I'm probably in over my head, but I came across the following sentence in a thesis by Evan Oliver entitled "Congruence Subarrangements of the Schmidt Arrangement":
"The modular group is the set of all matrices in P SL2(Z) whose elements reduce component-wise to the identity matrix over a prime ideal."
I know what a prime ideal is, and I have a basic understanding of the modular group, but the sentence above confounds me. His statement comes from a definition of PSL(2,Z) whereby
a ≡ d ≡ 1 (mod p) and b ≡ c ≡ 0 (mod p).
but that is not a definition of PSL(2,Z) that I'm familiar with. Is it related to the unit determinant?
Check out Wikipedia here under "Congruence Subgroups". Basically, when he talks about the "Modular Group" he is actually talking about $\Gamma(p)$, which is the kernel of the natural homomorphism $\phi_{p}: \mathrm{PSL}(2,\mathbb{Z}) \to \mathrm{PSL}(2,\mathbb{Z}_{p})$ (for some prime $p$).
Wikipedia is using the non-projective version of the group, so they mention that $\Gamma(N)$ is the set of matrices with $a \equiv d \equiv \pm 1 \bmod{N}$ and $b \equiv c \equiv 0$. But when we are working projectively, we can always scale the matrix so that $a \equiv d \equiv 1$.
(The more I read the Wikipedia article, the less I like it because $-I$ is not in the kernel when considering $\mathrm{SL}(2,\mathbb{Z}) \to \mathrm{SL}(2,\mathbb{Z}_{p})$ by reduction $\bmod{p}$, but this problem goes away when you consider the projective groups.)