Wednesday, September 13, 2006

An elementary proof of Krull's Intersection Theorem

To do analysis on commutative rings, we need a metric, and to prove that the distance function we have is actually a metric, we need to show that it satisfies the three conditions for a metric. Two are easy to prove, but the third requires Krull's Intersection Theorem. Standard proofs of this theorem require advanced knowledge and complicated lemmas, but I'll explain a new, simpler proof that requires only abstract algebra.
Above you will find the title and abstract for my colloquium on September 25 at 1 pm. The proof is not mine; I will merely be explaining it. Feel free to come.

4 comments:

Brent said...

Will an outline or video be posted for those of us who (sadly) are not in Williamstown anymore? (=

Becca said...

I second that opinion... A video would be nice!

Diana said...

Wow, I had no idea that I was so popular. Unfortuately, math colloquia are not videotaped. Wait, that's probably fortunate -- did I mention that I would be talking for 35-40 minutes on the subject of Krull's Intersection Theorem? In the meantime, here is a video to substitute for the absence of the colloquium video. This is actual footage of me, taken by one of the students in my geometry class this summer.

Anonymous said...

Hey it's my proof! :)

It seems that (two years ago) you liked it, I feel very proud ;)
HP