College Math Teaching

January 30, 2015

Nilpotent ring elements

Filed under: advanced mathematics, algebra, matrix algebra, ring theory — Tags: , — 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?

Then it dawned on me: x^n=0 \rightarrow x^{n+k}=0 for all k \geq 0 . Duh. Now it made sense. 🙂

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