About symmetric inner product space

Question

How lattice $L = V \otimes _k k[x] $ looks like? $L$ is free over $k[x]$ because $V$ is free over $k$ ; why this statement is true? Also I am trying to prove every unimodular $k[x]$ lattice has form $V \otimes _k k[x]$ how this fact is related to orthogonal basis ? How this statment looks like in matrix version?

Answer 1

Let $V$ be a vector space over $k$ equipped with a nondegenerate bilinear form $B$. Then $L=V\otimes_k k[x]$ is free and $B$ extends to a nondegenerate bilinear form $\tilde{B}$ on $L$ with the same determinant as $B$ (in particular it is an element of $k^\times$).

Proof.

$L$ is free because by choosing a basis $v_1,\ldots,v_n$ we get that $$V\cong k^n$$ for some $n$, so $$L=V\otimes_k k[x] \cong k^n\otimes_k k[x] \simeq k[x]^n.$$

Under this isomorphism the basis vectors for $L$ are $\ell_i=v_i\otimes 1$.

Now $B$ defines a bilinear form on $L$ by $\tilde{B}(v\otimes p,w\otimes q) = B(v,w)pq$.

Then $$\tilde{B}(\ell_i,\ell_j)=\tilde{B}(v_i\otimes 1,v_j\otimes_1)=B(v_i,v_j).$$ Thus $\det(\tilde{B}(\ell_i,\ell_j))=\det(B(v_i,v_j))$.

Every unimodular $k[x]$-lattice has this form

Edit: My prior "proof" had an error. This is what's proven in the Theorem in the OP.

Edit 2:

To show how the theorem in the question proves every unimodular lattice has this form we do the following.

Let $\ell_1,\ldots,\ell_n$ be the orthogonal $k[t]$ basis given by the theorem. Define $V=\{\sum_i c_i\ell_i : c_i\in k\}$ and define $B_0$ on $V$ by restricting $B$ to $V$. This is where we use orthogonality! The restriction is well-defined because the $\ell_i$ are orthogonal, so $B(\ell_i,\ell_j)=c_i\delta_{ij}$ for some constants $c_i\in k^\times$.

Then define $V\otimes_k k[t] \to L$ by $\ell_i\otimes p \mapsto p\ell_i$. Since the $\ell_i$ formed a $k[t]$ basis for $L$, this map is an isomorphism of $k[t]$ modules.

Then all we need to check is that the isomorphism sends $\tilde{B_0}$ to $B$. We do this by observing that $$\tilde{B_0}(\ell_i\otimes p,\ell_j\otimes q) = B_0(\ell_i,\ell_j)pq=B(\ell_i,\ell_j)pq=B(p\ell_i,q\ell_j),$$ as desired.