# Pre-image of a ring homomorphism

Let $\phi:R \rightarrow S$ be a ring homomorphism and let $J$ be an ideal in $S$, then it is quite easy to prove that $\phi^{-1}(J)$ is an ideal in $R$. But can someone help me with proving the opposite? i.e. if $\phi^{-1}(J)$ is an ideal in $R$ then $J$ is an ideal in $S$. (of course, it could be that the statement I am making is false in which case, can someone explain why that is so?)

• It's false, e.g., when $\phi$ is an inclusion map and $J$ is the union of an ideal of $R$ with a random subset of $S \setminus R$. – darij grinberg Apr 30 '15 at 14:56

The statement you want to prove is wrong, think for example of $\phi \colon \mathbf Z \to \mathbf Q$ the inklusion and $J = 2\mathbf Z$. Then $\phi^{-1}(J) = 2\mathbf Z \subseteq \mathbf Z$ is an ideal, but $J = 2\mathbf Z\subseteq \mathbf Q$ is not, as $\mathbf Q$ is a field and $2\mathbf Z \ne 0,\mathbf Q$.

The converse is not true.

Let $\phi:\mathbb{Z}\rightarrow \mathbb{Q}$ be the inclusion homomorphism. Then $\phi^{-1}(\mathbb{Z})= \mathbb{Z}$ is an ideal in $\mathbb{Z}$ but $\mathbb{Z}$ is not an ideal in $\mathbb{Q}$.

Ideals are kernels of ring homomorphisms. Viewing them like this might help understand why this only works one way: the pre-image along map $g$ of a kernel of map $f$ is the kernel of the composition $fg$ but knowing that the pre-image of a subset $X$ is the kernel of some homomorphism $g$ does not help to get a homomorphism $f$ that has $X$ as kernel.