Posted by Jason Polak on 22. February 2017 · Write a comment · Categories: commutative-algebra, homological-algebra · Tags:

Let $R$ be a commutative ring and $A$ and $R$-module. We say that $x_1,\dots,x_n\in R$ is a regular sequence on $A$ if $(x_1,\dots,x_n)A\not = A$ and $x_i$ is not a zero divisor on $A/(x_1,\dots,x_{i-1})A$ for all $i$. Regular sequences are a central theme in commutative algebra. Here’s a particularly interesting theorem about them that allows you to figure out a whole bunch of Ext-groups:

Theorem. Let $A$ and $B$ be $R$-modules and $x_1,\dots,x_n$ a regular sequence on $A$. If $(x_1,\dots,x_n)B = 0$ then
$$
{\rm Ext}_R^n(B,A) \cong {\rm Hom}_R(B,A/(x_1,\dots,x_n)A)$$

This theorem tells us we can calculate the Ext-group ${\rm Ext}_R^n(B,A)$ simply by finding a regular sequence of length $n$, and calculating a group of homomorphisms. We get two cool things out of this theorem: first, a corollary of this theorem is that any two maximal regular sequences on $A$ have the same length if they are both contained in some ideal $I$ such that $IA\not= A$, and second, it enapsulates a whole range of Ext-calculations in an easy package.

For example, let’s say we wanted to calculate ${\rm Ext}_\Z^1(\Z/2,\Z)$. Well, $2\in\Z$ is a regular sequence, and so the above theorem tells us that this Ext-group is just ${\rm Hom}_\Z(\Z/2,\Z/2) \cong\Z/2$.

Another example: is ${\rm Ext}_{\Z[x]}^1(\Z,\Z[x])\cong\Z$.

Of course, the above theorem is really just a special case of a Koszul complex calculation. However, it can be derived without constructing the Koszul complex in general, and so offers an instructive and minimalist way of seeing that for Noetherian rings and finitely generated modules, the notion of length of a maximal regular sequence is well-defined.

Posted by Jason Polak on 31. January 2017 · Write a comment · Categories: math

This post is a list of various books in commutative algebra (mostly called ‘Commutative Algebra’) with some comments. It is mainly geared towards students who might want to read about the subject on their own, though others might find it useful. It is not meant to be a comprehensive list, as it reflects the random process of my coming into contact with them.
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Posted by Jason Polak on 25. January 2017 · Write a comment · Categories: commutative-algebra, homological-algebra

We already saw that an abelian group with a $\Z$-direct summand is projective over its endomorphism ring. Finitely generated abelian groups are also projective over their endomorphism rings by essentially the same argument. What’s an example of an abelian group that is not projective over its endomorphism ring?

Here’s one: the multiplicative group $Z(p^\infty)$ of all $p$-power roots of unity. Another way to define this group is $\Z[p^{-1}]/\Z$. What is the endomorphism ring of this group? In fact it is the $p$-adic integers $\Z_p$. Indeed, an endomorphism $Z(p^\infty)\to\Z(p^\infty)$ has to send $1/p$ to an element $a_1$ such that $pa_1 = 0$. So we have the choice of the elements $0/p, 1/p,\cdots, (p-1)/p$, which form the cyclic subgroup $\Z/p$.

Similarly, $1/p^2$ has to be sent to an element $a_2$ such that $p^2a_2 = 0$, but also $pa_2 = a_1$. So $a_2$ has to be of the form $n/p^2$ where $n\in \Z$; in other words, $a_2$ can be in the cyclic subgroup $\Z/p^2$ generated by $1/p^2$. Hence, an endomorphism of $Z(p^\infty)$ is specified by an element of the inverse system $\cdots\to \Z/p^3\to \Z/p^2\to \Z/p$ where the transition maps are multiplication by $p$: in other words the $p$-adic integers $\Z_p$.

Now, we see that $Z(p^\infty)$ cannot be a projective $\Z_p$-module. Indeed, $\Z_p$ is a local ring and hence any projective $\Z_p$-module is in fact free (Kaplansky’s theorem) and in particular torsionfree. However, $Z(p^\infty)$ has nothing but torsion! In fact we can say more: since $\Z_p$ is a principal ideal domain, it has global dimension one, so the projective dimension of $Z(p^\infty)$ as a $\Z_p$-module is one.

Posted by Jason Polak on 25. January 2017 · Write a comment · Categories: math

I’ve spent countless hours thinking about associative rings. Yet, during my research today I read a paper by C.U. Jensen [1] and came across this elementary fact that I never thought about: if $R$ is an integral domain and $x,y$ are nonunits with $y$ divisible by every positive power of $x$ then $R$ is not Noetherian. Perhaps I’ve used this unconsciously, but I had to take a second to prove it.

Here’s the proof: if $y = z_1x = z_2x^2 = \cdots$ then the ideal $(z_1,z_2,\cdots)$ cannot be finitely generated, because $z_n = r_1z_1 + \cdots + r_{n-1}z_{n-1}$ and $y = z_ix^i$ for all $i$ implies that $x$ is actually invertible.

This fact is curious, because it plays into constructing some nonstandard models of the integers as I learned from Jensen’s paper: If $\Ucl$ is a nonprincipal ultrafilter on the natural numbers $\N$, then the product $\Z^\N/\Ucl$ is one such nonstandard model: it satisfies precisely the same first-order sentences that $\Z$ does. Yet, it’s not Noetherian, because the element represented by $(2,4,8,16,32,64,…)$ is not a unit and divisible by every power of $2$.

In fact, not only is the ultraproduct $\Z^\N/\Ucl$ not Noetherian, it actually has global dimension two. So, neither Noetherian nor having global dimension one is expressible in first-order logic.

It’s amazing what basic facts you can still come across in this era. Perhaps someone should write the book 100 Quick Facts You Didn’t Know About Rings?!

[1] Jensen. Peano rings of arbitrary global dimension. Journal of the London Mathematical Society, 1980, 2, 39-44

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 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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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.