1. ## Prime ideals

Let R be an integral domain where ideals are only of form aR={ar|r is an element of R} for some a in R. Prove that every nonzero prime ideal in R is a maximal ideal.

Besides knowing that every maximal ideal in a commutative ring with unity is a prime ideal, i do not know where to go.

How about this: assume that not every prime ideal in R is a maximal ideal. Then for some prime ideal I in R there is another prime ideal N such that I is contained in N and N is contained in R.
For some a,b in I then ab is in both I and N. This violates assumption that they are prime, so every prime must be maximal?

I have no idea if this is good and advice/help would be really great.

2. ## Every non zero prime ideal in a PID is a maximal Ideal

Originally Posted by curiousmuch
Let R be an integral domain where ideals are only of form aR={ar|r is an element of R} for some a in R. Prove that every nonzero prime ideal in R is a maximal ideal.

Besides knowing that every maximal ideal in a commutative ring with unity is a prime ideal, i do not know where to go.

How about this: assume that not every prime ideal in R is a maximal ideal. Then for some prime ideal I in R there is another prime ideal N such that I is contained in N and N is contained in R.
For some a,b in I then ab is in both I and N. This violates assumption that they are prime, so every prime must be maximal?

I have no idea if this is good and advice/help would be really great.

Let $\displaystyle (P)$ be a nontrivial prime ideal, and let $\displaystyle I$ be an Ideal containing $\displaystyle P$

So we have $\displaystyle P \subset I \subset R$ Now for P to be maximal either $\displaystyle P = I$ or $\displaystyle I = R$

Since I is generated by some element a of the Ring $\displaystyle (a) = I$

Now $\displaystyle p \in I$ so $\displaystyle p=ar$ for some $\displaystyle r \in R$ Now since P is a prime Ideal and $\displaystyle am \in (p)$ either $\displaystyle a \in p$ or $\displaystyle r \in p$.

So we have two cases.

Case I $\displaystyle a \in p$ then $\displaystyle (I)=(P)$.

Case II $\displaystyle r \in p$ then $\displaystyle r=ps$

now we know that $\displaystyle p=ar$ as well, so we sub this into above to get

$\displaystyle r=(ar)s \iff r-ars=0 \iff r(1-as)=0$ and Since we are in an integral domain and $\displaystyle r\ne 0$ we know that

$\displaystyle 1-as=0 \iff 1=as$ so $\displaystyle a,s$ are both units

so this implies that $\displaystyle I =R$ and we are done

3. Hey thanks for the great help, but I was a bit confused about one variable: where did the r=ps come from in analyzing the second case. Thanks.

4. Originally Posted by curiousmuch
Hey thanks for the great help, but I was a bit confused about one variable: where did the r=ps come from in analyzing the second case. Thanks.
Since r is in the prime ideal genterated by p it must be a multiple of p.

Remember every ideal is principle (generated by all multiples of an elemtent)

i.e there exists an $\displaystyle s \in R$ such that $\displaystyle sp=r$