# Monday April 6th

Note that most of the theory thus far has not relied on the properties of $\CC$, so Jantzen's strategy was to extend the base field to $K = \CC(T)$, rational functions in $T$, then setting $g_K \definedas K \tensor_\CC \lieg$.
The theory over $K$ adapts to $A = \CC[T]$, the PID of polynomials in one variable $T$ with $K$ as its fraction field and the "Lie algebra" $g_A = A \tensor_\CC \lieg$.

Setup:
Let $A$ be any PID, for example $\ZZ$ or $\CC[T]$, and $M$ a free $A\dash$module of finite rank $r$.
Let $e, f\in M$ and suppose $M$ has an $A\dash$valued symmetric bilinear form denoted $(\wait, \wait)$.
Since $M$ is finite rank, we can choose a basis $\theset{e_i}^r$, so the matrix $F$ of this form relative to this basis has nonzero determinant $D$ depending on the choice of basis.
A change of basis is realized by some $P \in \Gl(r, A)$, giving $F' = P^t F P$ (note that forms change by a transpose instead of an inverse) and $\det P \in A\units$.
Thus $D$ only changes by a unit $u = \qty{\det P}^2$.

We can define the dual module $M^* = \hom_A(M, A)$ which is also free of rank $r$, and contains a submodule $M\dual$ consisting of functions $e\dual: M \to A$ given by $e\dual(f) = (e, f)$ for any fixed $e\in M$.
Surprisingly, this doesn't give every hom: e.g. if the form only has even outputs.
Since $(\wait, \wait)$ is nondegenerate, the map $\phi: M\to M\dual$ sending $e$ to $e\dual$ is an isomorphism.

We'll now invoke the structure theory of modules over a PID: There exists a basis of $M^*$ given by $\theset{e+i^*}^r$ where $M\dual$ has a basis $\theset{d_i e_i^*}^r$ for some scalars $0\neq d_i \in A$.
We can define a dual basis of $M$ given by $\theset{e_i}^r$ where $e_i^*(e_j) = \delta_{ij}$.
We can similarly get a separate dual basis $\theset{f_i}$ where $f_i\dual = d_i e_i^*$.

We can compare these two bases:
$$
(e_i, f_j) = f_j\dual(e_i) = d_j e_j^* (e_i) = d_j \delta_{ij}
$$
(Formula 1)

Thus up to units, $D = \prod_{i=1}^r d_i$, so this hybrid matrix is one way to compute this determinant.

Fix a prime element in $A$, then there is an associated valuation $v_p: A \to \ZZ^+$ where $v_p(a) = n$ if $p^n \divides a$ but $p^{n+1}\notdivides a$.
Since $p$ is prime, $\bar M \definedas M/pM$ makes sense and is a finitely generated module over the field $\bar A = A/pA$; thus $\bar M$ is a vector space.

We'll now define a filtration: for $n\in \ZZ^+$, define $M(n) = \theset{e\in M\suchthat (e, M) \subset p^n A}$.
Then
$$
M = M(0) \supset M(1) \supset \cdots
$$
is a decreasing filtration, with corresponding images $\bar{M(n)}$ that are vector spaces.

Lemma
:   For $A$ a PID, $p\in A$ prime, $\bar A = A/pA$ with valuation $v_p$ and $M$ a free $A\dash$module with a nondegenerate symmetric bilinear form wrt some basis of $M$ having nonzero determinant $D$
    Then

    a. $v_p(D) = \sum_{n > 0} \dim_{\bar A} \bar{M(n)}$.
    b. For each $n$, the modified bilinear form $(e, f)_n \definedas p^{-n} (e, f)$ on $M(n)$ induces a nondegenerate form on $\bar{M(n)} / \bar{M(n+1)}$.

Proof (of (a))
:   \hfill

    1. For $f\in M$, write $f = \sum a_ij f_j$ in terms of the given basis, and $(e_i, f) = a_i d_i$.
      For $n> 0$, we have
      \begin{center}
      \begin{align*}
      f \in M(n) & \iff v_p((e_i, f)) \geq n \forall i \\
      & \iff v_p(a_i d_i) \geq n \\
      & \iff v_p(a_i) + v_p(d_i) \geq n \\
      & \iff v_p(a_i) \geq n - n_i \quad n_i \definedas v_p(d_i)
      \end{align*}
      \end{center}

      This $a_i$ must be divisibly by $p$ at least $n-n_i$ times.
      This $M(n)$ is spanned by $\theset{f_i \suchthat n_i \geq n} \union \theset{p^{n-n_i}f_i \suchthat n_i < n}$.

    2. So $\bar{M(n)}$ has basis $\theset{\bar f_i \suchthat n_i \geq n}$ and $\dim \bar{M(n)} = \# \theset{i \suchthat n_i \geq n}$.
      In particular, $\bar{M(n)} = 0$ for $n\gg 0$ since there are only finitely many $n_i$.
      Thus

      \begin{center}
      \begin{align*}
      \sum_{n > 0} \dim \bar{M(n)}
      &= \sum_{n > 0} \# \theset{i \suchthat n_i > n} \\
      &= \sum_{i=1}^r n_i \\
      &= \sum_{i=1}^r v_p (d_i) \\
      &= v_p \qty{ \prod_{i=1}^r d_i } \\
      &= v_p(D)
      \end{align*}
      \end{center}

Proof ( of (b) )
:   \hfill

    1. Note that $(e, f) \in p^n A \implies (e, f)_n \in A$, so this is well-defined on $M(n)$.
      To see that it's well-defined on $\bar{M(n)}$ we must show that $e \in p M(n)$.
      $$
      (e, pM(n))_n \subset pA \\
      \implies (e, M(n))_n \subset p^{-n}(pM, M(n)) \subset p^{-n+1}(M, M(n)) \subset p^{-n+1}p^n A = pA
      $$
      So there is an induced form $(\bar e, \bar f)_n$ on $\bar{M(n)}$.

      To show it's nondegenerate, need to compute the radical.

      - If $f\in M(n+1)$ then
        $$
        (f, M(n))_{n}=p^{-n}(f, M(n)) \in p^{-n}(f, M) e p^{-n} p^{n+1} A=p A
        $$
      
        so $\bar f \in \rad (\wait, \wait)_n$

      - See notes.