I'm stuck on the following problem:

Let $R$ be a commutative ring with $1$, and suppose $R$ is semilocal with maximal ideals $\mathfrak m_1, ... , \mathfrak m_s$. Let $\mathfrak a$ be an ideal of $R$. The problem is to show that under the usual surjection $\pi:R \rightarrow R/\mathfrak a$, the Jacobson radical of $R$ is sent to the Jacobson radical of $R/\mathfrak a$.

My attempt: suppose that $\mathfrak m_1, ... , \mathfrak m_t$ contain $\mathfrak a$, but $\mathfrak m_{t+1}, ... , \mathfrak m_s$ do not. Then $Jac(R/\mathfrak a) = (\mathfrak m_1/ \mathfrak a) \cap \cdots \cap (\mathfrak m_t/ \mathfrak a)$. Obviously $\pi Jac(R) \subseteq Jac(R/ \mathfrak a)$, and the converse inclusion is equivalent to the following claim: if $x \in \mathfrak m_1 \cap \cdots \cap \mathfrak m_t$, then there exists a $y \in \mathfrak m_1 \cap \cdots \cap \mathfrak m_s$ such that $x - y \in \mathfrak a$.

This should be some application of the Chinese remainder theorem, but I'm not seeing it. All I have to go on is that there exist $m_i \in \mathfrak m_i$ ($t+1 \leq i \leq s$) with $m_i \not\in I$. I also haven't found out a way to use the fact that $\mathfrak m_1 ,... , \mathfrak m_s$ are all the maximal ideals of $R$.

I was also thinking about looking at some explicit maps. If we let $\Gamma$ be the composition $$\frac{(R/\mathfrak a)}{Jac(R/ \mathfrak a)} \xrightarrow{\cong} \frac{(R/\mathfrak a)}{(\mathfrak m_1/\mathfrak a)} \oplus \cdots \oplus \frac{(R/\mathfrak a)}{(\mathfrak m_t/\mathfrak a)} \xrightarrow{\cong} R/\mathfrak m_1 \oplus \cdots \oplus R/\mathfrak m_t$$ then we have a commutative diagram $$\begin{matrix} R/Jac(R) & \rightarrow & \frac{(R/\mathfrak a)}{Jac(R/\mathfrak a)} \\ \downarrow & & \downarrow \Gamma \\ R/\mathfrak m_1 \oplus \cdots R/\mathfrak m_s & \rightarrow & R/\mathfrak m_1 \oplus \cdots \oplus R/\mathfrak m_t \end{matrix}$$ where the vertical arrows are isomorphisms, and the horizontal arrows are surjections.

You have to show $$\mathfrak m_1\cap\cdots\cap\mathfrak m_s+\mathfrak a=\mathfrak m_1\cap\cdots\cap\mathfrak m_t,$$ where $s\ge t$, $\mathfrak a\subseteq\mathfrak m_i$ for $1\le i\le t$, and $\mathfrak a\nsubseteq\mathfrak m_j$ for $t+1\le j\le s$.
The inclusion "$\subseteq$" is obvious.
For the converse notice that $\mathfrak m_{t+1}\cap\cdots\cap\mathfrak m_s+\mathfrak a=R$ hence $1=a+b$ with $a\in\mathfrak a$ and $b\in\mathfrak m_{t+1}\cap\cdots\cap\mathfrak m_s$. Let $x\in m_1\cap\cdots\cap m_t$. Then $x=ax+bx$ with $ax\in\mathfrak a$ and $bx\in\mathfrak m_1\cap\cdots\cap\mathfrak m_s$.
• Thanks for answering. Sorry, but why are $\mathfrak m_{t+1} \cap \cdots \mathfrak m_s$ and $\mathfrak a$ comaximal? – D_S Sep 7 '15 at 16:34
• @D_S Suppose it is contained in a maximal ideal. Now what's going on if the maximal ideal is $m_i$ with $1\le i\le t$ and what's going on if $t<i\le s$? – user26857 Sep 7 '15 at 16:36
• If $\mathfrak m_{t+1} \cap \cdots \cap \mathfrak m_s + \mathfrak a \subseteq \mathfrak m_{t+1}$, then also $\mathfrak a \subseteq \mathfrak m_{t+1}$, which is impossible. If $\mathfrak m_{t+1} \cap \cdots \cap \mathfrak m_s + \mathfrak a \subseteq \mathfrak m_1$, then $\mathfrak m_{t+1} \cap \cdots \cap \mathfrak m_s \subseteq \mathfrak m_1$. I can intersect on both sides to get that $$\mathfrak m_2 \cap \cdots \cap \mathfrak m_s \subseteq \mathfrak m_1 \cap \cdots \cap \mathfrak m_s$$ Hence those intersections are equal. But I'm not seeing the contradiction here. – D_S Sep 7 '15 at 17:01
• In the second case you get $m_{t+1}\cdots m_s\subseteq m_1$ and by using the definition of prime ideals some $m_i\subseteq m_1$ with $i>t$, a contradiction. – user26857 Sep 7 '15 at 17:05