Re: tensor product twister


Jannick Asmus wrote:
On 23.11.2006 02:12, magya_bloom@xxxxxxxxx wrote:
Jannick Asmus wrote:
On 10.11.2006 06:35, oferlock@xxxxxxxxx wrote:
Jannick Asmus wrote:
On 09.11.2006 05:08, oferlock@xxxxxxxxx wrote:
If A and B are algebras (over an algebraically closed field F) with no
nilpotent elements, can we say that A*B (tensor product) has no
nilpotent elements? I thought that should be easy, but can not say that
since linear combination of tensors can still be nilpotent. Is there a
theorem that rules that out?
Reduce to the case where A and B are finitely generated F-algebras. Then
use the assumption on F and the Hilbert Nullstellensatz that every
maximal ideal M of A*B has the unique representation m1*B+A*m2 where m1
and m2 are maximal ideals in A and B, respectively. Now try to deduce
the claim from the fact that in every finitely generated algebra over a
field the nilradical is the intersection of all maximal ideals (BTW:
This is some version of the Hilbert Nullstellensatz, too).
it's very interesting approach, and works, but i think you might have
to replace maximal with prime.
Right, this approach seems to be easier. Then it follows from the fact
that if K and L are field extensions of F the tensor product K*L (over
F) is reduced. The implication still holds if F is assumed to be perfect.


I don't see why the following is true:

use the assumption on F and the Hilbert Nullstellensatz that every
maximal ideal M of A*B has the unique representation m1*B+A*m2 where m1
and m2 are maximal ideals in A and B, respectively.

can someone enlighten?

Let A and B be finitely generated F-algebras and M a maximal ideal of
the tensor product A*B (over F).

Let m1 the inverse image of the natural ring homomorphism (of finitely
generated F-algebras!) A -> A*B. Then, by Hilbert's Nullstellensatz, m1
is maximal in A. In particular the induced ring homomorphism F -> A/m1
(*1) is an isomorphism, since F is algebraically closed (Hilbert's
Nullstellensatz again). Define m2 likewise wrt B -> A*B.

Now use (*1) & (*2) to deduce that M'=m1*B+A*m2 is maximal by showing
that F -> A*B/M' is an isomorphism. Since M' is contained in M, we have
M' = M, by maximality of M'.

HTH.

J.

not sure what "inverse image of the natural ring homomorphism" means in
this context. Wouldn't that be the whole algebra A? Or is it a
submodule of A*B whose inverse image is m1?

.

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