Proposition

Let $A \subset B$ be rings, $B$ integral over $A$.

If $x \in A$ is a unit in $B$, then it is a unit in $A$.
The Jacobson radical of $A$ is the contraction of the Jacobson radical of $B$.

Solution

1

Let $x \in A$ and suppose $x$ is a unit in $B$. Since $x^{-1} \in B$ and $B$ is integral over $A$, $x^{-n} + \sum_{i=0}^{n-1} a_ix^{-i} = 0$ for some $a_0, \cdots, a_{n - 1} \in A$. By multiplying $x^{n - 1}$ to each side, we obtain $x^{-1} = -\sum_{i=0}^{n-1} a_ix^{n-1-i} \in A$. Therefore, $x$ is a unit in $A$.

2

By Theorem 5.10[Atiyah], for any maximal ideal $m \subset A$, there exists a prime ideal $n \subset B$ such that $m = A \cap n$. By Corollary 5.8, such an $n$ must be a maximal ideal.

Therefore, for all maximal ideals $m \subset A$, there exists a maximal ideal $n \subset B$ such that $m = A \cap n$. Moreover, again by Corollary 5.8, for every maximal ideal $n \subset B$, $A \cap n$ is a maximal ideal in $A$.

In conclusion, every maximal ideal in $A$ is of the form $n \cap A$ where $n$ is a maximal ideal of $B$, and, for any maximal ideal $n$ of $B$, $n \cap A$ is a maximal ideal of $A$.

Since $J(A)$ and $J(B)$ are both the intersections of all the maximal ideals of $A$ and $B$, $J(A) = A \cap J(B)$.