1
$\begingroup$

Hi everyone,

let $A$ be a PID, let $\mathfrak{m}$ be a maximal ideal of $A[X]$.

I would like to find a direct simple proof of that fact that $\mathfrak{m}\cap A\neq 0$.

For the moment, I only know to prove it using the following fact : if $I$ is a prime ideal of $A[X]$ such that $I\cap A=0$ ($A$ PID), then one shows that $I=(f)$ for some $f\in A[X]$ .

But one can see easily that $I$ is never maximal in this case.

However, it seems quite intricate. Does anyone know a direct argument ?

Of course, the final goal is to prove that $\mathfrak{m}=(\pi, f)$, where $\pi\in A$ is a prime element and $f$ is irreducible modulo $\pi$, but it follows easily, once we know that $\mathfrak{m}\cap A\neq 0$.

Thanks!

Greg

$\endgroup$
3
  • 2
    $\begingroup$ Looks like homework. $\endgroup$ Sep 4, 2012 at 11:27
  • 1
    $\begingroup$ I don't think it is, he is far from being a student. $\endgroup$
    – Lierre
    Sep 4, 2012 at 13:25
  • 3
    $\begingroup$ This raises a frequent issue: One reason that a question would be a good homework problem is that it is a good question which can be answered by standard methods. Wouldn't it be depressing if the methods we teach in commutative algebra classes could never answer any good questions? $\endgroup$ Sep 4, 2012 at 14:31

2 Answers 2

6
$\begingroup$

As Inta's example shows, the problem as stated is not true. Here is a true restatement: Suppose that $A$ is a UFD with infinitely many primes. Then any maximal ideal $\mathfrak{m}$ of $A[x]$ has nontrivial intersection with $A$.

The morally right condition on $A$ is "for every nonzero $f \in A$, there is a nonzero prime ideal $\mathfrak{p}$ with $f \not \in \mathfrak{p}$." We will see that "UFD with infinitely many primes" is an easily checked condition which implies this.

Proof: Let $L=A[x]/\mathfrak{m}$. Since $\mathfrak{m}$ is assumed maximal, $L$ is a field. Suppose for the sake of contradiction that $A$ injects into $L$. Since $L$ is a field, $A$ must then be a domain; let $K=\mathrm{Frac}(A)$. Then $K$ injects into $L$ by the universal property of fraction fields.

Observe that $\mathfrak{m}$ is not $(0)$, as $A[x]$ is not a field. So there is some polynomial $\sum a_i x^i=0$ in $\mathfrak{m}$ and we deduce that $x$ is algebraic over $K$. So $L/K$ is a finite degree extension. Let the minimal polynomial of $x$ over $k$ be $$x^n + \frac{p_1}{q_1} x^{n-1} + \cdots + \frac{p_n}{q_n} = 0$$ where $p_i$ and $q_i \in A$. So $(1,x, \ldots, x^{n-1})$ is a $K$-basis for $L$.

Let $f = q_1 q_2 \cdots q_n$ and let $B = A[f^{-1}]$. Note that $x^{n}$ is in the $B$-linear span of $(1, x, \ldots, x^{n-1})$. By induction on $m$, we similarly have that $x^m$ is in the $B$-linear span of $(1, x, \ldots, x^{n-1})$. Since monomials in $x$ span $L$ over $A$, we get that $L$ is the $B$-linear span of $(1,x,\ldots, x^{n-1})$.

At this point, you should sense a contradiction is near. Since $(1,x, \ldots, x^{n-1})$ is a $K$-basis for $L$, every element of $L$ is uniquely of the form $\sum_{i=0}^{n-1} c_i x^i$ with $c_i \in K$. On the other hand, we have also just shown that every element of $L$ is of the form $\sum_{i=0}^{n-1} b_i x^i$ with $b_i \in B \subseteq K$. The only way these statements can both be true is if $B=K$. Indeed, if you trace through Inta's example, that's exactly what happens.

So this comes down to the problem: Find a condition which guarantees that $A[f^{-1}] \neq \mathrm{Frac}(A)$ for any nonzero $f$ in $A$. If $\mathfrak{p}$ is a nonzero prime ideal of $A$ with $f \not \in \mathfrak{p}$, and $g$ is a nonzero element of $\mathfrak{p}$, I claim that $1/g \not \in A[f^{-1}]$. Proof: Assume otherwise. Then $1/g = r/f^N$ and $f^N = gr$. But $f \not \in \mathfrak{p}$, so $f^n \not \in \mathfrak{p}$, which $g \in \mathfrak{p}$ implies $rg \in \mathfrak{p}$, a contradiction.

In particular, suppose that $A$ is a UFD with infinitely many primes. Let $f=p_1 p_2 \cdots p_n$ be the unique factorization of $f$ and take $\mathfrak{p} = (q)$ for any $q \neq p_1$, $p_2$, ..., $p_n$.


I modeled this proof on Zariski's proof of the Nullstellansatz; both arguments work by showing that fields don't want to be finitely generated modules over polynomial rings.

$\endgroup$
8
$\begingroup$

What about $A=\Bbb Z_{(p)}$ and $m=(pX-1)$?

$\endgroup$
2
  • 1
    $\begingroup$ You could even take $A$ to be a field, and $m = (X-1)$! $\endgroup$ Sep 4, 2012 at 16:31
  • 1
    $\begingroup$ Inta's example has the advantage of giving an instance where a closed point gets mapped to a non-closed (in fact, generic) point, a phenomenon that might not be so intuitive at first. $\endgroup$
    – Ramsey
    Sep 4, 2012 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.