Maximal ideals of $A[X],$ $A$ PID 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
 A: What about $A=\Bbb Z_{(p)}$ and $m=(pX-1)$?
A: 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.
