# Thursday, September 29

## Bilinear Forms

:::{.remark}
Recall that given $R\in\CRing, M\in \rmod^{\free,\fg}$ and a bilinear form $\ip{\wait}{\wait}:M\tensor_R M\to R$, chose a basis $\ts{e_i}$ of $M$ to form the Gram matric $g_e \in \Mat_{n\times n}(R)$ where $G_e(i, j) =\ip{e_i}{e_j}$.
This basis induces an isomorphism $M\cong R^n$ where $\ip{v}{w} = v^t G_e w$.
Changing basis to $\ts{f_i}$, there exists a $P\in \GL_n(R)$ such that $Pe_i = f_i$.
Moreover,
\[
\ip{f_i}{f_j} 
&= \ip{Pe_i}{Pe_j} \\
&= (Pe_i)^t G_e Pe_j \\
&= e_i^t P^t G_e P e_j \\
&\implies G_f = P^t G_e P
,\]
so any two Gram matrices are *similar* (note that this isn't conjugacy unless $P$ is orthogonal).
Also note that 
\[
\det(G_f) = \det(P^t G_e P) = \det(P^t) \det(G_e) \det(P) = \det(P)^2 \det(G_e)
.\]
:::

:::{.definition title="Discriminant"}
Define the **discriminant** $\delta$ of the bilinear module $(M, \ip\wait\wait)$ to be the class of $\det(G_e)$ in $(R, \cdot)/ (R\units)^\squares$.
:::

:::{.remark}
Note that if $R=\ZZ$, the units are $\pm 1$ and the squares are just $1$, so $\delta$ is a well-defined integer.
:::

:::{.remark}
Henceforth assume $\ip\wait\wait$ is symmetric and $M$ is finitely generated free.
This allows defining duals 
\[
\Phi: M &\to M\dual \da \Hom_R(M, R) \\
m &\mapsto \ip m\wait
.\]
We'll say the form is **nondegenerate** if $\Phi$ is injective, and **perfect** or **unimodular** if it is an isomorphism.
Note that these coincide over a field, but the latter is generally stronger.
:::

:::{.proposition title="?"}
Let $R$ be a domain and $M\in\rmod^{\fg,\free}$ with $\ip\wait\wait:M\tensorpower{R}{2}\to R$ and $\ts{e_i}$ an $R\dash$basis of $M$.

a. TFAE[^same_proof]
  1. $\Phi: M\injects M\dual$ is injective (so nondegenerate).
  2. $\det G_e\neq 0$.

b. TFAE:
  1. $\Phi: M\iso M\dual$ is an isomorphism (so perfect).
  2. $\det G_e \in R\units$ (so $G_e$ is **unimodular**).
    Equivalently, $\exists G_e\inv \in \Mat_n(R)$.[^adjugate]
  3. There exist a dual basis $\ts{e^i}$ for $M$ such that $\ip e_i e^j = \delta_{ij}$.


[^same_proof]: 
The proof of this equivalence is essentially the same, just base change to $\ff(R)$.

[^adjugate]: 
The last two are equivalent by looking at $A A^{\mathrm{adj}} = \det(A) I$.

:::

:::{.proof title="of part b"}
$1\implies 3$: 
If $\Phi$ is an isomorphism, take $e^j \da \Phi\inv(e_j\dual)$.

$3\implies 1$:
If $l \in M\dual$ and $x\in M$, take $l(x) = \ip{x}{\sum l(e_i)e^i}$.

$3\implies 2$:
Given $\ts{e^i}$, define $H = (e_1^t, \cdots, e_n^2) \in \Mat_{n\times n}(R)$ and one can check $H = G_e\inv$.

$2\implies 3$:
If $G_e$ is invertible, then take $e^j$ to be the $j$th column of $G_e\inv$.

:::

:::{.remark}
If $M\in \rmod^{\fg, \free}$ and $\ip\wait\wait\in \Sym\Bil(M\slice R)$, if $f \in \CRing(A, B)$ one can base change it to obtain $\ip{\wait}{\wait}_B \in \Sym\Bil(M\tensor_A B\slice R)$$ defined by 
\[
\ip{v\tensor b}{w\tensor b_2}_B \da b_1 b_2 f(\ip v w)
.\]
Picking a basis $\ts{e_i}$ one has $\ip{e_i\tensor 1}{e_j\tensor 1} = f(\ip{e_i}{e_j})$, so the new $G_e$ is obtain from the old one by applying $f$ to every entry.
Thus
\[
\disc(M\tensor_A B, \ip\wait\wait _B) = f(\disc(M, \ip\wait\wait ) )
.\]
:::

## Extensions

:::{.remark}
Let $V \da M\tensor_R K$ where $K = \ff(R)$, let $\inp\wait\wait\in \Sym\Bil(V\slice k)$.
Then there is a restriction $\ro{\inp\wait\wait}{M} \in \Sym\Bil(M\slice k)$ with $G_e\in \Mat_{n\times n}(K)$, and is in $\Mat_{n\times n}(R) \iff \inp{M}{M} \subseteq R$ (i.e. the form is **integral**).
Its discriminant satisfies $\delta\in K/(R\units)^\squares$, and if it's nondegenerate then $\delta \in K\units/(R\units)^\squares$ and yields a well-defined principal fractional ideal$\gens{\disc}$.
:::