Thread: Homomorphisms, fields and maximal ideals

Hi,

I was wondering if someone would be able to give me a hint on this problem I am working on. The problem sates: "If f: A-->B is a homomorphism from A onto B, and B is a field, then the kernel of f is a maximal ideal. "

I know that the kernel of f are all the elements of A that are carried to the "0" element of B. I also know that an ideal means that it must be "closed with respect to addition, negatives and B absorbs products of A".

So, I think that if I take an element x in A, and I apply f to it that means that the element x I started off with will be mapped to "0" which is in B. So the kern f = {f : f (x) = 0}.
Now I can show the following things are true:
1. Closes with respect to addition. f (x₁) = 0 , and f(x₂)=0, then f(x₁+x₂) = f (x₁)+f(x₂)=0.
2. Closed with respect to negatives. f(x) = 0, f(-x) = 0 = - f(x) = 0 = -0 = 0.
3. B absorbs products of A. If f(x) = 0, then f(x*y) = f(x) * f(y) = 0 * f(y) = 0.

I believe what I showed here is that the kernel of f is an ideal, but I don't know how to show it is a maximal ideal.

There was a hint given in the problem. It states that it may help if I : "Show that a field can have no non trivial ideals". I have proven that. But I don't know how that is going to help me with my proof. I also realize I didn't take into consideration that B is a field. Am I missing a step because of that as well?

Any help is greatly appreciated.

Thank you for your time!

I mean, I think that you should know the proposition

Theorem: Let be a commutative ring, and an ideal of . Then, is maximal if and only if is a field

This follows immediately from the lattice isomorphism theorem (fourth) because you know that the ideals of sitting above are in bijective correspondence with the ideals of , and as you have noted, there aren't a lot of those.

Now, apply this theorem to your problem, recalling another one of the isomorphism theorems.

you CAN prove this directly. suppose A is a commutative ring with identity, and that f:A→B is a surjective ring homomorphism onto a field B.

let J be any ideal of R containing ker(f) but not equal to ker(f). pick a in J - ker(f).

then f(a) ≠ 0, since a is not in ker(f), and since B is a field there exists b in B with f(a)b = 1_B.

since f is surjective, there exists x in A with f(x) = b.

by the fundamental isomorphism theorem, B is isomorphic to A/ker(f):

r + ker(f) ↔ f(r) for any r in A.

now consider ax in A. f(ax) = f(a)(f(x) = f(a)b = 1_B = f(1_A) ↔ 1_A + ker(f).

thus ax - 1_A is in ker(f). so ax - 1_A = k, for some k in ker(f).

this, in turn means: 1_A = ax - k. now k is in ker(f), which is contained in J, so k is in J. and ax is in J, since J is an ideal, x is in A, and a is in J.

thus 1_A is in J, so J = R. thus ker(f) is maximal.

Thank you Deveno,

I've been reading and going through my notes and looking for the Theorem Drexel28 mentioned, but I can't find anything in my notes about it. Since my professor assigned the problem to me, I figured there must be a different approach to the problem.

Thank you to both of you for the helpful information you both gave me.

one consequence of the fundamental isomorphism theorem is that there is a lattice-isomorphism (basically a order-preserving bijective set map, where the (partial) order on ideals is set inclusion) between the lattice of ideals of R that contain ker(f) and the ideals of S.

this isomorphism is:

J<-->f(J), where J contains ker(f) (f(J) is always an ideal of f(A) for any ring A, and since f is surjective on f(A), if K is an ideal of f(A), f^-1(K) is an ideal of A that contains ker(f). this is easy to prove, and instructive).

it turns out that this is true in "any structure that has kernels", including: groups, abelian groups, rings, modules, and vector spaces (where kernels are, respectively: normal subgroups, subgroups, ideals, submodules, and null spaces).

this is why homomorphisms are so important: given a homomorphism f:A-->B the sub-structure f(A) of B is usually "simpler" that that of A, but can give us a partial picture of the structure of A.

in groups, for example, if we have a homomorphism f:G-->G', and f(G) happens to be abelian (and the subgroups of an abelian group are "nicer" than subgroups in general, we get a (partial) list of normal subgroups of G for free: the pre-image of any subgroup of G'). in rings, one way we can prove a subset I of a ring R is an ideal, is to show that we have a ring-homomorphism f:R-->S, with f(I) an ideal in f(R). this can be useful if S is a "nice" ring, like Z, or F[x], where all the ideals are principal, or if S is finite, so there's only so many possibilities in the first place. rule #1 of algebra: good theorems let us be lazy.