I.1.5
12/4/2020
What a fucking word salad, but it's easy. It took me like 5 minutes. Regardless, that's all I'm doing for today, even
though I'm behind. Yes, yes, I'm a lazy slowpoke. Yes, yes, I always go too easy on myself. I self-deprecate a lot, but
if I had a clone, I'd pamper it.
So we have here a characterization of affine coordinate rings, eh? I've heard that said coordinate rings should be
thought of as "functions" (kn → k) on the corresponding zero set, so maybe I can sometimes think of the coordinate
ring as a "function space" or "function ring". In which case, I suppose the sense of this exercise advances slightly
past a merely dry algebra computation. For any An, we know that, on any zero set in it, the ring of kn → k
(polynomial) functions is just an *AHEM* "finitely generated k-algebra with no nilpotents"... Ouch, I guess it
degenerated into word salad again, but I guess words can only express so much *puckers lips*. No?
Okay *unpuckers lips* I've also heard that schemes and modern algebraic geometry "matter" partly
because they can "capture" nilpotents, and I guess what we're dealing with here is the nice, classical
case where we don't have nilpotents... or something. I'll keep an eye on those nilpotents, is all I'm
saying.
Anyway, here's the proof.
It's easy to check that
(1) B = A∕J has no nilpotents ⇐⇒ the J is radical.
(so, from strongest to weakest, maximal ⇐⇒ field, prime ⇐⇒ integral domain, radical ⇐⇒ no nilpotents)
So... to show the direction,
we have
B = A(Y ) where Y is some algebraic set in some An so A(Y ) = k[x1,...,xn]∕I(Y ) where I(Y ) is a radical ideal, but
this means that B is a finitely generated k-alg with no nilpotents (by (1))
So is done.
For the other direction, we are given,
B = k[x1,...,xn]∕J where J is radical (by (1)). Let Y = Z(J) in An so that
A(Y ) | = k[x1,...,xn]∕I(Y ) | ||
= k[x1,...,xn]∕I(Z(J)) | |||
= k[x1,...,xn]∕J | |||
= B | |||