Is it true that the intersection of a maximal ideal in $A[x]$ with $A$ is a maximal ideal in $A$?
Let's say A is Noetherian. I would be surprised if it isn't true but somehow I can't seem to show it. Any help or tip will be appreciated. Thanks!



Maximal ideal in polynomial ring
No, it's not true in general. E.g. the pricipal ideal generated by $px - 1$ is maximal
in $\mathbb Z\_p[x]$ (for any prime $p$); the quotient $\mathbb Z\_p[x]/(p x - 1)$ is precisely the field $\mathbb Q\_p$. However, the intersection of this ideal with $\mathbb Z\_p$ is equal to the zero
ideal, which is not maximal.


If the ring $A$ is Jacobson, then the result you want is true. (For in this case, if $\mathfrak m$ is a maximal ideal in $A[x]$, then
$A[x]/\mathfrak m$ is a finite type $A$-algebra which is a field, therefore is finite over $A$
by the Jacobson hypothesis, and so the image of $A$ in $A[x]/\mathfrak m$ (which is equal to
$A/A\cap \mathfrak m$) is itself a field, and so $A\cap \mathfrak m$ is maximal in $A$.)


