(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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Definition.
A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors.
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Proof.
We give two proofs.
Proof […]

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The set of torsion elements is denoted
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Let $R$ be a ring with $1$. Suppose that $R$ is an integral domain and an Artinian ring.
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Definition (Artinian ring).
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That is, whenever we have […]

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Let $R$ be a ring with $1\neq 0$. Let $a, b\in R$ such that $ab=1$.
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Definition.
An element $x\in R$ is called a zero divisor if there exists a […]

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Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal […]