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 ?

Thanks!

Greg

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