What is the relationship between prime ideals and their generators? All rings are assumed to have unity: $1\in R$.
In a ring $R$:

Definition: An element is a unit if it is invertible (it has a 2 sided multiplicative inverse).

 In a commutative ring $R$:

Definition: An ideal $P \ne R$ is a prime ideal if $ab \in P \implies a\in P \text{ or } b \in P$
Definition: An element $p\in R$ is a prime element if $(p)$ is a prime ideal.

 In an integral domain $R$:

Definition: For $a,b \in R$, we say $a$ is a factor or divisor of $b$ if $b\in (a)$.
Definition: For $a,b \in R$, we say $a$ is a proper factor of $b$ if $b \in (a)$ and $a \notin (b)$.
Definition: An element $a\in R$ is irreducible if it is not a unit and has no proper factors other than units.

 Known:
In an integral domain: 
$(p)$ is prime $\iff$ $p$ is prime $\implies$ $p$ is irreducible.
In a unique factorization domain: 
$(p)$ is prime $\iff$ $p$ is prime $\iff$ $p$ is irreducible.
 Question:
If I have a prime ideal, that is not principal (but finitely generated), say $(p,q)$, what can I conclude about $p$ and $q$? Can I say they are prime? In an integral domain, can I say they are irreducible? What additional criteria are needed to make these conclusions?
What about the other way, if $p$ and $q$ are prime, then can I conclude $(p,q)$ is a prime ideal? What additional criteria are needed?
I'm also interested in extending these ideas outside of domains and to noncommutative rings, but it's more important to understand them in an integral domain first.
 A: I'm skipping the questions about irreducibles. Too many questions here.
If you take the typical example: $R=\mathbb Z[\sqrt{-5}]$ where:
$$2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\tag{1}$$
Then the ideal, $I=(2,1+\sqrt{-5})$ is not principle, and it is prime - it is easy to show that $R/I\cong \mathbb Z/2\mathbb Z$, which is a field, so $I$ is maximal, hence prime. 
You also have that $2,3,1\pm \sqrt{-5}$ are irreducible, but not prime.
It is also easy to show that no element of $I$ is prime, since you can show that $R/(p)$ is finite for any $p\in I$, and if $(p)$ is prime, $R/(p)$ is an integral domain, and a finite integral domain is a field. Thus $(p)$ is maximal. But $(p)\subsetneq I\subsetneq R$, so $(p)$ is not maximal.
So there are times where you can't find any prime element of a prime ideal. 
In $\mathbb Z[x]$, $x^2+1$ and $5$ are prime, but $(x-2)(x+2)=(x^2+1)-5\in (5,x^2+1)$. So, no, if $p,q$ are prime, then it is not necessarily true that $(p,q)$ is prime.
According to Wikipedia, in UFDs, every irreducible element is prime (which practically goes without saying,) but not visa-versa. The converse is true when the ring satisfies the "ascending chain condition for principle ideals."
