Question in the proof of Theorem 7.4 from The Arithmetic of Elliptic Curves 
For $M\subseteq\mathrm{Hom}(E_1,E_2)$ finitely generated subgroup, the degree map, $\deg: M\rightarrow \mathbb{Z}$ is extended to $M\otimes \mathbb{R}$. It's probably obvious, but yet I'm unable to understand how it is extended. 

I would guess that it is defined as $\deg(\sum_i \phi_i\otimes r_i)=\sum_i r_i\deg(\phi_i)$. But I do not know how to claim that this map is well defined, as $\deg:M\rightarrow \mathbb{Z}$ was not a map of $\mathbb{Z}$-modules to begin with.
Moreover with this definition I would expect that $$n^2\deg(\phi)=\deg([n]\circ\phi)=\deg(n\phi)=\deg(\phi\otimes n)= n\deg(\phi)$$ which is not correct.
Apologies if it is obvious or I'm making some silly error, but how is the map being extended here?
EDIT: Following Quasicoherent's comment below, one can try defining $\deg(\phi\otimes r)=r^2 \deg(\phi)$, but I am still unable to check that this map is well defined. Any help would be appreciated.
 A: My professor told me this is how it is probably being extended:
Fact: $K$ be a field of characteristic different from 2. Then there is a 1-1 correspondence between quadratic forms and symmetric bilinear forms on a finite dimensional vector space over $K$ given by: $Q_B(x)=B(x,x)$ and $B_Q(x,y)=1/2(Q(x+y)-Q(x)-Q(y))$. Moreover this correspondence preserves the property of being positive definite.
Now $deg:M\rightarrow \mathbb{Z}$ is a quadratic form(from corollary III.6.3 of the book). We can define using this, a bilinear form $B=2B_{deg}:M\times M \rightarrow \mathbb{Z}$(multiplying by 2 so that it lands in $\mathbb{Z}$, no 1-1 correspondence being used here we are just defining a bilinear form using the above formula). 
Choosing a basis $\{e_i\}$ for $M$, we note that $B$ can be written as a matrix. Choosing $\{e_i\otimes 1\}$ as a basis for the vector space $V=M\otimes \mathbb{R}$, define a bilinear form on $V$ by the matrix $A=B/2$(can divide by 2 in $\mathbb{R}$). Then by the correspondence above, we get a quadratic form $Q_A$ on $V$ which restricts to the original quadratic form, deg, on $M\subseteq V$. Moreover $Q_A$ is continuous as it is given by $Q_A(x)=x^t(B/2)x$ for a column vector $x\in (V,\{e_i\})\simeq (\mathbb{R}^n,\{e_i\})$ and is thus a polynomial in the coordinates.
