Chapter V of Serre’s book A Course in Arithmetic is titled ‘Integral Quadratic Forms with Discriminant $\pm1$.’ In the first section, he defines an object $E$ in a category $S_n$ (indexed by integers $n>0$) as a free abelian group of rank $n$ together with a bilinear form $B\times B\rightarrow\mathbf Z$, denoted by $(x,y)\mapsto x.y$, such that the homomorphism of $E$ into $\operatorname{Hom}(E,\mathbf Z)$ defined by the form $x.y$ is an isomorphism; if $(e_i)$ is a base of $E$, and if $a_{ij}=e_i.e_j$, this is equivalent to the determinant of the matrix $A=(a_{ij})$ being equal to $\pm1$. He then goes on to say that if $E\in S_n$, the function $x\mapsto x.x$ makes $E$ a quadratic module over $\mathbf Z$ (with respect to the quadratic form $f(x)=x.x$), which he defines in the previous chapter. I have trouble seeing how $f$ is a quadratic form when the bilinear form we started with is not assumed symmetric. For $f$ to be a quadratic form, the function $(x,y)\mapsto f(x+y)-f(x)-f(y)=x.y+y.x$ must be a bilinear form. Indeed it is, but it need not coincide with $2x.y$ unless the bilinear form we started with were symmetric. He goes on to write that $f(x)=x.x=\sum_ia_{ii}x_i^2+2\sum_{i<j}a_{ij}x_ix_j$, where $a_{ij}=e_i.e_j$. Again I see why this is true when the bilinear form we started with is symmetric. But when it is not symmetric, it seems as though $a_{ij}$ need not equal $a_{ji}$.
(Of course, you can still write down a quadratic form starting from the bilinear form $(x,y)\mapsto \frac12(x.y+y.x)$, as is done in Chapter IV, but this need no longer be a $\mathbf Z$-valued form.)
My question is, is it true that the function $f$ always makes $E$ into a quadratic module over $\mathbf Z$ in the above way, when the bilinear form $E\times E\rightarrow\mathbf Z$ is not assumed symmetric?