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 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.

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 05. October 2016 · 2 comments · Categories: commutative-algebra, homological-algebra · Tags:

For finite commutative rings, integral domains are the same as fields. This isn’t too surprising, because an integral domain $R$ is a ring such that for every nonzero $a\in R$ the $R$-module homomorphism $R\to R$ given by $r\mapsto ra$ is injective. Fields are those rings for which all these maps are surjective. But injective and surjective coincide for endofunctions of finite sets. Therefore, domains are the same thing as fields for finite rings.

But did you know that there is another class of commutative rings for which fields are the same as integral domains? Indeed, for self-injective rings, fields are the same as domains. By definition, a commutative ring $R$ is self-injective if $R$ is injective as an $R$-module. Note: for noncommutative rings, which we don’t consider here, there is a difference between left and right self-injective; that is, an arbitrary ring may be injective as a left module over itself, but not right self-injective, and vice-versa.

In other words, self-injective integral domains are fields. And, the proof is sort of along the lines of the one for finite rings:

Proof. Let $a\in R$ be nonzero. Then the multiplication map $R\xrightarrow{a} R$ is injective, and fits into a diagram

2016-10-01-fdomain

Where the dotted arrow exists because $R$ is injective as an $R$-module; since it is a map $R\to R$ it is given by multiplication by some $b\in R$. Therefore $1 = ab$. QED.

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Let $R$ be a ring and $M$ and $R$-module. If every finitely generated submodule of $M$ is flat, then so is $M$, because direct limits commute with the $\mathrm{Tor}$-functor. What about the converse? If $M$ is flat, are all its finitely generated submodules flat too?

Not necessarily! In fact, here’s a roundabout argument without an actual counterexample: we’ve already seen that the weak dimension of a ring is less than or equal to one iff every ideal is flat. And, for Noetherian rings, the weak dimension is the same as the global dimension. For a field, the global dimension of $k[X]:=k[x_1,\dots,x_n]$ is $n$ and so if $n\geq 2$ then $k[X]$ must have ideals that are not flat, and yet each ideal is finitely generated. Hence $k[X]$ as a $k[X]$-module is flat (as it’s free) but has finitely generated $k[X]$-submodules that cannot be flat.

Amusingly, this counterexample is also a counterexample to the statement that to any conjecture one should give either a proof or an explicit counterexample!

Hint: for an actual counterexample, $(x,y)$ in $k[x,y]$ works!

Given an idempotent $e$ in a ring $R$, the right ideal $eR$ is projective as a right $R$-module. In fact, $eR + (1-e)R$ is actually a direct sum decomposition of $R$ as a right $R$-module. An easy nontrivial example is $\Z\oplus\Z$ with $e = (1,0)$.

Fix an $a\in R$. If $aR$ is a projective right $R$-module, however, that doesn’t mean that $a$ is an idempotent. In fact $aR$ is projective whenever $a$ is a nonzerodivisor, and in this case $aR$ is just isomorphic to $R$ itself as a right $R$-module.

So how do idempotents come into play in general? It turns out we have to look at annihilators! The right annihilator of $e$ is the right ideal $(1-e)R$. Indeed, $e(1-e) = 0$. And, if $er = 0$, then $(1 – e)r = r$, so anything that annihilates $e$ is a multiple of $(1-e)$. So we see that the annihilator of $eR$ is $(1-e)R$.

What about in general? It turns out that if $aR$ is projective, the right annihilator of $a$ must be of the form $eR$ for an idempotent $e$. Indeed, if $aR$ is projective, then the map $R\to aR$ given by $r\mapsto ar$ has a splitting $\varphi:aR\to R$. I’ll leave it as an exercise to show that the right annihilator of $a$ is $(1 – \varphi(a))R$, and that $1 – \varphi(a)$ is in fact an idempotent.

Conversely, if the right annihilator of an $a\in R$ is of the form $eR$ for some idempotent, then multiplication by $1-e$ gives the splitting of the natural map $R\to aR$, so $aR$ must be projective.

Posted by Jason Polak on 16. June 2016 · Write a comment · Categories: commutative-algebra, homological-algebra, modules

Projective modules are the algebraic analogues of vector bundles, and they satisfy some strong properties. To state one we will first introduce the notation $P^* := {\rm Hom}_R(P,R)$ for any right $R$-module $P$. (Working with right $R$-modules is just a convention)

Here’s one property that projective modules satisfy: if $P$ is a right projective module over a ring $R$ then the natural map
$$
e:P\to P^{**}$$
given by $e(p)(f) = f(p)$ is a monomorphism—which, in the category of $R$-modules, just means that $e$ is injective. The first question should be: is it ever not an isomorphism? The lack of surjectivity for $e$ can already be found when $R = k$ is a field.

Here, if $P = \oplus_I k$ then ${\rm Hom}_k(\oplus_I k,k) = \prod_I k$ so the dual has strictly greater cardinality as soon as $I$ is an infinite set. In fact, this same argument shows that the $P$ cannot be isomorphic to $P^{*}$, let alone $P^{**}$ whenever $P$ is not finitely generated.

But $e$ is always a monomorphism whenever $P$ is projective. If $P$ is arbitrary, then $e$ may not be a monomorphism. For example if $R = \Z$ then $P=\Z/2$ is a counterexample. ${\rm Hom}_\Z(\Z/2,\Z) = 0$. Another more striking example is $P = \Q$, the rational numbers. So, $e$ may fail to be a monomorphism even when $P$ is flat.

Can you give any examples of $e$ being a monomorphism even when $P$ is not projective?

Posted by Jason Polak on 22. December 2015 · Write a comment · Categories: homological-algebra · Tags: ,

For modules one has the isomorphism theorem $(A/C)/(B/C) \cong A/B$ for $C\leq B\leq A$. One way to remember it is through analogy with canceling of fractions. Another way to remember and prove it is to put all the modules in a 3×3 commutative diagram
$$
\begin{matrix}
C & \to & B & \to & B/C\\
\downarrow & ~ & \downarrow & ~ & \downarrow\\
C & \to & A & \to & A/C\\
\downarrow & ~ & \downarrow & ~ & \downarrow\\
0 & \to & A/B & \to & (A/C)/(B/C)
\end{matrix}
$$
where there are also zero arrows on the edges of the diagram, but I have omitted them for ease of typesetting. All the columns are exact, and the first two rows are exact, so the remaining row is exact giving the required isomorphism via the 3×3-lemma.

Posted by Jason Polak on 19. December 2015 · Write a comment · Categories: group-theory, homological-algebra · Tags:

Let $A$ be an abelian group. We call an element $a\in A$ torsion if there exists a natural number $n$ such that $na = 0$. The set of all torsion elements $T(A)$ of $A$ form a subgroup of $A$, and we can think of $T$ as an endofunctor on the category of abelian groups. Here are some examples:

  • $T(\Z) = 0$
  • $T(\Z\oplus \Z/n) = 0\oplus\Z/n\cong\Z/n$
  • $T(\Q) = 0$
  • $T(\Q/\Z) = \Q/\Z$

Finding the set of torsion points of an abelian group isn’t always easy as in these examples, since abelian groups may not always be written out in such an explicit way. A fascinating and nontrivial result of Barry Mazur is that the torsion subgroup of $E(\Q)$ for an elliptic curve $E$ over $\Q$ is one of fifteen possibilities: $\Z/n$ for $1\leq n\leq 10$ or $n=12$ or $\Z/2\times \Z/2n$ for $1\leq n\leq 4$. Such strange numerology!
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Let $R$ be any commutative ring. The content of a polynomial $f\in R[x]$ is by definition the two-sided ideal in $R$ generated by the coefficients of $f$. If $f,g\in R[x]$, then $c(fg)\subseteq c(f)c(g)$, because each coefficient of $fg$ is a linear combination of elements of $c(f)c(g)$. Sometimes, however, this inclusion is strict. For example, if $k$ is a field of characteristic two, and $R = k[u,v]$ then $f = u + vX$ satisfies $c(f^2)\subset c(f)^2$, where the inclusion is strict. Indeed, $f^2 = u^2 + v^2X^2$ so $c(f^2) = (u^2,v^2)$, whereas $c(f)c(f) = (u^2,v^2,uv)$. A ring $R$ in which $c(fg) = c(f)c(g)$ for all $f,g\in R[x]$ is called Gaussian. We have just seen that $k[u,v]$ is not Gaussian, and in fact, we didn’t even have to specify that $k$ is characteristic two. What about a polynomial ring $k[u]$ over a field $k$ in one variable? Since $k$ is a field, $k[u]$ is a principal ideal domain (PID), and PIDs are always Gaussian. These observations can be clarified by looking at the concept of weak dimension.
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