# Every Ideal of the Direct Product of Rings is the Direct Product of Ideals

## Problem 536

Let $R$ and $S$ be rings with $1\neq 0$.

Prove that every ideal of the direct product $R\times S$ is of the form $I\times J$, where $I$ is an ideal of $R$, and $J$ is an ideal of $S$.

Contents

## Proof.

Let $K$ be an ideal of the direct product $R\times S$.
Define
$I=\{a\in R \mid (a,b)\in K \text{ for some } b\in S\}$ and
$J=\{b\in S \mid (a, b)\in K \text{ for some } a\in R\}.$

We claim that $I$ and $J$ are ideals of $R$ and $S$, respectively.

Let $a, a’\in I$. Then there exist $b, b’\in S$ such that $(a, b), (a’, b’)\in K$.
Since $K$ is an ideal we have
$(a,b)+(a’,b’)=(a+a’, b+b)\in k.$

It follows that $a+a’\in I$.
Also, for any $r\in R$ we have
$(r,0)(a,b)=(ra,0)\in K$ because $K$ is an ideal.

Thus, $ra\in I$, and hence $I$ is an ideal of $R$.
Similarly, $J$ is an ideal of $S$.

Next, we prove that $K=I \times J$.
Let $(a,b)\in K$. Then by definitions of $I$ and $J$ we have $a\in I$ and $b\in J$.
Thus $(a,b)\in I\times J$. So we have $K\subset I\times J$.

On the other hand, consider $(a,b)\in I \times J$.
Since $a\in I$, there exists $b’\in S$ such that $(a, b’)\in K$.
Also since $b\in J$, there exists $a’\in R$ such that $(a’, b)\in K$.

As $K$ is an ideal of $R\times S$, we have
$(1,0)(a,b’)=(a,0)\in K \text{ and } (0, 1)(a’,b)=(0, b)\in K.$ It yields that
$(a,b)=(a,0)+(0,b)\in K.$ Hence $I\times J \subset K$.

Putting these inclusions together gives $k=I\times J$ as required.

## Remark.

The ideals $I$ and $J$ defined in the proof can be alternatively defined as follows.
Consider the natural projections
$\pi_1: R\times S \to R \text{ and } \pi_2:R\times S \to S.$ Define
$I=\pi_1(K) \text{ and } J=\pi_2(K).$

Since the natural projections are surjective ring homomorphisms, the images $I$ and $J$ are ideals in $R$ and $S$, respectively.
(see the post The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal.)

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