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 $S$.

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.)

The post Every Ideal of the Direct Product of Rings is the Direct Product of Ideals first appeared on Problems in Mathematics.

(a) Let $F$ be a field. Show that $F$ does not have a nonzero zero divisor.

(b) Let $R$ and $S$ be nonzero rings with identities.
Prove that the direct product $R\times S$ cannot be a field.

Proof.

(a) Show that $F$ does not have a nonzero zero divisor.

Seeking a contradiction, suppose that $x$ is a nonzero zero divisor of the field $F$. This means that there exists a nonzero element $y\in F$ such that
\[yx=0.\] Since $y$ is a nonzero element in $F$, we have the inverse $y^{-1}$ in $F$.

Hence we have
\begin{align*}
0=y^{-1}\cdot 0=y^{-1}(yx)=(y^{-1}y)x=x.
\end{align*}
This is a contradiction because $x$ is a nonzero element.

We conclude that the field $F$ does not have a nonzero zero divisor.

(Remark that it follows that a field is an integral domain.)

(b) Prove that the direct product $R\times S$ cannot be a field.

Since $R$ and $S$ have identities, the direct product $R\times S$ contains nonzero elements $(1,0)$ and $(0,1)$.

The product of these elements is
\[(1,0)\cdot (0,1)=(1\cdot 0, \, 0\cdot 1)=(0,0).\] Similarly we also have
\[(0,1)\cdot (1,0)=(0,0).\]

It follows that $(1,0)$ is a nonzero zero divisor of $R\times S$. By part (a), a field does not have a nonzero zero divisor.
Hence $R\times S$ is never a field.

The post No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field first appeared on Problems in Mathematics.