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!