- #1

Kreizhn

- 743

- 1

## Homework Statement

Let I,J be ideals in a commutative ring R. Assume that R/(IJ) is reduced (so that it contains no nilpotent elements). Prove that [itex] IJ = I \cap J [/itex]

## Homework Equations

For two ideals I and J, we have

[tex] IJ = \left\{ a_1 b_1 + \cdots + a_n b_n : a_i \in I, b_i \in J, i=1,\ldots,n \ n \in \mathbb N \right\} [/tex]

## The Attempt at a Solution

I've been thinking about this problem for some time. I think I've narrowed it down to show a small result, but perhaps someone could comment on whether I'm thinking in the right direction.

It seems unlikely that we will be able to use the fact that R/(IJ) is reduced to create a constructive proof, so we move to prove via contradiction. Since it is always true that [itex] IJ \subseteq I \cap J [/itex] let us assume for the sake of contradiction that the inclusion is strict, and show that there is a nilpotent element in R/(IJ).

Now by looking at the definition of the product ideal, it seems that the best choice for a candidate element would be to find two elements [itex] s,t \in I \cap J [/itex] such that [itex] (s+t) \in (I \cap J)\setminus IJ [/itex] if such an element exists (this is what I have yet to prove). Then the projection map [itex] \pi: R \to R/(IJ) [/itex] would yield

[tex] \pi(s+t) = (s+t) + IJ \neq IJ [/tex]

because we chose s+t to not be in IJ. On the other hand

[tex] [(s+t) + IJ]^2 = (s+t)^2 +IJ = s^2 + st + ts + t^2 + IJ = IJ [/tex]

where the equality to IJ follows because s and t are both in I and J, and the ring is commutative. Hence [itex] (s+t)^2 + IJ = IJ [/itex] and so this is a nonzero nilpotent element which is a contradiction.

I think this is all correct. However, I'm a bit stuck on showing that [itex] \exists s, t \in I \cap J, (s+t) \in (I\cap J)\setminus IJ [/itex]. Certainly, we know there exists at least one element, so it remains to show that there is a second, and that we can choose them such that the sum is not in the product ideal. Any thoughts? I'm also open to other suggestions.