# Commutative Algebra - Going-up theorem

• Jul 21st 2010, 01:05 AM
AtticusRyan
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 $\displaystyle \mathfrak{p}$ 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.
• Jul 21st 2010, 03:40 AM
NonCommAlg
Quote:

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.

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

in 2, suppose that $\displaystyle q'$ is prime. the map $\displaystyle f:A/p \rightarrow B/q'$ defined by $\displaystyle f(a+p)=a+q'$ is a well-defined injective ring homomorphism and so we may assume that $\displaystyle p=q'=\{0\}, \ q \neq \{0\},$ and $\displaystyle B$ is an integral domain. now let $\displaystyle 0 \neq x \in q.$ then $\displaystyle x^n + a_1x^{n-1} + \cdots + a_n=0,$ for some $\displaystyle a_i \in A$ and $\displaystyle n \in \mathbb{N}$. we'll assume that $\displaystyle n$ is as small as possible. clearly $\displaystyle a_n \in q \cap A = \{0\}.$ thus $\displaystyle x(x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1})=0,$ which gives us $\displaystyle x^{n-1} + a_1x^{n-2} + \cdots + a_{n-1}=0$, because $\displaystyle B$ is a domain and $\displaystyle x \neq 0.$ that contradicts the minimality of $\displaystyle n$ and we're done.
• Jul 21st 2010, 03:55 AM
AtticusRyan
Quote:

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

Why may we assume $\displaystyle p=q'=\{0\},\; q\neq \{0\}, B$ an integral domain? Could you give your proof without making this reduction so I can try to see why you made it? Thanks heaps.
• Jul 21st 2010, 04:45 AM
NonCommAlg
Quote:

Originally Posted by AtticusRyan
Why may we assume $\displaystyle p=q'=\{0\},\; q\neq \{0\}, B$ 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 $\displaystyle A,B$ with $\displaystyle A/p$ and $\displaystyle B/q'$ respectively. then $\displaystyle p,q'$ in $\displaystyle A,B$ become $\displaystyle \{0\}$ in $\displaystyle A/p$ and $\displaystyle B/q'.$ also the condition $\displaystyle q' \subset q$ in $\displaystyle B$ becomes $\displaystyle q \neq \{0\}$ in $\displaystyle B/q'.$
• Jul 21st 2010, 05:32 AM
AtticusRyan
For 3:

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

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

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

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