maximal ideal is prime

Proof.  Let $\mathfrak{m}$ be a maximal ideal of such a ring $R$ and let the ring product $rs$ belong to $\mathfrak{m}$ but e.g.  $r\notin\mathfrak{m}$. The maximality of $\mathfrak{m}$ implies that  $\mathfrak{m}\!+\!(r)=R=(1)$.  Thus there exists an element  $m\in\mathfrak{m}$  and an element  $x\in R$  such that  $m\!+\!xr=1$.  Now $m$ and $rs$ belong to $\mathfrak{m}$, whence

 $s=1s=(m\!+\!xr)s=sm\!+\!x(rs)\in\mathfrak{m}.$

So we can say that along with $rs$, at least one of its factors (http://planetmath.org/Product) belongs to $\mathfrak{m}$, and therefore $\mathfrak{m}$ is a prime ideal of $R$.

Title maximal ideal is prime MaximalIdealIsPrime 2013-03-22 17:37:59 2013-03-22 17:37:59 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 16D25 msc 13A15 SumOfIdeals MaximumIdealIsPrimeGeneralCase CriterionForMaximalIdeal