Given distinct maximal ideals $M_1,...,M_n$, is $M_1\cdots M_n$ radical?

Question

Let $R$ be a commutative ring with $1$ and $M_1,...,M_n$ be distinct maximal ideals in $R$. What I want to show is $M_1\cap\cdots \cap M_n=M_1\cdots M_n$. If I can show that $M_1\cdots M_n$ is radical, then the proof ends since $M_1\cap\cdots\cap M_n=rad(M_1\cdots M_n)$. How can I prove this?

Answer 1

I do not know what radical is so here is a proof of this without using the fact stated in the question

First we observe that $M_i+M_j=R$. Since this is a maximal Ideal containing $M_i$. Now we show this by Induction. Proving for $n=2$:

There exists $x \in M_1,y \in M_2$ such that $x+y=1$. Say $a \in M_1\cap M_2$, then $ax+ay=a$ but $ax \in M_1M_2$ and $ay \in M_1 M_2$. Therefore $M_1\cap M_2 \subset M_1M_2$.

Now for the general case we have to show that $M_1$ and $M_2...M_n$ are co maximal . But this is true since $M_1$ is a maximal Ideal.

Answer 2

Here's a lemma one would hope for in order to use induction:

Lemma: If $I$ and $J$ are radical ideals, and $I+J=R$, then $IJ$ is radical.

Proof: Suppose $x^n\in IJ\subseteq I\cap J$. By the definition of radical ideals, we have that $x\in I\cap J$, but we still need to show it's in $IJ$. Here is where the comaximality condition comes in: let $i+j=1$ with $i\in I,j\in J$. Then $x=ix+xj\in IJ$. Thus $x^n\in IJ$ implies $x\in IJ$, and $IJ$ is radical.

Obviously the base case of induction for your proof is easy: $M_1+M_2=R$, and both ideals are prime, so $M_1M_2$ is radical. Now we need to justify the inductive step.

If we've already shown $I=\prod_{i=1}^kM_i$ is radical, then why should $I+M_{k+1}=R$? Here's a sketch to show you what to try:

1. If $I+M_{k+1}\neq R$, it's contained in a maximal idea.
2. That maximal ideal has to be $M_{k+1}$
3. Since $M_{k+1}$ is prime and $I\subseteq M_{k+1}$, one of the previous $M_i$ for $1\leq i\leq k$ is contained in $M_{k+1}$. Explain why this contradicts the hypotheses.
4. Conclude $I+M_{k+1}=R$.
5. Apply the lemma to carry out the inductive step.

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Bonus exercise: show that $IJ$ need not be radical if the $I+J=R$ condition is dropped.