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 $r s$ must belong to $\\mathfrak{p}$ . Since $\\mathfrak{p}$ is , 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 $r s$ 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.

单词	QuotientRingModuloPrimeIdeal
释义	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 $r s$ must belong to $\\mathfrak{p}$ . Since $\\mathfrak{p}$ is , 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 $r s$ 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.
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