In the section "Over discrete rings" in this Wikipedia page, it is remarked that
Similarly, a homography of P(Q) corresponds to an element of the modular group, the automorphisms of P(Z).
I asked someone who knows more than me and they said they could not find a reliable source for this remark-- that it might not be true. In addition, I found a question asking about this line; in the answers, it said that the entire page was written strangely and that the line was later deleted (but I guess it was re-added to the page?).
Specifically, I'm asking about the last part. Is $\text{SL}(2,\mathbf Z)$ isomorphic to the automorphism group of the projective line over the integers?
The projective line $\mathbb{P}^1\mathbb{Z}$ over the integers is defined as the set of all pairs of integers modulo the equivalence relation induced by by nonzero integer scalar multiples. Its automorphism group is the modular group, meaning $PSL_2(\mathbb{Z}),$ with trivial center, not $SL_2(\mathbb{Z}),$ which has a center of size 2.
To see this, consider the images $(a;c)$ and $(b;d)$ of $(1;0)$ and $(0;1),$ respectively, under an automorphism of $\mathbb{P}^1\mathbb{Z}.$ $$\begin{pmatrix}a & b\\\ c & d\end{pmatrix}$$ must have determinant 1 if I have written both $(a;c)$ and $(b;d)$ in lowest terms: otherwise, surjectivity would fail; one can show this by showing that $(1;0)$ and $(0;1)$ would not end up in the range by the Euclidean algorithm. Furthermore, the whole automorphism is carried out by this matrix.
One can use the matrices $$S=\begin{pmatrix}0 & -1\\\ 1 & 0\end{pmatrix}\mathrm{\quad\quad and\quad\quad}T=\begin{pmatrix}1 & 1\\\ 0 & 1\end{pmatrix}$$ and their inverses to generate all of $SL_2(\mathbb{Z}),$ or alternatively to carry out the Euclidean algorithm to find the greatest common factor of two numbers. Finally, $-I$ has a trivial action on the projective line, so we quotient out by the center of $SL_2(\mathbb{Z})$ to get a faithful action.