Thread: Commutative Algebra - Going-up theorem

1. Commutative Algebra - Going-up theorem

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

1. Let $\mathfrak{q}$ be an ideal of B which is maximal subject to $\mathfrak{q}\cap A\subseteq \mathfrak{p}$. Show that
- $\mathfrak{q}$ is a maximal ideal of B
- $\mathfrak{q}\cap S=\emptyset$, where $S=A\setminus \mathfrak{p}$ (solved)

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

2. Let $\mathfrak{q}'\subsetneq \mathfrak{q}$ be ideals of B such that $\mathfrak{q}'\cap A=\mathfrak{p}=\mathfrak{q}\cap A$. Show that
- $\mathfrak{q}'$ is NOT a prime ideal of B [HINT: Consider integrality in $B/\mathfrak{q}'$].

3. Suppose that $\mathfrak{q}$ is a maximal ideal of B, such that $\mathfrak{q}\cap A\subseteq\mathfrak{p}$. Show that
- $\mathfrak{q}\cap A=\mathfrak{p}$. (solved)

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

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.

2. 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 $\mathfrak{p}$ be a prime ideal of A

1. Let $\mathfrak{q}$ be an ideal of B which is maximal subject to $\mathfrak{q}\cap A\subseteq \mathfrak{p}$. Show that
- $\mathfrak{q}$ is a maximal ideal of B
- $\mathfrak{q}\cap S=\emptyset$, where $S=A\setminus \mathfrak{p}$ (solved)

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

2. Let $\mathfrak{q}'\subsetneq \mathfrak{q}$ be ideals of B such that $\mathfrak{q}'\cap A=\mathfrak{p}=\mathfrak{q}\cap A$. Show that
- $\mathfrak{q}'$ is NOT a prime ideal of B [HINT: Consider integrality in $B/\mathfrak{q}'$].

3. Suppose that $\mathfrak{q}$ is a maximal ideal of B, such that $\mathfrak{q}\cap A\subseteq\mathfrak{p}$. Show that
- $\mathfrak{q}\cap A=\mathfrak{p}$. (solved)

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

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.
in 1 we can only say that q is a prime (not maximal) ideal of B. check the question again!

in 2, suppose that $q'$ is prime. the map $f:A/p \rightarrow B/q'$ defined by $f(a+p)=a+q'$ is a well-defined injective ring homomorphism and so we may assume that $p=q'=\{0\}, \ q \neq \{0\},$ and $B$ is an integral domain. now let $0 \neq x \in q.$ then $x^n + a_1x^{n-1} + \cdots + a_n=0,$ for some $a_i \in A$ and $n \in \mathbb{N}$. we'll assume that $n$ is as small as possible. clearly $a_n \in q \cap A = \{0\}.$ thus $x(x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1})=0,$ which gives us $x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1}=0$, because $B$ is a domain and $x \neq 0.$ that contradicts the minimality of $n$ and we're done.

3. Originally Posted by NonCommAlg
in 1 we can only say that q is a prime (not maximal) ideal of B. check the question again!

in 2, suppose that $q'$ is prime. the map $f:A/p \rightarrow B/q'$ defined by $f(a+p)=a+q'$ is a well-defined injective ring homomorphism and so we may assume that $p=q'=\{0\}, \ q \neq \{0\},$ and $B$ is an integral domain. now let $0 \neq x \in q.$ then $x^n + a_1x^{n-1} + \cdots + a_n=0,$ for some $a_i \in A$ and $n \in \mathbb{N}$. we'll assume that $n$ is as small as possible. clearly $a_n \in q \cap A = \{0\}.$ thus $x(x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1})=0,$ which gives us $x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1}=0$, because $B$ is a domain and $x \neq 0.$ that contradicts the minimality of $n$ and we're done.
Why may we assume $p=q'=\{0\},\; q\neq \{0\}, B$ an integral domain? Could you give your proof without making this reduction so I can try to see why you made it? Thanks heaps.

4. Originally Posted by AtticusRyan
Why may we assume $p=q'=\{0\},\; q\neq \{0\}, B$ an integral domain? Could you give your proof without making this reduction so I can try to see why you made it? Thanks heaps.
in my solution i replaced $A,B$ with $A/p$ and $B/q'$ respectively. then $p,q'$ in $A,B$ become $\{0\}$ in $A/p$ and $B/q'.$ also the condition $q' \subset q$ in $B$ becomes $q \neq \{0\}$ in $B/q'.$

5. For 3:

The solution I had was only partial. Assume that $\mathfrak{q}\cap A\subsetneq \mathfrak{p}$. Then it's not too hard to show that $Ba \cap A\subset\mathfrak{p}$ for all $a\in \mathfrak{p}\setminus (\mathfrak{q}\cap A)$.

How do we then use this to derive a contradiction? Consider the ideal $\mathfrak{q}+Ba\subsetneq\mathfrak{q}$. Then since $\mathfrak{q}$ is maximal with respect to the given property we must have $(\mathfrak{q}+Ba)\cap A\subsetneq\mathfrak{p}$. We want to use this to derive a contradiction (I'm thinking the nonsense $\mathfrak{p}\subsetneq \mathfrak{p}$).

Is it true in this case that $(\mathfrak{q}+Ba)\cap A=\mathfrak{q}\cap A+Ba\cap A$?

And then, having done this, how do we put them all together to get 4.?