Characteristic-zero scalar extension of geometric semisimplicity

Let $A\subseteq {\mathrm{End}}_{\mathbb{Q}}(V)$ be the $\mathbb{Q}$-span of the monodromy matrices. It contains the identity and is closed under multiplication, because a product of monodromy matrices is a monodromy matrix, so it is a finite-dimensional unital $\mathbb{Q}$-algebra. For every field $E$ over $\mathbb{Q}$ the canonical map $A{\otimes }_{\mathbb{Q}}E\to {\mathrm{End}}_E(V{\otimes }_{\mathbb{Q}}E)$ is injective: rationally independent matrices stay independent after extending scalars, as one sees by comparing coordinates in a rational basis of $V$. Its image is the $E$-span of the monodromy matrices, so submodules for this image are exactly the monodromy-invariant subspaces of $V{\otimes }_{\mathbb{Q}}E$; in particular the action of $A_{\mathbb{C}}=A{\otimes }_{\mathbb{Q}}\mathbb{C}$ on $V_{\mathbb{C}}$ is faithful.

Assume $V_{\mathbb{C}}$ is semisimple. Let $J=J(A_{\mathbb{C}})$. If $T\subseteq V_{\mathbb{C}}$ is a simple submodule, then $JT$ is a submodule equal to $0$ or $T$, and $JT=T$ would give $J^{k}T=T$ for every $k\ge 1$, contradicting the nilpotence of $J$ from [F1]. Hence $J$ kills every simple summand of $V_{\mathbb{C}}$ and therefore kills all of $V_{\mathbb{C}}$, since a semisimple module is a direct sum of simple submodules by [F4]. Faithfulness of the action, established in step 1.1, then forces $J=0$: an element of $J$ acts as zero on $V_{\mathbb{C}}$. Therefore $A_{\mathbb{C}}$ is semisimple.

Choose the rational basis $a_1,\dots ,a_d$ of $A$ and let $T=(\mathrm{tr}(L_{a_i{a}_j}){)}_{i,j}$ be the Gram matrix of the regular trace form of $A$ in this basis. Its entries are rational: passing from $A$ to $A_{\mathbb{C}}$ extends the rational matrix of left multiplication by $a_i{a}_j$ to a complex matrix with the same trace. By step 2.1 and [F2] the regular trace form of $A_{\mathbb{C}}$ is nondegenerate, so $\det T\mathrm{\ne }0$ as a complex number, and since $\det T\in \mathbb{Q}$ it is a nonzero rational number. For every field $k$ the scalar extension $\mathbb{Q}\hookrightarrow k$ therefore makes $T$, viewed over $k$, a nondegenerate matrix. No embedding of $k$ into $\mathbb{C}$ is used.

Let $k$ be any characteristic-zero field and put $A_k=A{\otimes }_{\mathbb{Q}}k$. For $x\in J(A_k)$ and $b\in A_k$ the product $xb$ lies in the nilpotent ideal $J(A_k)$ by [F1], so $(xb{)}^n=0$ for some $n$ and the endomorphism $L_{xb}$ of $A_k$ is nilpotent; by [F3] its trace vanishes. Nondegeneracy of the extended matrix $T$ from step 3.1 then forces $x=0$, so $J(A_k)=0$. By [F1] the quotient $A_k\mathrm{/}J(A_k)=A_k$ is a semisimple ring.

Finally $V_k=V{\otimes }_{\mathbb{Q}}k$ is semisimple over $A_k$. Choose a $\mathbb{Q}$-basis $v_1,\dots ,v_d$ of $V$. The map $(A_k{)}^d\to V_k$, $(x_1,\dots ,x_d)\mapsto {\sum }_i{x}_i{v}_i$, is a surjective $A_k$-module homomorphism, and the regular module of the semisimple ring $A_k$ is a direct sum of simple modules by [F1]. Hence $V_k$ is a sum of images of simple modules, each of which is simple or zero, and $V_k$ has finite length over $k$; by [F4] it is a finite direct sum of simple modules, that is, semisimple. By step 1.1 this is exactly semisimplicity of the monodromy module, equivalently every invariant subspace has an invariant complement by [F4], which is semisimplicity of the local system ${\mathcal{V}}_{\mathbb{Q}}\otimes k$. The zero fiber is the empty direct sum and is included.