If you find any mistakes, please make a comment! Thank you.

In a finite commutative ring, all prime ideals are maximal


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.19

Solution:

We begin with a lemma.

Lemma: If R is a finite integral domain, then R is a field.

Proof: Let uR be nonzero, and define φu:RR by φu(r)=ur. We claim that φu is injective. To see this, suppose φu(r)=φu(s); then ur=us, so that u(rs)=0. Since u0, rs=0, so that r=s. Hence φu is injective. Since R is finite, φu is also surjective. In particular, there exists vR such that φu(v)=uv=1, so that u is a unit. Thus every nonzero element of R is a unit, and R is a field. ◻

Now if PR is prime, then R/P is a finite integral domain. Then R/P is a field, and thus PR is maximal.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu
Close Menu