characterization of field

Proposition 1.

Let Rโ‰ 0 be a commutative ring with identityPlanetmathPlanetmath. The ring R (as above) is a field if and only if R has exactly two ideals: (0),R.


(โ‡’) Suppose โ„› is a field and let ๐’œ be a non-zero ideal of โ„›. Then there exists rโˆˆ๐’œโŠ†โ„› with rโ‰ 0. Since โ„› is a field and r is a non-zero element, there exists sโˆˆโ„› such that


Moreover, ๐’œ is an ideal, rโˆˆ๐’œ,sโˆˆ๐’ฎ, so sโ‹…r=1โˆˆ๐’œ. Hence ๐’œ=โ„›. We have proved that the only ideals of โ„› are (0) and โ„› as desired.

(โ‡) Suppose the ring โ„› has only two ideals, namely (0),โ„›. Let aโˆˆโ„› be a non-zero element; we would like to prove the existence of a multiplicative inverseMathworldPlanetmath for a in โ„›. Define the following set:

๐’œ=(a)={rโˆˆโ„›โˆฃr=sโ‹…a,ย for someย โขsโˆˆโ„›}

This is clearly an ideal, the ideal generated by the element a. Moreover, this ideal is not the zero idealMathworldPlanetmathPlanetmath because aโˆˆ๐’œ and a was assumed to be non-zero. Thus, since there are only two ideals, we conclude ๐’œ=โ„›. Therefore 1โˆˆ๐’œ=โ„› so there exists an element sโˆˆโ„› such that


Hence for all non-zero aโˆˆโ„›, a has a multiplicative inverse in โ„›, so โ„› is, in fact, a field. โˆŽ

Title characterization of field
Canonical name CharacterizationOfField
Date of creation 2013-03-22 13:57:03
Last modified on 2013-03-22 13:57:03
Owner alozano (2414)
Last modified by alozano (2414)
Numerical id 7
Author alozano (2414)
Entry type Theorem
Classification msc 12E99
Synonym a field only has two ideals
Related topic Field
Related topic Ring
Related topic Ideal