Theorem 19.9

Return to Theorems, Glossary, Homework Problems.

Every field $F$ is an integral domain.

Let $a,b \in F,$ and suppose that $a\not= 0$. Then if $ab=0,$ we have

\begin{align} \bigg(\frac{1}{a}\bigg)(ab) = \bigg(\frac{1}{a}\bigg)0 = 0. \end{align}

But then

\begin{align} 0 = \bigg(\frac{1}{a}\bigg)(ab) = \bigg[\bigg(\frac{1}{a}\bigg)a\bigg]b = 1b = b. \end{align}

We have shown that $ab=0$ with $a\not=0$ implies that $b=0$ in $F$, so there are no divisors of 0 in $F$. Of course, $F$ is a commutative ring with unity, so our theorem is proved.

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