1. Maximal and Prime Ideals

Letting R be a finite commutative ring with unity, show that all prime ideals of R are also maximal ideals of R.

I know that if an ideal, I, of R is maximal then R/I is a field and therefore an ID, which would make I a prime ideal of R; but, I'm unsure how to approach going the other way..

2. that is good that you are unsure how to go the other way because the other direction is simply not true. It is only true if your ring is a principal ideal domain.

For a counterexample consider polynomials over the integers. Consider the ideal (x). $\mathbb{Z}[x]/(x)\cong \mathbb{Z}$ which is an integral domain and is in particular not a field. The fact that it is an integral domain shows it is prime, and the fact that it is not a field shows that it is not maximal. So there is an example of a prime ideal which is not maximal in a commutative ring with 1.

3. R is not an PID, so how would I show that all prime ideals of R are also maximal ideals of R?

4. Originally Posted by Gamma
that is good that you are unsure how to go the other way because the other direction is simply not true. It is only true if your ring is a principal ideal domain.

For a counterexample consider polynomials over the integers. Consider the ideal (x). $\mathbb{Z}[x]/(x)\cong \mathbb{Z}$ which is an integral domain and is in particular not a field. The fact that it is an integral domain shows it is prime, and the fact that it is not a field shows that it is not maximal. So there is an example of a prime ideal which is not maximal in a commutative ring with 1.
in Coda202's question the ring R is finite. if P is a prime ideal of R, then R/P is a finite integral domain. we know that a finite integral domain is a field and thus P is maximal.

5. Righto, missed the finite part when reading it. Do what we said, you know it is an Integral Domain because the ideal is prime, call this F. Now you know it is a finite integral domain, and we can prove that is a field.

All you gotta do is show every nonzero element has a multiplicative inverse. Let a be such an element, and consider the map $\phi: F \rightarrow F$ by $\phi(x)=ax$.

This map is clearly well defined and is injective. suppose $\phi(x)=\phi(y)\Rightarrow ax=ay \Rightarrow a(x-y)=0$. a is assumed to be nonzero, and F is an integral domain with no zero divisors, so in particular $x-y=0 \Rightarrow x=y$.

Because F is finite, injectivity implies surjectivity, so there must in fact be some x which gets sent to 1, this is the inverse under multiplication of a. This is true for all nonzero a in F, so it is indeed a field