# Lying Over and Going Up

If you haven’t heard the terms in the title of this post, then you are probably bracing yourself for this to be some weird post on innuendos or something. Let’s first do some motivation (something I’m not often good at…remember that Jacobson radical series of posts? What is that even used for? Maybe at a later date we’ll return to such questions). We can do ring extensions just as we do field extensions, but they tend to be messier for obvious reasons. So we want some sort of property that will force an extension to be with respect to prime ideals. Two such properties are “lying over” and “going up.”

Let $R^*/R$ be a ring extension. Then we say it satisfies “lying over” if for every prime ideal $\mathfrak{p}\subset R$ in the base, there is a prime ideal $\mathfrak{p}^*\subset R^*$ in the extension such that $\mathfrak{p}^*\cap R=\mathfrak{p}$. We say that $R^*/R$ satisfies “going up” if in the base ring $\mathfrak{p}\subset\mathfrak{q}$ are prime ideals, and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then there is a prime ideal $\mathfrak{q}^*\supset \mathfrak{p}^*$ which lies over $\mathfrak{q}$. (Remember that Spec is a contravariant functor).

Note that if we are lucky a whole bunch of posts of mine will finally be tied together and this was completely unplanned (spec, primality, localization, even *gasp* the Jacobson radical). First, let’s lay down a Lemma we will need:

Let $R^*$ be an integral extension of R. Then
i) If $\mathfrak{p}$ a prime ideal of R and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then $R^*/\frak{p}^*$ is integral over $R/\mathfrak{p}$.
ii) If $S\subset R$, then $S^{-1}R^*$ is integral over $S^{-1}R$.

Proof: By the second iso theorem $R/\frak{p}=R/(\frak{p}^*\cap R)\cong (R+\frak{p}^*)/\frak{p}^*\subset R^*/\frak{p}^*$, so we can consider $R/\frak{p}$ as a subring of $R^*/\frak{p}^*$. Take any element $a+\frak{p}^*\in R^*/\frak{p}^*$. By integrality there is an equation $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$ with the $r_i\in R$. Now just take everything $\mod \frak{p}^*$ to get that $a+\frak{p}^*$ integral over $R/\frak{p}$. This yields part (i).

For part (ii), let $a^*\in S^{-1}R^*$, then $a^*=a/b$, where $a\in R^*$ and $b\in\overline{S}$. By integrality again we have that $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$, so we multiply through by $1/b^n$ in the ring of quotients to get $(a/b)^n+(r_{n-1}/b)(a/b)^{n-1}+\cdots +r_0/b^n=0$. Thus $a/b$ is integral over $S^{-1}R$.

I’ll do two quick results from here that will hopefully put us in a place to tackle the two big results of Cohen and Seidenberg next time.

First: If $R^*/R$ is an integral ring extension, then $R^*$ is a field if and only if $R$ is a field. If you want to prove this, there are no new techniques from what was done above, but you won’t explicitly use the above result, so I won’t go through it.

Second: If $R^*/R$ is an integral ring extension, ten if $\frak{p}$ is a prime ideal in R and $\frak{p}^*$ is a prime ideal lying over $\frak{p}$, then $\frak{p}$ is maximal if and only if $\frak{p}^*$ is maximal.

Proof: By part (i) of above, $R^*/\frak{p}^*$ is integral over $R/\frak{p}$ and so as a corollary to “First” we have one is a field if and only if the other is. This is precisely the statement that $\frak{p}$ is maximal iff $\frak{p}^*$ is maximal.