Let $A$ be a Dedekind domain, $K= Frac(A)$ its field of fractions and
$V$ a $n$-dimensional vector space over $K$.
A lattice of $V$ (with respect to ring $A$) is a sub-$A$-module $X$ of $V$
that is finitely generated and spans $V$.
Let $n= \dim_K V$ and assume $V$ is endowed with a non-degenerate bilinear form $T(v, w)$. It is known that $T$ extends to a non-degenerate bilinear form (again denoted by $T$) on the exterior algebra of $V$, and, in particular, on $W:= \wedge^n V$. This form induces by non-degeneracy and dimensional reasons an isomorphism
$$ T: W \otimes_K W \to K. $$
Let again $X$ be a lattice of $V$, and let $X_W$ be it' s $n$-th exterior power, identified with a lattice of $W$. The image of $X_W \otimes_A X_W$ under $T$ is a non-zero fractional ideal of $K$, which is called the discriminant of $X$ with respect to $T$; we denote it by $D_{X,T}$.
In Serre's 'Local Fields' at page 49 is remarked:
The above definition shows that $D_{X,T}$ is isomorphic as an $A$-module to $X_W \otimes_A X_W$; its ideal class (modulo the principal ideals) is thus a square.
I not understand it. Why is the class of $D_{X,T}$ in the https://en.wikipedia.org/wiki/Ideal_class_group
a square. The ideal class group is the quotient group $J_K/P_K$ where $J_K$ is the group of fractional ideals of the ring $A$, and $P_K$ is its subgroup of principal ideals with operation = ideal multiplication. Therefore the claim is equivalent to existence of an nonzero element $a \in A$ and a fractional ideal $\frak{f}$ such that $D_{X,T}= \frak{f}^2 \cdot (a)$ where $(a)= a \cdot A$.
The short answer is that the ideal class group of $A$ is equivalently its Picard group. This is to say, it is the group of $A$-modules which are locally free of rank $1$ (i.e. modules $M$ s.t. $M_{\mathfrak p}\simeq A_{\mathfrak p}$ for all prime ideal $\mathfrak p\subset A$) with group operation given by tensor products. Hence, saying that $D_{X,T}\simeq X_W\otimes_AX_W$ exactly says that is is a square in the Picard group.
In terms of fractional ideals, fix some nonzero $x\in X_W\subset W$, and let $\mathfrak f=T(X_W\otimes x)\subset K$. Note that $\mathfrak f$ is a fractional ideal. I claim that $\mathfrak f^2=D_{X,T}\cdot(a)$, where $a:=T(x\otimes x)$, inside $K$. It suffices to check this equality locally, i.e. to check that $(\mathfrak fA_{\mathfrak p})^2=D_{X,T}\cdot aA_{\mathfrak p}\subset K$ for all primes $\mathfrak p\subset A$.
Hence, we may assume that $A$ is a dvr (and in particular a PID). Then, $X_W$ is finitely generated and torsion free over the PID $A$, so free as an $A$-module; now, $X_W\otimes_AK=W$ is rank $1$ over $K$, so we conclude that $X_W$ is rank one over $A$. Hence, $\mathfrak f^2$ is generated by elements of the form $$T(\lambda x\otimes x)T(\mu x\otimes x)=\lambda\mu T(x\otimes x)^2=\lambda\mu T(x\otimes x)\cdot a\in D_{X,T}\cdot(a)$$ with $\lambda,\mu\in A$, so $\mathfrak f^2\subset D_{X,T}\cdot(a)$. In the other direction, $D_{X,T}\cdot(a)$ is generated by elements of the form $$T(\lambda x\otimes\mu x)\cdot a=T(\lambda x\otimes\mu x)T(x\otimes x)=\lambda\mu T(x,x)^2=T(\lambda x\otimes x)T(\mu x\otimes x)\in\mathfrak f^2,$$ with $\lambda,\mu\in A$, so also $D_{X,T}\cdot(a)\subset\mathfrak f^2$.