I have a question on a remark of Serre in chapter III §2 of his book Local Fields (p. 49). In this section he wants to define the discriminant of a lattice with respect to a bilinear form.
The setting is as follows: we have a Dedekind domain $A$ and its fraction field $K$. We suppose that $V$ is a $K$-vector space of finite dimension $n$, and we put $W = \bigwedge^n_K V$. Now let $X$ be a lattice of $V$, i.e., an $A$-submodule that is finitely generated and that spans $V$ (note that $X$ does not need to be a free $A$-module).
Serre claims that the $n$-th exterior power $\bigwedge_A^n X$ can be identified with a lattice of $W$. For me it is not clear why the map $\bigwedge_A^n X\to \bigwedge_K^n V$ is injective.
To be clear, the latter map is constructed in the following way. We have the map $X\to V\to \bigwedge_K V$ that is the composition of the inclusion map and the canonical injection into the exterior algebra. Then using the universal property of tensor algebras we find that this map factorises through $X\to \bigwedge_A X$. The obtained map is a homomorphism of graded algebras so taking the $n$-th component we get the desired map.
In fact, this is an example where base change is applied: from $A$ to its localisation $S^{-1}A = K$ (with $S = A-\{0\}$). One can proof that $\bigwedge_K^n V$ is the localisation of $\bigwedge_A^n X$ with respect to the same multiplicative set. So it seems to me that an equivalent statement is that $\bigwedge_A^n X$ has no $A$-torsion. But why this is the case?
Any explanation about why Serre's identification is allowed will be appreciated. Thanks in advance!