maximal ideals of the Rad A

Question

Let $A$ a semilocal ring and $M_1,M_2,\dots,M_n$ are all of its maximal ideals then the Jacobson radical $rad(A) =M_1M_2\cdots M_n = M_1\cap M_2\cap \cdots \cap M_n$

I tried to use the Chinese Remainder Theorem, How can I solve this?

Thanks for your time and help.

Answer 1

The only thing to prove is $M_1 \dotsb M_n \supset M_1 \cap \dotsb \cap M_n$.

Let $N_i$ be the product of all $M_j$ with $i$ omitted. It is easy to see that $M_i$ does not contain $N_i$, thus $M_i+N_i=R$. We obtain:

$$M_1 \cap \dotsb \cap M_n = (M_1 \cap \dotsb \cap M_n)(M_1+N_1)\dotsb(M_n+N_n)$$ and now you see that if you expand this, each summand is contained in $M_1 \dotsb M_n$.