ring theory | College Math Teaching

January 30, 2015

Nilpotent ring elements

Filed under: advanced mathematics, algebra, matrix algebra, ring theory — Tags: nilpotent elements, ring theory — collegemathteaching @ 3:12 am

I’ve been trying to brush up on ring theory; it has been a long time since I studied rings in any depth and I need some ring theory to do some work in topology. In a previous post, I talked about ideal topologies and I might discuss divisor toplogies (starting with the ring of integers).

So, I grabbed an old text, skimmed the first part and came across an exercise:

an element $x \in R$ is nilpotent if there is some positive integer $n$ such that $x^n = 0$ . So, given $x, y$ nilpotent in a commutative ring $R$ one has to show that $x+y$ is also nilpotent and that this result might not hold if $R$ is not a commutative ring.

Examples: in the ring $Z_9, 3^2 =0$ so $3$ is nilpotent. In the matrix ring of 2 by 2 matrices,

$\left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right)$ and $\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$ are both nilpotent elements, though their sum:

$\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$ is not; the square of this matrix is the identity matrix.

Immediately I thought to let $m, n$ be the smallest integers for $x^m =y^n = 0$ and thought to apply the binomial theorem to $(x+y)^{mn}$ (of course that is overkill; it is simpler to use $(x+y)^{m+n}$ . Lets use $(x+y)^{m+n}$ . I could easily see why $x^{m+n} = y^{m+n} =0$ but why were the middle terms ${m+n \choose k} x^{(m+n)-k}y^k$ also zero?