Re: hilbert's nullstellensatz

Hagen 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.

The essential point in your statement is that if in
the field extension L|K the field L is a finitely
generated K-algebra, then L|K is algebraic.
This fact is sometimes called the >weak Hilbert Null-
stellensatz< - see for example Zariski-Samuel, Commutative
Algebra. It is equivalent to Hilbert's Nullstellensatz.

Sorry to nit-pick, but what exactly do you mean
by saying that two propositions are "equivalent"?
Logically, I suppose, any two correct propositions are equivalent.

In this case, one proposition - the one I used -
seems to me more or less trivial,
while the other one - the nullstellensatz -
is rather difficult.

Doubtless there is a chain of reasoning
which goes through the first proposition
and winds up with the second,
but that hardly shows the propositions are equivalent.

The argument you gave assumed, among other results,
that the radical of an ideal I is the intersection
or the prime ideals containing I,
which is not at all obvious.


--
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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