How do I take the contraction of an ideal which is not in the image of the given morphism?

Question

If I have a morphism of rings $\phi: A \to B$ which is not surjective, how should I take the preimage of an ideal not contained in the image of $\phi$?

Answer 1

Let $\phi:A\to B$ be any function between sets and $I$ any subset of $B$. Then $\phi^{-1}(I)$ is defined as the set of all $a\in A$ such that $\phi (a) \in I$.

In your case you just need to show that if $A$ and $B$ are rings, $I$ an ideal of $B$, and $\phi$ a ring homomorphism, that $\phi^{-1}(I)$ is an ideal of $A$. Then the definition makes sense.