Let $\phi: R \rightarrow S$ be a homomorphism of Noetherian rings with prime spectra $X, Y$, and suppose the contraction map $\phi^{\ast}: Y \rightarrow X$ is surjective. I'm trying to show that for any ideal $I$ of $R$, the height of $IS$ (that is, $\phi(I)S$) is $\leq$ the height of $I$.
It's obvious in the special case that $\phi$ satisfies incomparability: that is, if $Q_1 \subset Q_2$ are primes of $S$, then $\phi^{-1} Q_1 \subset \phi^{-1} Q_2$ (proper inclusion). It follows from here that if $Q$ is a prime of $S$, then $$\textrm{ht } Q \leq \textrm{ht } \phi^{-1}Q$$ So, choose a minimal prime ideal $P$ of $I$ whose height is that of $I$, let $Q$ be a prime of $S$ contracting to $P$. Then $IS \subseteq Q$, so $$\textrm{ht } IS \leq \textrm{ht } Q \leq \textrm{ht } \phi^{-1}Q = \textrm{ht } I $$
$\blacksquare$
If you don't have incomparability, you can have $\textrm{ht } Q > \textrm{ht } \phi^{-1}Q$, even if lying over holds. For example, let $A$ be a ring of dimension $> 1$ containing a field $k$. Then the contraction of any height one prime of $A$ under the inclusion map $k \rightarrow A$ has height zero.
Let $J$ be the kernel of $\phi$. I had thought to split $\phi$ up as a surjection $R \rightarrow R/J$ followed by an injection $\overline{\phi}: R/J \rightarrow S$. Now the injection $\overline{\phi}$ inherits the lying over property, and extension of ideals is transitive. So it suffices to consider the cases where $\phi$ is surjective and injective separately.
In the surjective case, the contraction map satisfies incomparability, so we can apply the argument I mentioned. Now I'm left with proving the result when $\phi$ is injective and I haven't gotten too far. I thought Krull's altitude theorem might come in somewhere, since the minimal number of generators of $IS$ is $\leq$ the minimal number of generators of $I$.
I would appreciate a small hint (please don't give me the answer).
