# Product ideals that yield reduced quotients

• Kreizhn
In summary, the conversation discusses proving that IJ = I ∩ J for ideals I and J in a commutative ring R, given that R/(IJ) is reduced. The conversation also touches on proving an equality of ideals involving polynomials and a specific element a in R. The first approach of proving the initial statement is via contradiction while the second approach involves showing an isomorphism between two quotient rings.
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.

Hi Kreizhn!

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

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.

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?

Kreizhn said:
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).

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.

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).

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

## 1. What are "product ideals that yield reduced quotients"?

"Product ideals that yield reduced quotients" refers to a concept in mathematics where a product of two or more ideals in a ring results in a quotient ring that has a smaller number of elements compared to the original ring. This can be useful in simplifying calculations and understanding the structure of the ring.

## 2. How do you determine if a product ideal will yield a reduced quotient?

In general, it is not easy to determine if a product ideal will yield a reduced quotient. However, there are some known cases where this is true, such as when the ideals are generated by elements that commute with each other. In other cases, it may require more advanced techniques and analysis.

## 3. What are the applications of product ideals that yield reduced quotients?

Product ideals that yield reduced quotients have various applications in mathematics, particularly in the study of rings and their properties. They can also be used in algebraic geometry to understand the structure of algebraic varieties and their ideals.

## 4. Can product ideals that yield reduced quotients be generalized to other mathematical structures?

Yes, the concept of product ideals that yield reduced quotients can be extended to other algebraic structures such as modules and algebras. In these cases, the reduced quotient is defined as the quotient of the product structure by the product ideal.

## 5. Are there any limitations or drawbacks to using product ideals that yield reduced quotients?

One limitation of using product ideals that yield reduced quotients is that it may not always be possible to determine if the quotient is reduced. It also requires a good understanding of the structure of the ring and the ideals involved. Additionally, the reduced quotient may not always provide useful information or simplification in all cases.

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