# 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 productβ $(r+\mathfrak{p})(s+\mathfrak{p})$ of two residue classes vanishes, i.e. equals $\mathfrak{p}$, then we haveβ $rs+\mathfrak{p}=\mathfrak{p}$,β and therefore $rs$ must belong to $\mathfrak{p}$. Since $\mathfrak{p}$ is , either $r$ or $s$ belongs to $\mathfrak{p}$, i.e.β $r+\mathfrak{p}=\mathfrak{p}$β orβ $s+\mathfrak{p}=\mathfrak{p}$.β 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 thatβ $(r+\mathfrak{p})(s+\mathfrak{p})=rs+\mathfrak{p}=\mathfrak{p}$. Since $R/\mathfrak{p}$ has no zero divisors,β $r+\mathfrak{p}=\mathfrak{p}$β orβ $s+\mathfrak{p}=\mathfrak{p}$. Thus, $r$ or $s$ belongs to $\mathfrak{p}$, i.e. $\mathfrak{p}$ is a prime ideal.

Title quotient ring modulo prime ideal QuotientRingModuloPrimeIdeal 2013-03-22 17:37:09 2013-03-22 17:37:09 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 13C99 CharacterisationOfPrimeIdeals QuotientRing