# Homework Help: Product ideals that yield reduced quotients

1. Jun 17, 2011

### Kreizhn

1. The problem statement, all variables and given/known data

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 $IJ = I \cap J$

2. Relevant equations

For two ideals I and J, we have
$$IJ = \left\{ a_1 b_1 + \cdots + a_n b_n : a_i \in I, b_i \in J, i=1,\ldots,n \ n \in \mathbb N \right\}$$

3. 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 $IJ \subseteq I \cap J$ 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 $s,t \in I \cap J$ such that $(s+t) \in (I \cap J)\setminus IJ$ if such an element exists (this is what I have yet to prove). Then the projection map $\pi: R \to R/(IJ)$ would yield
$$\pi(s+t) = (s+t) + IJ \neq IJ$$
because we chose s+t to not be in IJ. On the other hand
$$[(s+t) + IJ]^2 = (s+t)^2 +IJ = s^2 + st + ts + t^2 + IJ = IJ$$
where the equality to IJ follows because s and t are both in I and J, and the ring is commutative. Hence $(s+t)^2 + IJ = IJ$ 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 $\exists s, t \in I \cap J, (s+t) \in (I\cap J)\setminus IJ$. 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.

2. Jun 17, 2011

### micromass

Hi Kreizhn!

If $a\in I\cap J\setminus IJ$, isn't a+IJ nilpotent then?

3. Jun 17, 2011

### Kreizhn

Oh yes, quite obviously. Since then $(a +IJ)^2 = a^2 + IJ = a\cdot a + IJ$ and $a \cdot a \in IJ$ since $a \in I$ and $a \in J$. How did I miss that? Thanks.

4. Jun 17, 2011

### Kreizhn

While I'm here, I have another quick question.

I want to show that if R is a commutative ring, a in R, and $f_1(x),\ldots, f_r(x) \in R[x]$ then the following ideals are equal
$$(f_1(x),\ldots, f_r(x),x-a) = (f_1(a),\ldots, f_r(a),x-a)$$
Now this is fairly easy if R[x] is a Euclidean domain, since I can then write
$$f_i(x) = q_i(x)(x-a) + f_i(a)$$
and make appropriate substitutions. However, I am only told that R is commutative, which if memory serves, is not sufficient for R[x] to be Euclidean. Is there another way of doing this, or is the statement incorrect in that R needs to be a field?

5. Jun 17, 2011

### micromass

Hmm, you can't use the division algorithm here. But to show that

$$(f_1(x),\ldots, f_r(x),x-a) = (f_1(a),\ldots, f_r(a),x-a)$$

It suffices to show that

$$R[x]/(f_1(x),\ldots, f_r(x),x-a) = R[x]/(f_1(a),\ldots, f_r(a),x-a)$$

which shouldn't be hard to show... (intuitively, R[x]/(x-a) is just exchanging all the x's with a's).

6. Jun 17, 2011

### Kreizhn

That's actually what I'm asked to show in the next part of the question, so it seems like I'm supposed to do it a different way.

Edit: Well, the next question actually asks me to go one step further and show that
$$R[x]/(f_1(x),\ldots, f_r(x),x-a) \cong R/(f_1(a),\ldots, f_r(a))$$
which is just using a corollary to the third iso theorem.

7. Jun 17, 2011

### micromass

Ah, I see. Well, maybe you can show that

$$f_i(x)=q_i(x)(x-a)+f_i(a)$$

holds in general (so without referencing the division algorithm). This comes down to showing that if g(a)=0, then (x-a) divides g(x).

8. Jun 17, 2011

### Kreizhn

I'll give both ways a shot, just for the practice. Thanks.