College Math Teaching

March 24, 2015

Google Doodle

Filed under: advanced mathematics, algebra, famous mathematicians — Tags: , — collegemathteaching @ 2:52 am

Screen shot 2015-03-23 at 9.47.32 PM

For the prize: what is the significance of the two chains, and which one directly applies to the human subject of this doodle?

The human subject of the doodle is Emmy Noether.


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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