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



Maximal ideals of $A[X],$ $A$ PID
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 $q \neq p\_1$, $p\_2$, ..., $p\_n$.




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I modeled this proof on [Zariski's proof of the Nullstellansatz](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183510605); both arguments work by showing that fields don't want to be finitely generated modules over polynomial rings.


