Re: commuative algebra ?

Fedor wrote:
yep I realized that too, and I think that this is because it uses Krull
theorem that says that every element belongs to a maximal ideal, and
that theorem only works with commutative rings ?

Below is my try at answering your original question
about the Gelfand map in case A is a
non-commutative Banach algebra ....

The text I'm using as a reference on Banach algebras is
Ronald Douglas' "Banach Algebra Techniques in Operator Theory".

If we have a Banach algebra A with unity 1, 1 and 0
are different because of the axiom || 1 || = 1.

To get a maximal two-sided ideal from a non-zero
character phi of A, it suffices to take the kernel of phi.
(cf. Proposition 2.33 of Douglas).

In the other direction, from a (proper) maximal two-sided
ideal M of A, we would like to construct a character phi
such that M = ker(phi).

If M is maximal, M is closed (proof omitted).
The problem that I see is in trying to get
A/M to be a division algebra (when A is non-commutative).

If A is commutative, then M is maximal ideal
in the usual sense and A/M is a division algebra, (*)
a field even. This is then used to get the
wanted character phi, in Proposition 2.33.

[It's instructive to consult the proof of (*),
e.g. Niederreiter's "Introduction to finite fields"]

If a is not in M, how to get an inverse for a (mod M)?
In the non-commutative case, we can try to enlarge
M to a bigger ideal:
J = { s.a.r + m : s in A, r in A, m in M}.
Then J properly contains M, and also is both a
left-ideal and a right-ideal of A. So
by the maximality of M, J = A.

So, for some x, y, z, with x,y in A and z in M,
x.a.y + z = 1.

But this doesn't give us the desired inverse of a (mod M).

So, if A is a non-commutative Banach algebra and M
is a maximal two-sided ideal of A, A/M is just
some Banach algebra, but not necessarily a
division algebra and the argument used by
Douglas in his Proposition 2.33 to get a character
phi with M = ker(phi) doesn't seem to work without
the commutativity of A assumption.

David Bernier
.