Nullstellensatzsatz
Oct. 8th, 2008 03:42 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoyEDIT 13 October: Turns out it was finitely generated division algebra, not finite dimensional. In which case I do see how to prove the Nullstellensatz from it, it's not so different from Murthy's "HN1", and a lot of my complaint is nullified.
There are too many statements referred to as the Nullstellensatz. (And I just mean the ordinary Hilbert's Nullstellensatz - according to Wikipedia there's also a "differential Nullstellensatz", a "combinatorial Nullstellensatz" (actually I think I may have seen that once somewhere, but I don't remember it) and "Stengle's Positivstellensatz" (not quite the same, but...))
I mean, we all know the full, strong Nullstellensatz - k algebraically closed, J an ideal in k[x_1,...,x_n], then I(V(J))=rad J. And the weak Nullstellensatz, that if k, J as above, J proper, then V(J)≠∅.
Well last year when I was taking commutative algebra, Murthy also referred to the statement that if k a field, A a finitely generated algebra over k which is itself a field, then k finite dimensional over A, as the Nullstellensatz. And with a bit of prodding from Planetmath, I was able to remember how we got from there to the actual Nullstellensatz (well, the weak one, anyway), but it's pretty nonobvious.
And now today, Ginzburg referred to the statment that if k algebraically closed, A a finite dimensional division algebra over it, then A=k, as the Nullstellensatz. I have no idea how you prove it from that, though.
Gah. It is not good to have so many statements running around under the same name when the relation between them is so totally nonobvious!
-Harry
There are too many statements referred to as the Nullstellensatz. (And I just mean the ordinary Hilbert's Nullstellensatz - according to Wikipedia there's also a "differential Nullstellensatz", a "combinatorial Nullstellensatz" (actually I think I may have seen that once somewhere, but I don't remember it) and "Stengle's Positivstellensatz" (not quite the same, but...))
I mean, we all know the full, strong Nullstellensatz - k algebraically closed, J an ideal in k[x_1,...,x_n], then I(V(J))=rad J. And the weak Nullstellensatz, that if k, J as above, J proper, then V(J)≠∅.
Well last year when I was taking commutative algebra, Murthy also referred to the statement that if k a field, A a finitely generated algebra over k which is itself a field, then k finite dimensional over A, as the Nullstellensatz. And with a bit of prodding from Planetmath, I was able to remember how we got from there to the actual Nullstellensatz (well, the weak one, anyway), but it's pretty nonobvious.
And now today, Ginzburg referred to the statment that if k algebraically closed, A a finite dimensional division algebra over it, then A=k, as the Nullstellensatz. I have no idea how you prove it from that, though.
Gah. It is not good to have so many statements running around under the same name when the relation between them is so totally nonobvious!
-Harry
