I am reviewing some algebraic number theory, and found myself confused about two seemingly distinct notions of discriminant that are used.
Let $K$ be a field, and $L$ a finite separable extension. We can define the trace pairing $\mathrm{Tr}_{L/K}:L \times L \rightarrow K$ which sends $(\alpha, \beta) \to \alpha \beta$, which is nondegenerate. Then the discriminant of $L/K$ is the discriminant of the trace pairing as a symmetric bilinear form.
For any $\beta_1, \dots, \beta_n \in L$ we can define $$\mathrm{D}(\beta_1, \dots, \beta_n) = \det((\mathrm{Tr}_{L/K}(\beta_i \beta_j))_{ij})$$
and if the $\beta_i$ are a basis for $L$ as a $K$-vector space, then we recover the discriminant.
We can also define for an extension $B/A$ of rings, where $B \cong A^n$ is a finitely generated free $A$-module (and I think one can even define this when $B$ is finite projective), a trace pairing $\mathrm{Tr}_{B/A}:B \times B \rightarrow A$ analogously to the above. This too has a discriminant, relative to an $A$-basis of $B$.
When $A$ be a PID with quotient field $K$, $L$ a finite separable extension of $K$, and $B$ the integral closure of $A$ in $L$, then any finitely generated, nonzero $B$-submodule of $L$ is a free $A$-module of rank $[L:K]$.
My question is the following: how do we relate $\mathrm{disc}_{L/K}$ to $\mathrm{disc}_{B/A}$ in the above situation? In particular, the discriminant of a number field $K$ is defined to be the discriminant of $\mathscr{O}_K$ over $\mathbb{Z}$ relative to any basis. How does this compare to $\mathrm{disc}_{K/\mathbb{Q}}$, and to what extent can these constructions be generalized (e.g. to a relative setting)?
Best Answer
Your discriminant concepts depend on the choice of a basis.
Let $B$ be a commutative ring with a subring $A$ such that $B$ is a finite-free $A$-module. When $x_1,\ldots,x_n$ and $y_1,\ldots, y_n$ are two $A$-bases of $B$, let $y_j = \sum_{i=1}^n a_{ij}x_i$, so $(a_{ij})$ is the change of basis matrix from the $x$'s to the $y$'s. Then we have the matrix equation $$ ({\rm Tr}_{B/A}(y_iy_j)) = (a_{ij})^{\top}({\rm Tr}_{B/A}(x_ix_j))(a_{ij}), $$ so taking determinants tells us $$ {\rm disc}_{B/A}(y_1,\ldots,y_n) = (\det(a_{ij}))^2{\rm disc}_{B/A}(x_1,\ldots,x_n). $$ The number $\det(a_{ij})$ is a unit in $A$ since a change of basis matrix always has a determinant in $A^\times$. Thus the discriminants of two $A$-bases of $B$ are equal up to a unit square factor in $A$.
Example. For a field $F$, a nonconstant polynomial $f(x)$ in $F[x]$ has a discriminant, defined in terms of its roots, and this is a special case of the above construction using a power basis: when $B = F[x]/(f(x))$, $$ {\rm disc}_{B/F}(1,x,\ldots,x^{n-1}) = {\rm disc}(f(x)). $$
Returning to the general discussion, when we have a finite extension of fields $L/K$ and use $A = K$ and $B = L$, two $K$-bases of $L$ have discriminants that are equal up to a square factor in $K^\times$. That's why the discriminant of $L/K$, defined as the discriminant of any $K$-basis of $L$, is well-defined only modulo $(K^\times)^2$.
When $A = \mathbf Z$, its only unit square is $1$. Thus in a commutative ring $R$ that is a finite-free $\mathbf Z$-module (like the integers of a number field), all $\mathbf Z$-bases of $R$ have the same discriminant. This is why the definition of the discriminant of a number field $K$ as the discriminant of a $\mathbf Z$-basis of $\mathcal O_K$ is independent of the choice of basis.
In algebraic number theory, discriminants can be extended to the relative setting $L/K$, but something additional has to be done in the definition due to the bottom ring of integers $\mathcal O_K$ possibly not being a PID. In particular, the discriminant $\mathfrak d_{L/K}$ is an ideal in the bottom ring $\mathcal O_K$ rather than an element there.
One approach is to define the ideal $\mathfrak d_{L/K}$ by the multiplicities in its prime ideal factors. For a nonzero prime ideal $\mathfrak p$ in $\mathcal O_K$, the localization $\mathcal O_{K,\mathfrak p}$ is a DVR (hence PID) with fraction field $K$, so its integral closure in $L$ is a finite free $\mathcal O_{K,\mathfrak p}$-module of rank $[L:K]$. All $\mathcal O_{K,\mathfrak p}$-bases of the integral closure have the same discriminant up to a unit square factor, so they all generate the same ideal in $\mathcal O_{K,\mathfrak p}$, say $(\mathfrak p\mathcal O_{K,\mathfrak p})^{a_\mathfrak p}$. We have $a_{\mathfrak p} = 0$ when $\mathfrak p$ is unramified in $L$, so the product $$ \prod_{\mathfrak p} \mathfrak p^{a_{\mathfrak p}} $$ makes sense as an ideal and that is defined to be $\mathfrak d_{L/K}$. In other words, this approach is defining the global discriminant to be the product of local discriminants.
Lang's Algebraic Number Theory defines $\mathfrak d_{L/K}$ in another way in the chapter on the different and discriminant, as the ideal in $\mathcal O_K$ generated by discriminants of $K$-bases of $L$ that are contained in $\mathcal O_L$. He then proves his definition agrees with the one I gave above, so my definition above is a theorem in Lang’s approach.