how to prove a=r(a)<=>a is a intersection of prime ideals.if a is an ideal≠(1) in commutative ring A.

Question

how to prove a=r(a)<=>a is a intersection of prime ideals.if a is an ideal≠(1) in commutative ring A.r mean radical.

I can't prove it,here is what I did.

a is a intersection of prime ideals mean a is nilradical.

0∈a because a is an ideal,so all nilradical is a subset of a.

If we can prove "x isn't a nilpotent => x∉a" or "a is a subset of nilradical" ,then we can prove (=>).

I can't prove "x isn't a nilpotent => x∉a".

"a is a subset of nilradical"<=>"x is a nilpotent if x is a element of a"

failed again.

then I consider assumed "there exist a element a such that it's not a nilpotent" to prove 1∈a,if it works,then a=(1).

but failed again.

How to prove it?

Answer 1

This is the proof that you can find on Eisenbud's Commutative algebra, p. 70:

First, we proof the following proposition:

Let $U \subset R$ be a multiplicatively closed set, and $p \subset R$ an ideal maximal among those not meeting $U$. Then $p$ is prime.

Proof: Let $f,g \notin p$. We want to show that $fg \notin p$. By maximality of $p$, both $p +(f)$, $p+(g)$ meet $U$. Hence, there are $a,b \in R$ and $i,j \in p$ s.t. both $i+af, j+bg$ are on $U$. Since $U$ is multiplicatively closed, the product is in $U$. Now assume $fg \in p$. Then the product is also in $p$, but then we have a contradiction because we were assuming that $p \cap U = \emptyset$. Therefore $fg \notin p$ and this means that $p$ is prime.

Now, we want to prove that $$r(I) = \bigcap_{p \supset I} p$$ i.e. that $r(I)$ is the intersection of all the prime ideals containing $I$. Note that this proves you question.

For $\subset$, note that $f \in r(I) \Rightarrow f^n \in I$ for some $n$. But if $p$ contains $I$, then it contains $f^n$. But $p$ is prime, so $f \in p$.

For $\supset$, we will use the proposition. We will show that $f \notin r(I)$ implies $f$ not in the intersection. So let $f \notin r(I)$. Then, $U= \{ f^n, n > 0 \}$ is disjoint to $I$, and $U$ is a multiplicatively closed set. Therefore we can choose an ideal maximal among those not meeting $U$ that contains $I$, say $p$. By the previous proposition, $p$ is prime, and since it contains $I$, it belongs to the intersection. But $p$ is disjoint to $U$, so $f$ is not in the intersection, and we are done.