## You are here

Homemaximal ideal is prime

## Primary tabs

# 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 $rs$ 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 belongs to $\mathfrak{m}$, and therefore $\mathfrak{m}$ is a prime ideal of $R$.

Related:

SumOfIdeals, MaximumIdealIsPrimeGeneralCase, CriterionForMaximalIdeal

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

16D25*no label found*13A15

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections