This is a problem in Atiyah Macdonald, Commutative Algebra. Problem 5.22: $S$ is a subring of an integral domain $R$. $R$ is a finitely generated $S$ algebra. If the Jacobson radical of $S$ is 0, then the Jacobson radical of $R$ is 0.

I followed the hint provided in the book as follows.

The purpose is to show for each element of $R$, there is a maximal ideal of $R$ avoiding it. Pick a $0\neq v\in R$. Since we have injective maps $S\to R$ and $R\to R_v$, we embed $S$ in $R_v$ as a subring and this $R_v$ is finitely generated as well over $S$.

It is clear that there is $0\neq s\in A$ such that we can extend the ring homomorphism $\phi:S\to\Omega$ to $\tilde{\phi}:R_v\to\Omega$ where $\Omega$ is some algebraically closed field. Here $\Omega$ is chosen to be the algebraically closed field of $S/m$ where $s\not\in m$ as jacobson radical of $S$ is 0. It is clear that $v$ under the map $R_v\to\Omega$ is not zero or we will have $1$ being sent to $0$.

However, I am stuck at showing $Ker(R_v\to\Omega)\subset R_v$ is maximal ideal since this will lead to $Ker(R_v\to\Omega)\cap B$ is maximal. It is not even clear that this will be maximal. However, I can show if $S$'s Jacobson radical is trivial, then $R=S[x]$'s is trivial. It is possible that $R=S[x]/Q$ where $Q$ is some prime ideal and Jacobson radical of $R$ is the intersection of maximal ideals containing $Q$. I also noticed that quotient and localization do not necessarily commute with infinite intersection. What other options do I have here?


You have a non-trivial ring homomorphism $\tilde{\phi}: R_v \to \Omega$ with $\ker(\tilde{\phi}\mid S)=m$. Hence there are ring extensions $$k=S/m \le R_v/\ker(\tilde{\phi}) \le \Omega$$ Now, by the exercise below, $R_v/\ker(\tilde{\phi})$ is a field, i.e. $\ker(\tilde{\phi}) \le R_v$ is maximal.

$$$$ Exercise: Let $k$ be a field and let $\Omega$ be an algebraic extension of $k$. Then each subring $k \le A \le \Omega$ is a field.

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