Re: tensor product twister

On 23.11.2006 22:46, magya_bloom@xxxxxxxxx wrote:
Jannick Asmus wrote:
On 23.11.2006 18:49, magya_bloom@xxxxxxxxx wrote:
magya_bloom@xxxxxxxxx wrote:
Jannick Asmus wrote:
On 23.11.2006 14:59, magya_bloom@xxxxxxxxx wrote:

not sure what "inverse image of the natural ring homomorphism" means in
this context.
Sorry for being not very clear. I meant "preimage": If f: G -> H is a
map of sets, then the preimage of a subset H' of H is f^-1(H') = { g in
G | f(g) in H' }.
I understood that part. I just don't know what that subset H' is. When
you say preimage of a natural homomorphism, which sub-algebra of A*B
are we pulling back?
of course we are pulling back M. Sorry about my stupid question.
Ok, my fault that I was not noting what we are taking the preimage of. :-(

But why do we need nullstellensatz to conclude that m1 is maximal?
We know that m1 is prime, as preimages of prime ideals under unitary
ring homomorphisms are prime again. But how do you conclude that m1 is
maximal?

that is exactly what I have trouble with. I know that we can define
some non-zero algebra homomorphism form A-> F (where F is our alg
closed field), for every a in A (since they are non-nilpotent),

such that the image of a under A -> F is non-zero,

but from that is there a direct conclusion of maximality? Is m1 an
intersection of kernels of these maps? lost.

Use, e.g., the version given in
http://groups.google.com/group/sci.math/msg/5eba8d12f1b78dc9. It is part
of a thread on different versions of the Hilbert Nullstellensatz one
should be familiar with in algebraic geometry.
Cf. for a list of different versions of Hilbert's theorem:
http://groups.google.com/group/sci.math/msg/60c967c1371c747c.






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