Jacobson radical - Integral extension Let $A\subseteq B$ be an integral extension. 
Question is to prove that :

Jacobson radical of $A$ is the contraction of Jacobson radical of $B$.

Let $a\in J(B)\cap A$ and $y\in A$. To prove that $1+ay$ is a unit in $A$.
As $a\in J(B)$ and $y\in A\subseteq B$, $1+ay$ is a unit in $B$.
As $a\in A,y\in A$, we have $1+ay\in A$. 
$1+ay\in A$ is a unit in $B$. As $A\subseteq B$ is integral, this means that $1+ay$ is a unit in $A$.
Thus, $1+ay$ is a unit in $A$ for all $y\in A$. Thus, $a\in J(A)$. So, $J(B)\cap A\subseteq J(A)$.
It remains to prove that $J(A)\subseteq J(B)\cap A$. I was expecting that 
for any ring extension $A\subseteq B$, we have $J(A)\subseteq J(B)$. But then, I realized it is not true in general.
Let $R$ be an integral domain whose Jacobson radical is non zero and $K$ be its field of fractions. Then $R\subseteq K$ is a ring extension and $0\neq J(R)$ where as $J(K)=0$ so,  $J(R)\nsubseteq J(K)$.
Let $a\in J(A)$ and I have to prove that $a\in J(B)$ i.e., $1+ay$ is a unit in $B$ for each $y\in B$.
Let $y\in B$. As $B$ is integral over $A$ there exists $a_1,\cdots,a_n\in A$ such that $$y^n+a_1y^{n-1}+\cdots+a_n=0.$$
We  know that $1+a_ix$ is a unit in $A$ for every $1\leq i\leq n$. I am not able to use this to conclude that $1+ay$ is a unit in $B$.
Any hint is welcome. Please do not post full answer.
 A: Sorry, I found it easiest to prove without considering elements.
Recall that
$$
J(B) = \bigcap_{\mathfrak{m} \trianglelefteq B \ \text{maximal}} \mathfrak{m} \, .
$$
We can prove the claim using two basic results about integral extensions.  I refer to their numbers in Atiyah-MacDonald.
Corollary 5.8. Let $A \subseteq B$ be rings, $B$ integral over $A$; let $\mathfrak{q}$ be a prime ideal of $B$ and let $\mathfrak{p} = \mathfrak{q}^c = \mathfrak{q} \cap A$.  Then $\mathfrak{q}$ is maximal iff $\mathfrak{p}$ is maximal.
Moreover, every maximal ideal of $A$ is the contraction of a maximal ideal of $B$ by Lying Over (Theorem 5.10) and the above corollary.  Can you take it from here?

 Letting $i: A \hookrightarrow B$ be the inclusion, then \begin{align*} J(B)^c = i^{-1}\left(\bigcap_{\mathfrak{m} \trianglelefteq B \ \text{maximal}} \mathfrak{m}\right) = \bigcap_{\mathfrak{m} \trianglelefteq B \ \text{maximal}} i^{-1}\mathfrak{m} = \bigcap_{\mathfrak{n} \trianglelefteq A \ \text{maximal}} \mathfrak{n} = J(A) \, . \end{align*}

Perhaps you can translate the proof above into a statement about elements, since Corollary 5.8 is equivalent to your statement $1 + ay$ is a unit in $B$ iff it is a unit in $A$.
