maximal ideal is prime

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 $R$ and let the ring product $r s$ 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 $r s$ belong to $\mathfrak{m}$ , whence

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

So we can say that along with $r s$ , 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
Canonical name	MaximalIdealIsPrime
Date of creation	2013-03-22 17:37:59
Last modified on	2013-03-22 17:37:59
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Theorem
Classification	msc 16D25
Classification	msc 13A15
Related topic	SumOfIdeals
Related topic	MaximumIdealIsPrimeGeneralCase
Related topic	CriterionForMaximalIdeal