Intersection of maximal ideals of $\mathbb{Q}[x]/(f(x))$ are nilpotent elements

Question

I'm studying for an algebra qualifying exam (here is the practice exam to prove its not a HW problem) and I was trying to do problem 5(c). Here it is again:

Let $R=\mathbb{Q}[x]/(f(x))$ where $f\in \mathbb{Q}[x]$ is a non-constant polynomial. Show that the intersection of all the maximal ideals of $R$ is exactly the nilpotent elements in $R$.

Here is my attempt at the solution: Let $M$ be a maximal ideal of $R$. I know that $\mathbb{Q}[x]/(f(x))/M=\mathbb{Q}[x]/(f(x),M)$ and that $\mathbb{Q}[x]$ is a PID so $(f(x), M)=(p(x))$ for some $p\in \mathbb{Q}[x]$. But since $M$ is maximal, $\mathbb{Q}[x]/(f(x),M)=\mathbb{Q}[x]/(p(x))$ should be a field and therefore $p(x)=x-a$ for some $a\in \mathbb{Q}$. Therefore $\cap_{M \text{ maximal}} M=0$ which is clearly a nilpotent element.

I am not sure if what I have done is correct and if it is, I am unsure how to prove the other inclusion (that there are no non-trivial nilpotent elements). Thanks in advance for the help!

Answer 1

• If $a$ is nilpotent then it is in every maximal ideal.

• Factorize $f\in \Bbb{Q}[x]$ in irreducibles. Deduce the maximal ideals of $\Bbb{Q}[x]/(f)$. If $a$ is in every maximal ideal then it is in $\bigcap_j \mathfrak{m}_j=\prod_j \mathfrak{m}_j$ and $a^{\deg(f)}$ is in $(f)= 0$.

Answer 2

I have to recall that

1. the Nilradical (i.e. the ideal of nilpotent elements) of a unital commutative ring $R$ is the intersection of all prime ideal of $R$

2. there is a bijection between maximal ideals of $R/I$ and maximal ideals of $R$ containing $I$

Now, $\mathbb{Q}[x]$ is a PID and then all non zero prime ideals are maximal ideals.

It turns out that the prime ideals in the quotient are maximal and then the nilradical concides to the intersection of the maximal ideals (that usually is called the Jacobson Radical)