings over division rings. Note the subtle detail that “semisimple” is used here instead of “left semisimple” because left semisimple is the same thing as right semsimple.

In the commutative world, the story for Krull dimension zero is not so simple. For example, every finite commutative ring has Krull dimension zero. Indeed, if $R$ is a ring with Krull dimension greater than zero, then there would exist two distinct primes $P\subset Q$ so that $R/P$ is an integral domain that is not a field. Thus, $R$ is infinite, as every finite integral domain is a field.

The story becomes simpler if we require $R$ to have no nilpotent elements: i.e., that $R$ is reduced. In this case, a commutative ring is reduced and of Krull dimension zero if and only if every principal ideal is idempotent. Every principal ideal being idempotent means that for every $x\in R$ there is an $a\in R$ such that $xax = x$. Rings, commutative or not, satisfying this latter condition are called von Neumann regular. So:

Theorem. A commutative ring has Krull dimension zero and is reduced if and only if it is von Neumann regular.

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Posted by Jason Polak on 29. December 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring. We say that an $R$-algebra $A$ is separable if it is projective as an $A\otimes_R A^{\rm op}$-module. Examples include full matrix rings over $R$, finite separable field extensions, and $\Z[\tfrac 12,i]$ as a $\Z[\tfrac 12]$-algebra.

The 1970 classic Separable Algebras by deMeyer and Ingraham acquaints the reader with this important class of algebras from two viewpoints: the noncommutative one through structure theory and the Brauer group, and the commutative one through Galois theory.

This book accomplished the rare feat of keeping me interested; throughout its pages I found I could apply its results to familiar situations: Why are the only automorphisms of full matrix rings over fields inner? Why are such rings simple? What makes Galois theory tick? Separable Algebras explains with clarity how familiar algebra works through the lens of separable algebras.
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Posted by Jason Polak on 29. December 2016 · Write a comment · Categories: commutative-algebra

Let $R$ be an integral domain and $a,b\in R$. Suppose that $a | b$ and $b | a$. By definition, this means that $ax = b$ and $by = a$ for some $x,y\in R$, and so $b(1 – yx) = 0$. If $b\not = 0$ then $yx = 1$ and so $x$ and $y$ are units. Of course, if $b$ is zero, then $a = 0$, and in either case $b = ax$ where $x$ is a unit.

Concluding, we see that in an integral domain, if $a | b$ and $b | a$ then $b = ax$ where $x$ is a unit of $R$. Did you know that this implication can fail when $R$ is not an integral domain?

Here’s a counterexample that I learned in a paper of Anderson and Valdez-Leon. Let $R = k[X,Y,Z]/(X – XYZ)$ where $k$ is a field, and let $x,y,z$ denote respectively the images of $X,Y,Z$ in $R$. Then $x | xy$ and $xy | x$, the latter because $xyz = x$. However, it is not possible to write $xy$ as a unit multiple of $x$.

Why is this? Well, first of all, $y$ is definitely not a unit in $R$ because in the quotient ring obtained by setting $x=z=0$, this would mean that $y$ would be a unit in the polynomial ring $k[y]$. On the other hand, this argument does not preclude the existence of an $f\in R$ such that $fx = xy$.

So suppose such an $f$ did exist. Then in $k[X,Y,Z]$, we would have the relation $fX – YX = gX(1 – YZ)$ for some $g\in k[X,Y,Z]$. Since $k[X,Y,Z]$, is an integral domain, we have $f = Y + g(1 – YZ)$. If $f$ were a unit in $R$, then by setting $X = 0$ we see that $Y + g(1 – YZ)$ would be a unit in $k[Y,Z]$. Setting $Y = Z$ shows that $Z + g(1 – Z^2) = Z + g – gZ^2$ would be a unit in $k[Z]$, which is impossible since the term $gZ^2$ shows $Z + g – gZ^2$ is not a constant polynomial.

I like this example not only because it’s neat, but because it illustrates the principle that a good place to look for counterexamples in commutative ring theory is in the land of quotients of polynomial rings.

Posted by Jason Polak on 26. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring and let $A$ be an $R$-algebra. We say that $A$ is separable if $A$ is projective as an $A\otimes_RA^{\rm op}$-module. There is a multiplication map $\mu:A\otimes_RA^{\rm op}\to A$ given by $a\otimes a’\mapsto aa’$, whose kernel we’ll call $J$.

It’s a fact that $A$ is separable if and only if there exists an idempotent $e\in A\otimes_RA^{\rm op}$ such that $\mu(e) = 1$ and $Je = 0$. The most familiar examples of separable algebras to many readers are probably the finite separable field extensions of a given field $R$. So let’s see an example of the separability idempotent!

For example, $\Q(\sqrt{2})$ is a separable $\Q$-algebra. One can then write down explicitly the above definitions. For example, $\Q(\sqrt{2})\otimes_\Q\Q(\sqrt{2})\cong \Q(\sqrt{2})\oplus \Q(\sqrt{2})$ via the map given on pure tensors by $(a,b)\mapsto (ab,a\sigma(b))$ where $\sigma:\Q(\sqrt{2})\to\Q(\sqrt{2})$ is the $\Q$-algebra isomorphism given by $\sqrt{2}\mapsto -\sqrt{2}$. The inverse of this map is given by
$$ (u,v)\mapsto \frac{u+v}{2}\otimes 1 + \frac{u-v}{2\sqrt{2}}\otimes\sqrt{2} $$
The multiplication map under this isomorphism translates to a map
$$\Q(\sqrt{2})\oplus\Q(\sqrt{2})\to Q(\sqrt{2})\\
(a,b)\mapsto a
Therefore, the kernel of the multiplication map is $J = 0\oplus\Q(\sqrt{2})$ and the separability idempotent is just $e = (1,0)$. Of course, it’s naturally to want to observe this idempotent in its natural habitat of the tensor product $\Q(\sqrt{2})\otimes_Q\Q(\sqrt{2})$. Here it is folks:
e = \frac{1}{2}\otimes 1 + \frac{1}{2\sqrt{2}}\otimes\sqrt{2}
It’s easy to see that in general $J$ is the ideal generated by the set $\{ 1\otimes a – a\otimes 1 : a\in A\}$. Now just try verifying $Je = 0$ directly! Or not.

Posted by Jason Polak on 26. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $A\to B\to C$ be ring homomorphisms. Consider the following statements:

  1. $B$ is a projective $A$-algebra,
  2. $C$ is a projective $B$-algebra,
  3. $C$ is a projective $A$-algebra.
Question. Which pairs of these statements imply the third?

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Posted by Jason Polak on 22. November 2016 · Write a comment · Categories: commutative-algebra · Tags: ,

Let $k$ be a field. The ring $M_n(k)$ of $n\times n$ matrices over $k$ has some automorphisms, given by conjugation by elements of $\GL_n(k)$. These are inner automorphisms, and this action happens to be the adjoint action of $\GL_n$ on its Lie algebra. Are there any other automorphisms? The answer is no, and the reason fits in the framework of automorphisms of central separable algebras. In this post we’ll sketch how this works.

Let $R$ be a commutative ring. An $R$ algebra $A$ is called separable if $A$ is projective as an $A\otimes_R A^{\rm op}$-module and $A$ is called central separable if it is separable and the center of $A$ coincides with $R$.

A typical example is a matrix ring $M_n(R)$ over $R$. In fact, whenever $M$ is a finitely generated projective and faithful module then ${\rm Hom}_R(M,M)$ is a central separable $R$-algebra. So even if you don’t have much intuition for central separable algebras in general, everything we say about them will certainly apply to algebras of the form ${\rm Hom}_R(M,M)$.

For a commutative ring, let’s call a finitely generated projective module of constant rank one a line bundle. The Picard group of a commutative ring $R$ is the group of isomorphism classes of line bundles. The group operation is the tensor product. If $A$ is a central separable algebra, then $A^A = \{ b\in A : ab = ba \forall a\in A\} = R$ is a trivial example of a line bundle.

More generally, $(A_\sigma)^A = \{ b\in A : ab = b\sigma(a)\forall a\in A\}$ is also a line bundle. This gives a well-defined map
{\rm Aut}_R(A)\to {\rm Pic}(R)
that happens to fit into an exact sequence
1\to {\rm Inn}_R(A)\to {\rm Aut}_R(A)\to {\rm Pic}(R).
Consequently, if ${\rm Pic}(R)$ is trivial, then every $R$-algebra automorphism of $A$ must in fact be inner. For example, this works for fields and principal ideal domains, and hence answers our query that we started with: why is every algebra automorphism of a full matrix ring inner? The construction of the exact sequence, which I will not explain at this point, actually gives a somewhat constructive answer: if $\sigma$ is a given automorphism then $(A_\sigma)^A$ is a free $R$-module on a generator $v$ that is a unit in $A$, and $\sigma(x) = v^{-1}xv$.