Re: hilbert's nullstellensatz

Jannick Asmus wrote:

If M is a maximal ideal in R then M/R is a field.
This is a finitely-generated extension of C, and so is C itself.
Hence one has a homomorphism f: R -> C with kernel M.
For each coordinate x_i, let f(x_i) = a_i.
Then x_i - a_i is in M for i = 1,...,n,
and it follows easily enough that these linear polynomials generate M.
You are using another version of Hilbert's Nullstellensatz (the
algebraic one). It translates to the version given above by the OP.

I don't think so.
The argument I used is much simpler than the Nullstellensatz.
If you think they are equivalent can you explain
how to deduce the Nullstellensatz from the result I gave?

I didn't say they are equivalent (but I think they are), but here is a
sketch of proof of implication you wanted me to give:

You said my argument used another version of the Nullstellensatz,
which suggests to me that the two versions are equivalent.
Incidentally, you said my version was the "algebraic" one.
How do you describe the other version?

Let I be an ideal of polynomial ring A=k[X1,...,Xn] of the algebraically
closed field k. Let V(I) the variety of zeros of all polynomials in I
and Z(V(I)) the ideal of all polynomials in A vanishing on V(I). We
claim that rad(I)=Z(V(I)) where rad(I) denotes the radical of I.

It is easy to see that rad(I) is contained in Z(V(I)). For the converse
inclusion we take f not in rad(I) and show that f is not in Z(V(I)):
Since rad(I) is the intersection of all prime ideals in R containing I,
there is a prime P not containing f. Then the image ff of f in B=A/P is
not zero. Let M be a maximal ideal in the finitely generated k-algebra
C=B[X]/(ff.X-1). By your version of Hilbert's Nullstellensatz, the
homomorphism k -> C/M is an isomorphism, since the field C/M is a
finitely generated k-algebra. Let x1,...,xn in k be the corresponding
images of X1,...,Xn in k via C/M. Then, by construction, x=(x1,...xn) in
k^n is in V(I) with f(x)=/=0.

Note that we are dealing with so called weak and strong versions of
Hilbert's Nullstellensatz here.

I've only heard of one version of the theorem.

The fact is, it is much easier to prove that a maximal ideal
is the set of polynomials vanishing at a point
than it is to prove the Nullstellensatz (as I understand it).
In particular your proof of the "equivalence" above
(which uses a different definition of rad(I))
seems to me much more difficult than the proof of the maximal ideal result.

However, I guess this is only a matter of opinion.




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Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
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s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
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Relevant Pages

  • Re: hilberts nullstellensatz
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  • Re: hilberts nullstellensatz
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