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).
