Product ideals that yield reduced quotients

[Kreizhn]

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 IJ = I \cap J

Homework 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\}

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.

 

[micromass]

Hi Kreizhn!

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

 

[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.

 

[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?

 

[micromass]

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?

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).

 

[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.

 

[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).

 

[Kreizhn]

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