Originally Posted by

**AtticusRyan** I'm trying to prove the Going-Up theorem from Commutative Algebra using a different method to that given in the classic reference Atiyah and Macdonald. There's a couple of parts I'm having trouble with.

All rings are commutative.

- Let A be a subring of B

- Let B be integral over A

- Let $\displaystyle \mathfrak{p}$ be a prime ideal of A

1. Let $\displaystyle \mathfrak{q}$ be an ideal of B which is maximal subject to $\displaystyle \mathfrak{q}\cap A\subseteq \mathfrak{p}$. Show that

- $\displaystyle \mathfrak{q}$ is a maximal ideal of B

- $\displaystyle \mathfrak{q}\cap S=\emptyset$, where $\displaystyle S=A\setminus \mathfrak{p}$ (solved)

NOTE: Not sure if this should be "a maximal ideal of B", or "an ideal of B, maximal subject to $\displaystyle \mathfrak{q}\cap S=\emptyset$".

2. Let $\displaystyle \mathfrak{q}'\subsetneq \mathfrak{q}$ be ideals of B such that $\displaystyle \mathfrak{q}'\cap A=\mathfrak{p}=\mathfrak{q}\cap A$. Show that

- $\displaystyle \mathfrak{q}'$ is NOT a prime ideal of B [HINT: Consider integrality in $\displaystyle B/\mathfrak{q}'$].

3. Suppose that $\displaystyle \mathfrak{q}$ is a maximal ideal of B, such that $\displaystyle \mathfrak{q}\cap A\subseteq\mathfrak{p}$. Show that

- $\displaystyle \mathfrak{q}\cap A=\mathfrak{p}$. (solved)

4. Using 1-3, deduce that if $\displaystyle \mathfrak{p}\subseteq \mathfrak{p}'$ are prime ideals of A and $\displaystyle \mathfrak{q}$ is a prime ideal of B such that $\displaystyle \mathfrak{p}=A\cap \mathfrak{q}$ then there exists a prime ideal $\displaystyle \mathfrak{q}'$ of B such that

- $\displaystyle \mathfrak{q}\subseteq \mathfrak{q}'$

- $\displaystyle p'=A\cap \mathfrak{q}'$.

Any help would be greatly appreciated - I have made some progress, for example in Q1 I am fairly sure you have to suppose for a contradiction you can find an ideal $\mathfrak{r}\supsetneq\mathfrak{q}$ and then let $x\in \mathfrak{r}\setminus \mathfrak{q}$ and somehow show $(\mathfrak{q}+Bx)\cap A\subseteq \mathfrak{p}$. Not sure how to do this.

In 2 it may be helpful to know two theorems:

- If B is integral over A, q is an ideal of B, and P = A int q, then B/q is integral over A/P.

- if B is integral over A and both are integral domains, then A is a field if and only if B is a field.