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Why is the degree of a linear holomorphism $f:X\rightarrow X$ such that $[z] \mapsto [cz]$ from the complex torus $X=\mathbb{C}/\Lambda$ to itself equal to |c|^2? (here we assume that $\Lambda=\mathbb{Z}+\tau \mathbb{Z}$ with $\operatorname{Im}(\tau)>0$, $c\Lambda \subseteq \Lambda$, and $[z]$ designs the equivalence class of $z$ in $X$).

This is mentioned on this answer, but I haven’t been able to find the reason by myself.

I’ve tried to compute it by using $\deg(f)=\sum_{p\in f^{-1}({z})}\operatorname{mult}_p(f)$ where $f([z])=[cz]$ and hence $f^{-1}([z])=[c^{-1}z]$, but I’m stuck at trying to find $\operatorname{mult}_p(f)$ for the points in $f^{-1}([z])$ here.

Edit: I already understand why $\operatorname{mult}_p(f)=1$ for all $p\in f^{-1}([z])$ but I don’t see why should $|f^{-1}([z])|=|c|^2$.

Gokimo
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    Well, if it actually were a biholomorphism, its degree would be one… however, your map is a local biholomorphism, so its multiplicity at any point is one. – Aphelli Apr 27 '23 at 21:27
  • @Aphelli Sorry, it wasn’t a biholomorphism, just a holomorphism. I edited the question. – Gokimo Apr 27 '23 at 22:45
  • @Aphelli But then, I should find that $f^{-1}([z])$ has cardinal $|c|^2$? – Gokimo Apr 27 '23 at 22:54
  • Yes. You should start by showing that $f^{-1}([z])$ is the determinant of $c$ as an endomorphism of $\Lambda$. – Aphelli Apr 28 '23 at 06:33
  • @Aphelli Sorry, but I don’t really see how to do that… I guess the endomorphism would be represented by the matrix ${{c,0},{0,c}}$ with respect to the $\mathbb{C}$ basis given by ${1,0}:=1$, ${0,1}:=i$, and its determinant would then be $|c|^2$? But I don’t see how that’s related to $f^{-1}([z])$… – Gokimo Apr 28 '23 at 09:09
  • $f^{-1}(0)$ is exactly $c^{-1}\Lambda/\Lambda$, which is isomorphic to $\Lambda/c\Lambda$, whose cardinality is exactly the determinant of $c$ acting on $\Lambda$. Now consider the characteristic polynomial $\chi(x)$ of $c$ acting on $\Lambda$… – Aphelli Apr 28 '23 at 09:57
  • @Aphelli I get $\chi(x)=(x-c)^2$, but I’m not sure what to do now… – Gokimo Apr 28 '23 at 10:47

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First, note that $f$ is a local biholomorphism. Therefore, its degree is the cardinality of $f^{-1}([0])$.

Now, $f^{-1}([0])=c^{-1}\Lambda/\Lambda \cong \Lambda/c\Lambda$.

Let $\chi(x)=x^2+ax+d$ be the characteristic polynomial of the endomorphism $c$ acting on $\Lambda$: then clearly $\Lambda/c\Lambda$ has cardinality $d$, so we need to show that $d=|c|^2$.

This result is clear if $c \in \mathbb{Z}$, so let’s assume that it’s not the case. Since $\chi(c)=0$, $c$ is an algebraic integer contained in a quadratic extension $K$ of $\mathbb{Q}$.

Let $(u,v)$ be a basis of $\Lambda$, then it’s easy to see that $\mathbb{Q}(v/u)=\mathbb{Q}(c)$, hence that $c \notin \mathbb{R}$.

Thus, $d$ is the product of $c$ and its only conjugate over $\mathbb{Q}$ – but its conjugates include $\overline{c}$, so that $d=c\overline{c}=|c|^2$.

Aphelli
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  • Thank you! I’m trying to follow step by step. Why is the cardinality of $\Lambda/c\Lambda$ clearly $d$…? – Gokimo Apr 28 '23 at 15:30
  • Let $M \in \mathcal{M}_2(\mathbb{Z})$ have positive determinant. Then there are $P,Q \in GL_2(\mathbb{Z})$ such that $D=PMQ$ is diagonal with coefficients $a\mid b$ with $a,b >0$. Then it’s clear that $\mathbb{Z}^2/M\mathbb{Z}^2 \cong \mathbb{Z}^2/D\mathbb{Z}^2$ has cardinality $ab=\det{M}$. – Aphelli Apr 28 '23 at 15:57
  • Thank you, I will try to understand the rest of the steps. – Gokimo Apr 28 '23 at 19:57