No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field

Problems and solutions of ring theory in abstract algebra

Problem 531

(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.

 
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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.


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