Let $f : R \to R$ be an injective unitary endomorphism of a commutative ring with 1. Let $I$ be an ideal of $R$. I have several related questions concerning the image of $I$ under $f$:

1) Under which condition is $f(I)$ already an ideal of $R$ (i.e. when do we have $f(I) = f(I) R$)?

Certainly this is true if $f$ is surjective (see this question for instance). Is this condition also necessary?

2) Suppose furthermore that $I$ is prime. Under which condition is $f(I)$ a prime ideal?

A partial answer is given for instance here (the image of prime ideal under a surjective homomorphism is a prime ideal). But I am particularly interested in the situation where the homomorphism is not surjective.

3) Suppose that $I$ is a minimal prime ideal and that $f^{-1}(I) = I$. Under which condition does $f(I) = I$ (or at least $f(I) R = I$) hold?

For (possibly non-minimal) prime ideals this does not hold. But I cannot think of a counterexample of a minimal prime ideal.

One always has $f(I) \subseteq I$.

4) Is there a geometric interpretation of these conditions? (The interpretation of the extension of an ideal was discussed here).

  • $\begingroup$ The questions are too general to have good answers and indicate that you haven't looked at a single example before asking. Better do that and then ask a better or more focussed question. It also seems to me that you actually are interested in something else, and this seems to be only the preparation for it. $\endgroup$ – Martin Brandenburg Nov 23 '14 at 23:07

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