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# quotient ring modulo prime ideal

Theorem. Let $R$ be a commutative ring with non-zero unity 1 and $\mathfrak{p}$ an ideal of $R$. The quotient ring $R/\mathfrak{p}$ is an integral domain if and only if $\mathfrak{p}$ is a prime ideal.

Proof. $1^{{\underline{o}}}$. First, let $\mathfrak{p}$ be a prime ideal of $R$. Then $R/\mathfrak{p}$ is of course a commutative ring and has the unity $1+\mathfrak{p}$. If the product $(r+\mathfrak{p})(s+\mathfrak{p})$ of two residue classes vanishes, i.e. equals $\mathfrak{p}$, then we have $rs+\mathfrak{p}=\mathfrak{p}$, and therefore $rs$ must belong to $\mathfrak{p}$. Since $\mathfrak{p}$ is prime, either $r$ or $s$ belongs to $\mathfrak{p}$, i.e. $r+\mathfrak{p}=\mathfrak{p}$ or $s+\mathfrak{p}=\mathfrak{p}$. Accordingly, $R/\mathfrak{p}$ has no zero divisors and is an integral domain.

$2^{{\underline{o}}}$. Conversely, let $R/\mathfrak{p}$ be an integral domain and let the product $rs$ of two elements of $R$ belong to $\mathfrak{p}$. It follows that $(r+\mathfrak{p})(s+\mathfrak{p})=rs+\mathfrak{p}=\mathfrak{p}$. Since $R/\mathfrak{p}$ has no zero divisors, $r+\mathfrak{p}=\mathfrak{p}$ or $s+\mathfrak{p}=\mathfrak{p}$. Thus, $r$ or $s$ belongs to $\mathfrak{p}$, i.e. $\mathfrak{p}$ is a prime ideal.

## Mathematics Subject Classification

13C99*no label found*

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