Posted by Jason Polak on 27. August 2017 · Write a comment · Categories: math, modules · Tags: , ,

Let $R$ be an associative ring with identity. The Jacobson radical ${\rm Jac}(R)$ of $R$ is the intersection of all the left maximal ideals of $R$. So, ${\rm Jac}(R)$ is a left ideal of $R$. It turns out that the Jacobson radical of $R$ is also the intersection of all the right maximal ideals of $R$, and so ${\rm Jac}(R)$ is also an ideal!

The idea behind the Jacobson radical is that one might be able to explore the properties of a ring $R$ by first looking at the less complicated ring $R/{\rm Jac}(R)$. Since the ideals of $R$ containing ${\rm Jac}(R)$ correspond to the ideals of $R/{\rm Jac}(R)$, the ring $R/{\rm Jac}(R)$ has zero Jacobson radical. Often the rings $R$ for which ${\rm Jac}(R) = 0$ are called Jacobson semisimple.

This terminology might be a tad bit confusing because typically, a ring $R$ is called semisimple if every left $R$-module is projective, or equivalently, if every left $R$-module is injective. How does the notion of semisimple differ from Jacobson semisimple? The Wedderburn-Artin theorem gives a classic characterisation of semisimple rings: they are exactly the rings that are finite direct products of full matrix rings over division rings. Since a full matrix ring over a division ring has no nontrivial ideals, the product of such rings must have trivial Jacobson radical. Thus:

A semisimple ring is Jacobson semisimple.

The converse is false: there exists a ring that is Jacobson semisimple but not semisimple. For example, let $R$ be an infinite product of fields. Then ${\rm Jac}(R) = 0$. However, $R$ is not semisimple. Why not? If it were, by Wedderburn-Artin it could also be written as a finite product of full matrix rings over division rings, which must be a finite product of fields because $R$ is commutative. But a finite product of fields only has finitely many pairwise orthogonal idempotents, whereas $R$ has infinitely many.

Incidentally, because $R$ is not semisimple, there must exist $R$-modules that are not projective. However, $R$ does have the property that every $R$-module is flat!

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 28. January 2012 · Write a comment · Categories: books, modules · Tags: , ,

As it happens every so often, I browse the mathematical library pseudorandomly, and look out for interesting titles; usually a prerequisite for interesting is that they have something to do with the realm of algebra. This is exactly how I found Faith’s book, with its captivating title urging me to borrow it.

Now, inevitably in mathematical research, one has to efficiently skim through papers and books to find specific ideas and facts. The unfortunate thing is that sometimes it is easy to neglect the stimulation of the idle curiosity that probably brought most mathematicians into their fields in the first place, and so I try to combat this neglect by my idle browsing and blogging.

I try not to spend too much time on this so that I progress with my degree, but I try to nurture my curiosity through reading anything that looks interesting. Returning to books, I do believe there are few worse literary follies than a graduate algebra textbook that lacks imagination in its examples and theorems and passion in its explication. I only fear that such books will tend to promote in the learning of higher algebra what most institutions have done with calculus, and that is to make it a tiresome mechanical effort, washing away the once vibrant and fanciful colours from the gentle tendrils of the mind.

But fear not! Should the mental dessication start to occur in a young algebraist’s mind; should the flames of passion dim for the wonders of the injective module, she can always turn to the entire object of this post, videlicet Faith’s “Rings and Things and a Fine Array of Twentieth Century Associative Algebra”
. I refer to the second edition, incidentally, which corrects many errors from the 1st edition.

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Posted by Jason Polak on 16. June 2011 · 4 comments · Categories: homological-algebra · Tags: ,

I’m going to talk a bit about flatness. I created the following exercise for myself while studying flatness: describe some good examples of flat modules that are not projective.

We call a left $ R$-module $ F$ flat if the functor $ -\otimes F$ is exact, and of course the analogue can be made for right R-modules. As the tensor functor $ -\otimes F$ is always right exact, this is a rather natural definition to make. Recall that there is a similar definition for the $ \mathrm{Hom}(P,-)$ functor: we call an $ R$-module $ P$ projective if $ \mathrm{Hom}(P,-)$ is exact.

Projective modules have a rather nice characterisation. That is, a module $ P$ is projective if and only if it is the direct summand of a free module. In particular free modules are projective. Mac Lane in his book “Homology” gives an example of a projective module that is not free: take the ring $ \mathbb{Z}\oplus\mathbb{Z}$ and consider the submodule $ \mathbb{Z}\oplus 0$ of it.
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