I've been talking a little about abelian categories these days. That's because I've been going over Weibel's An Introduction to Homological Algebra. It's a book I read before, and I still feel pretty confident about the material. This time, though, I think I'm going to explore a few different paths that I haven't really given much thought to before, such as diagram proofs in abelian categories, group cohomology (more in-depth), and Hochschild homology.

Back to abelian categories. An abelian category is a category with the following properties:
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Posted by Jason Polak on 16. January 2018 · Write a comment · Categories: commutative-algebra, ring-theory · Tags:

Let $\Z[\Z/n]$ denote the integral group ring of the cyclic group $\Z/n$. How would you create $\Z[\Z/n]$ in Sage so that you could easily multiply elements?

First, if you've already assigned a group to the variable 'A', then

will give you the corresponding group ring and store it in the variable 'R'. The first argument of 'GroupAlgebra(-,-)' is the group and the second is the coefficient ring. Sage uses 'ZZ' to denote the integers, 'QQ' to denote the rationals, etc.

So how do you specify the cyclic group $A$? The first posibility is to use the construction:

where you'd replace 'n' by the actual number that you want there. This is useful if you want to work with other permutation groups, because the elements of 'A' are stored as permutations:

The output to this snippet is:

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Posted by Jason Polak on 01. September 2017 · Write a comment · Categories: ring-theory · Tags:

In the previous post we saw the following definition for a ring $R$: An element $r\in R$ is called strongly nilpotent if every sequence $r = r_0,r_1,r_2,\dots$ such that $r_{n+1}\in r_nRr_n$ is eventually zero. Why introduce this notion?

Well, did you know that every finite integral domain is a field? If $R$ is an integral domain and $a\in R$ is nonzero, then the multiplication map $R\to R$ given by $x\mapsto ax$ is injective. If $R$ is finite, then it must also be surjective so $a$ is invertible!

Another way of stating this neat fact is that if $R$ is any ring and $P$ is a prime ideal of $R$ such that $R/P$ is finite, then $P$ is also a maximal ideal. A variation of this idea is that every prime ideal in a finite commutative ring is actually maximal. Yet another is that finite commutative rings have Krull dimension zero.
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Posted by Jason Polak on 31. August 2017 · Write a comment · Categories: ring-theory · Tags:

Let $R$ be an associative ring. An element $r\in R$ is called nilpotent if $r^n = 0$ for some $n$. There is a stronger notion: an element $r\in R$ is called strongly nilpotent if every sequence $r = r_0,r_1,r_2,\dots$ such that $r_{n+1}\in r_nRr_n$ is eventually zero.

How are these two related? It is always the case that a strongly nilpotent element is nilpotent, because if $r$ is strongly nilpotent then the sequence $r,r^2,r^4,r^8,\dots$ vanishes. However, the element
$$\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$$
in any $2\times 2$ matrix ring is nilpotent but not strongly nilpotent. Notice how we had to use a noncommutative ring here—that's because for commutative rings, a nilpotent element is strongly nilpotent!