Say I have a ring R, is there any general way to find out how many ideals it has? I know that if it's a field then there are only 2 ideals, namely (0) and (1), however what if the ring is not a field, then how can I determine how many ideals it contains?
EDIT: Simplify the ring into $\mathbb{Z}/n\mathbb{Z}$ then look at the divisors and that is the number of ideals.
So, if I have a ring $\mathbb{Z}[x]/(6,2x-1)$. Then I can reduce this down into $\mathbb{Z}/n\mathbb{Z}$ and find the number of ideals from this (which will then be 2 and be a field), but how does $\mathbb{Z}[x]/(6,2x-1)$ become $\mathbb{Z}[x]/(3,2x-1)$ and then how does this become $\mathbb{Z}/3\mathbb{Z}$? I then see once it is $\mathbb{Z}/3\mathbb{Z}$ that it will have 2 ideals, (0) and (1), so it will be a field. But I am lost on the simplification of the ring (field) into $\mathbb{Z}/3\mathbb{Z}$.
As I mentioned in the comments, the ideals of the quotient ring $R/I$ are in one to one correspondence with the ideals of $R$ containing $I$.
In your example of $\mathbb Z[x]/(6,2x-1)$, You now have to find the ideals of $\mathbb Z[x]$ containing $(6,2x-1)$. Those are just $\mathbb Z[x]$, $(2,2x-1)$ , $(3,2x-1)$ and $(6,2x-1)$ because $2x-1$ is irreducible. But $(2,2x-1)$ is the entire ring. Similarly, $(6,2x-1)=(3,2x-1)$. So your quotient ring has just two ideals, which is the bare minimum. So the quotient is in fact a field!