In Zagier's 1-2-3 of Modular Forms, the following is written. For context, $\mathfrak{z}$ is a CM point in the upper-half plane, though this is not relevant for the question.
By definition, $\mathfrak{z}$ satisfies a quadratic equation over $\mathbb{Z}$, say $A \mathfrak{z}^2 + B\mathfrak{z} + C = 0$. There is then always a matrix $M \in M(2,\mathbb{Z})$, not proportional to the identity, which fixes $\mathfrak{z}$. (For instance, we can take $M = \big(\begin{smallmatrix}B & C \\ -A & 0 \end{smallmatrix}\big)$.)
This seems obviously false for me. Take $\mathfrak{z}$ to be $i$. Then $i^2 + 1 = 0$, yet obviously $i$ is not a fixed point of $\big(\begin{smallmatrix}0 & 1 \\ -1 & 0 \end{smallmatrix} \big)$. What am I missing?
Recall that the group $\operatorname{SL}_2(\mathbb{Z})$ acts on the upper half-plane by Möbius transformations: $$ \begin{pmatrix} a & b\\ c & d \end{pmatrix} \cdot z = \frac{az+b}{cz+d} \, . $$ If $A {z}^2 + B{z} + C = 0$, then \begin{align*} 0 = A {z}^2 + B{z} + C \iff -Az^2 = Bz+C \iff z = \frac{Bz+C}{-Az} = \begin{pmatrix} B & C\\ -A & 0 \end{pmatrix} \cdot z \end{align*} so $z$ is fixed by the given matrix as claimed. ($z \neq 0$ since it's in the upper half-plane.) For your example, we have $$ i = -\frac{1}{i} = \frac{0 \cdot i + 1}{-1 \cdot i + 0} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \cdot i $$ so $i$ is a fixed point of $\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$.