3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $bx\in\mathfrak m_1\cap\cdots\cap\mathfrak m_s$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for answering. Sorry, but why are $\mathfrak m_{t+1} \cap \cdots \mathfrak m_s$ and $ \mathfrak a$ comaximal? $\endgroup$ – D_S Sep 7 '15 at 16:34
  • $\begingroup$ @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 $t<i\le s$? $\endgroup$ – user26857 Sep 7 '15 at 16:36
  • $\begingroup$ 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. $\endgroup$ – D_S Sep 7 '15 at 17:01
  • $\begingroup$ 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. $\endgroup$ – user26857 Sep 7 '15 at 17:05
  • $\begingroup$ Ohhhhhhhhh. lol $\endgroup$ – D_S Sep 7 '15 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.