Proving a prime ideal is maximal in a PID Let R be a principal ideal domain that has no zero-divisors but $0$. Show that every
prime ideal I ≤ R with I $\neq$ ($0$) is a maximal ideal.
I'm new to rings and ideals and don't know how to wrap my head around this question.
 A: Let $(p)$ be the prime ideal generated by $p$ and $(m)$ an ideal generated by $m$ which contains $p$, since $p\in (m)$, there exists $a\in R$ such that $p=am$. This implies that $am\in (p)$, and since $(p)$ is a prime, $m\in (p)$ or $a\in (p)$.
Suppose that $a\in (p)$. Write $a=up$. Then we have $p=upm$, which implies that $p(1-um)=0$. Since PID does not have zero divisors and $p\neq 0$, we deduce that $1-um=0$ an $um=1$. This implies that $m$ is invertible and $(m)=R$.
Suppose that $m\in (p)$. Write $m=bp$. Then for every $x\in (m)$, we have $x=um=ubp$ and we deduce that $x\in (p)$ and $(m)\subset (p)$. This implies that $(p)$ is maximal since an ideal which contains $(p)$ is either $R$ or $(p)$.
A: Let S = (a), a≠0 be a non-zero prime ideal in R (PID)
Let T = (b), be an ideal such that S ⊆ T
Now T ⊆ R

=> (b) ⊆ R

=> a ∈ S

=> a ∈ T

Let a = bm for some m ∈ R
bm ∈ S
either m ∈ S or b ∈ S      (since S is a prime ideal)

If b ∈ S
     all multiples of b would be in S
     T ⊆ S
     T = S (since we assumed S ⊆ T)

If m ∈ S
   m = an for some n ∈ R
   m = (bm)n
   m = bnm
   m(1-bn) = 0
   bn = 1 => 1 ∈ T

T = R (if 1 ∈ T)
 Since we have shown T = R or T = S whenever S ⊆ T
 This proves
 S is an maximal ideal

