PlanetMath: maximal ideal is prime

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maximal ideal is prime

(Theorem)

Theorem. In a commutative ring with non-zero unity, any maximal ideal is a prime ideal.

Proof. Let $\mathfrak{m}$ be a maximal ideal of such a ring and let the ring product 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 and belong to $\mathfrak{m}$ , whence

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

So we can say that along with

, at least one of its factors belongs to $\mathfrak{m}$ , and therefore $\mathfrak{m}$ is a prime ideal of

"maximal ideal is prime" is owned by pahio.

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See Also: sum of ideals, maximal ideal is prime (general case)

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Cross-references: implies, ring product, ring, prime ideal, maximal ideal, non-zero unity, commutative ring
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This is version 5 of maximal ideal is prime, born on 2007-11-23, modified 2007-11-24.
Object id is 10054, canonical name is MaximalIdealIsPrime.
Accessed 601 times total.

Classification:

AMS MSC:	13A15 (Commutative rings and algebras :: General commutative ring theory :: Ideals; multiplicative ideal theory)
	16D25 (Associative rings and algebras :: Modules, bimodules and ideals :: Ideals)

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