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Two Questions on ring homomrphisms and ideals

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Two Questions on ring homomrphisms and ideals

Hello,

Let R,S be commutative rings with unit, f: R -> S be a ring homomorphism. Then it can be shown that the inverse image of any prime ideal I in S is again prime ideal in R. However, I was told that this result fails when replacing prime ideals with maximal ideals, but I could not find a counterexample by myself. So could someone provide an example where I is maximal ideal in S, but f^(-1) (I) is not a maximal ideal in R?

Secondly, what is the most general condition on f so that the image f(I') of any prime ideal I' in R is again a prime ideal in S?


Hi Joking,
Excuse moi, but I didn't understand your reply.
1.- What do you want with R = z? An automorphism? f: Z --> Z.
2.- f(x) = (x,x) is wrong because the range of f is in Z, the ring of integers.
perucho

Oh, there's a mistake. It should be "f:R->S", i.e. f:Z->ZxQ.

Thanks for noticing. :)

joking

No friend, thanks to you. I have learned from you a few things.
perucho

> So could someone provide an example where I is maximal ideal in S,
> but f^(-1) (I) is not a maximal ideal in R?

Let Z be the ring of integers and and let Q be the ring of rationals. Consider the product ring S=ZxQ and let R=Z. Let f:R->Z be such that f(x)=(x,x). Now it is clear that Zx0 is a maximal ideal in S, but f^{-1}(Zx0)=0 is not maximal in R.

> Secondly, what is the most general condition on f so that the image
> f(I') of any prime ideal I' in R is again a prime ideal in S?

I don't think there is a useful one (other then being an isomorphism). Even for a quotient map q:R->R/I we have the correspondence between (prime) ideals in R/I and (prime) ideals in R containing I. For example if R=Z, I=2Z and I'=3Z, then q(I') is a whole ring which is not prime (by definition prime ideals are proper subsets).

joking

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