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.
随便看	WLOG WLOG WolfPrize Wolf Prize WolstenholmesTheorem Wolstenholme’s theorem WoodallNumber Woodall number Word word WordProblem word problem WordsForNumbersInSlavicLanguages words for numbers in Slavic languages Works on Polynomials WorksOnPolynomials1 WorldRecordsInMathematics world records in mathematics wpfunction WreathProduct wreath product Writhe writhe WronskianDeterminant Wronskian determinant