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in this site 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 $\displaystyle s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$ So we can say that along with $ rs$, at least one of its factors belongs to $ \mathfrak{M}$, and therefore $ \mathfrak{M}$ is a prime ideal of $ R$.

I dont understand this part: s=1s=(m+xr)s=sm+x(rs)$\in m$. Why is x(rs)$\in m$? isnt it supposed to x(rs)$\in (m)$? and why sm $\in m$ since we dont know if s $\in m$ or not, yet.

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You’re failing to distinguish the ideal $\mathfrak m$ (Fraktur ‘m’) from the element $m$ (Roman italic ‘m’) of that ideal: $x(rs)\in\mathfrak m$ because by hypothesis $rs\in\mathfrak m$, and $\mathfrak m$ is an ideal; $sm\in\mathfrak m$ because $m\in\mathfrak m$ and $\mathfrak m$ is an ideal.

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oh it comes from ideals' def.. $m\subset R$ ring if $x\in R$ and $rs\in m$ then $xrs\in m$,$rsx \in m$.thanks. –  emmett Dec 10 '12 at 3:55

You should use M to implie an ideal of R. so m is an element of ideal M. So on, sm $\in$ M as a property of an ideal, so does x(rs) $\in$ M with rs $\in$ M as hypothesis

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ok done it. thank you –  emmett Dec 10 '12 at 3:55

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