# Maximal ideals of polynomial ring containing a fixed element

We know that for a field $k$ and $f\in k [x]$, the only maximal ideals of $k [x]$ containing $f$ are the ideals generated by prime factors of $f$. Now, I want to know that if $R$ is an arbitrary commutative ring with identity, is there any description for maximal ideals containing $f$?Or, can we find $\cap_{f\in m\in Max (R [x])}m$?

• Consider the case when $f$ has degree zero. Then you see that in your question you can replace $R[x]$ be an arbitrary ring. So then you are asking for the max-Spectrum of the ring $R/fR$. That is, your question is whther there is a description of Max-Spec(R) for an arbitrary ring R. – user1688 Dec 18 '15 at 17:13
• $f$ is not a constant term in general, so $max (R [x]/f)\not=max (R)$ in general case. – Tedi Fox Dec 18 '15 at 18:48
• Right -- so your question is harder than an impossibly general question. The intersection of the max ideals of A containing f "is" the intersection of the max ideals of A/(f) which is the Jacobson Radical of A/(f). If you want more then you can look up properties of the Jacobson radical. – eric Dec 18 '15 at 20:49

## 1 Answer

Let's think about this...

To get out of the issue raised by Corbennick, let us assume that $f\in R[x]\setminus R$.

Next let $\mathfrak m\subset R$ be an ideal and consider $I=\mathfrak m+(f)\subset R[x]$ and observe that $R[x]/I\simeq (R/\mathfrak m)[x]/\left(\bar f\right)$, so if $\mathfrak m$ is a maximal ideal in $R$, then the maximal ideals in $R[x]$ containing $\mathfrak m$ and $f$ are exactly those corresponding to maximal ideals of $(R/\mathfrak m)[x]$ containing $f$, so this is the case covered in the question.

Next let's consider a maximal ideal $\mathfrak M\subset R[x]$ and let $\mathfrak m=\mathfrak M\cap R$. In order to get some traction, we'd like this to be maximal, so let's assume that $R$ is a Jacobson ring. (Finitely generated algebras over a Jacobson ring are also Jacobson, and so are fields, and $\mathbb Z$, so many of the most obvious rings are such.) In that case $\mathfrak m$ is also a maximal ideal and we're winning. Obviously, if $\mathfrak M$ contains $f$, then it also contains $\mathfrak m+ (f)$, so we're in the situation described above.

So we can conclude the following:

Claim: Let $R$ be a Jacobson ring and $f\in R[x]\setminus R$. Then the maximal ideals $\mathfrak M$ of $R[x]$ containing $f$ are exactly those for which the image of $\mathfrak M$ in $\left(R/(\mathfrak M\cap R)\right)[x]$ is generated by a prime factor of the image of $f$ in that ring.

Note:

1. I don't think one can say more than this.
2. One can get a similar description without the Jacobson assumption, assuming instead that $R$ has finite Krull dimension and using the above process to give a recursive description of these ideals based on the dimension.