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Revision difference : maximal ideal is prime |
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| \textbf{Theorem.} In a commutative ring with non-zero unity, any maximal ideal is a prime ideal. |
\textbf{Theorem.} In a commutative ring with non-zero unity, any maximal ideal is a prime ideal. |
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| {\em 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\, |
{\em 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 |
$\mathfrak{m}\!+\!(r) = R = (1)$.\, 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 |
| $$s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$$ |
$$s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$$ |
| So we can say that along with $rs$, at least one of its \PMlinkname{factors}{Product} belongs to $\mathfrak{m}$, and therefore $\mathfrak{m}$ is a prime ideal of $R$. |
So we can say that along with $rs$, at least one of its \PMlinkname{factors}{Product} belongs to $\mathfrak{m}$, and therefore $\mathfrak{m}$ is a prime ideal of $R$. |
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