FI-modules over a Noetherian $\mathbb{Q}$-algebra form a Noetherian category

By [F2] the $k$-module $V_n$ is finitely generated for every $n\ge 0$, and since $k$ is Noetherian and $W_n\subseteq V_n$, each $W_n$ is a finitely generated $k$-module as well.

The graded $k[T]$-module ${\mathrm{\Phi }}_a(V)$ is finitely generated: by [F3] the map $T\text{\hspace{0.17em}⁣}:{\mathrm{\Phi }}_a(V{)}_n\to {\mathrm{\Phi }}_a(V{)}_{n+1}$ is surjective for every $n\ge a$, so every element of ${\mathrm{\Phi }}_a(V)$ lies in the $k[T]$-submodule generated by ${\mathrm{\Phi }}_a(V{)}_0\oplus \cdots \oplus {\mathrm{\Phi }}_a(V{)}_a$; each ${\mathrm{\Phi }}_a(V{)}_i$ with $i\le a$ is a quotient of $V_{a+i}$, hence a finitely generated $k$-module by step 1.1, so it is generated by finitely many elements.

Since ${\mathrm{\Phi }}_a(W)$ is a $k[T]$-submodule of the finitely generated module ${\mathrm{\Phi }}_a(V)$ over the Noetherian ring $k[T]$, it is a finitely generated $k[T]$-module; writing a finite generating set as finite sums of homogeneous components, the homogeneous components of the generators form a finite generating set consisting of elements $x_1,\dots ,x_r$ with $x_i\in {\mathrm{\Phi }}_a(W{)}_{n_i}$.

For each $i$ choose a lift $w_i\in W_{a+n_i}$ of $x_i$ (possible because ${\mathrm{\Phi }}_a(W{)}_{n_i}$ is a quotient of $W_{a+n_i}$), and for each $n<a$ choose a finite generating set of the $k$-module $W_n$, which is possible by step 1.1; let $\tilde{W}\subseteq W$ be the sub-FI-module generated by all these elements. Then ${\mathrm{\Phi }}_a(\tilde{W})={\mathrm{\Phi }}_a(W)$: the inclusion gives ${\mathrm{\Phi }}_a(\tilde{W})\subseteq {\mathrm{\Phi }}_a(W)$, while each $x_i$ is the image of $w_i\in \tilde{W}$, so ${\mathrm{\Phi }}_a(\tilde{W})$ contains a generating set of ${\mathrm{\Phi }}_a(W)$. By exactness [F1], applied to $0\to \tilde{W}\to W\to W\mathrm{/}\tilde{W}\to 0$, we get ${\mathrm{\Phi }}_a(W\mathrm{/}\tilde{W})=0$.

Let $U:=W\mathrm{/}\tilde{W}$. Restricting scalars to $\mathbb{Q}$ gives an FI-module $U^{\mathbb{Q}}$ over $\mathbb{Q}$, a subquotient of $V^{\mathbb{Q}}$ by construction. The evaluation surjection ${⨁}_{m\le a}M(V_m)\twoheadrightarrow V$ remains a surjection of $\mathbb{Q}$-vector spaces after restriction, and the restriction of the free $k$-FI-module $M(V_m)$ is the free $\mathbb{Q}$-FI-module on $V_m^{\mathbb{Q}}$, because both are the same direct sum of the same coefficient spaces with the same transition maps; hence $V^{\mathbb{Q}}$ is generated in degree $\le a$ over $\mathbb{Q}$ by [F2]. Therefore $\mathrm{weight}(V^{\mathbb{Q}})\le a$ by [F3] applied over the characteristic-zero field $\mathbb{Q}$, and $\mathrm{weight}(U^{\mathbb{Q}})\le a$ because weight is monotone under subquotients [F4].

By [F6] the coinvariants of $U^{\mathbb{Q}}$ are the restriction of the coinvariants of $U$, so ${\mathrm{\Phi }}_a(U^{\mathbb{Q}})=0$ by step 4.1. Hence for every $n\ge a$ the coinvariants $(U_n^{\mathbb{Q}}{)}_{S_{n-a}}={\mathrm{\Phi }}_a(U^{\mathbb{Q}}{)}_{n-a}$ vanish, and step 5.1 with [F4] gives $U_n^{\mathbb{Q}}=0$, which forces $U_n=0$; for $n<a$ the module $U_n$ vanishes by the choice of generators of $W_n$ in step 4.1. Thus $U=0$, that is $W=\tilde{W}$ is generated by the finite set of elements chosen in step 4.1, so $W$ is finitely generated.

Steps 1.1–6.1 prove that every sub-FI-module $W$ of a finitely generated FI-module $V$ over a commutative Noetherian ring $k$ containing $\mathbb{Q}$ is finitely generated, so the category of FI-modules over $k$ is Noetherian. All generating sets and lifts chosen are finite, the Hilbert basis theorem enters only through its choice-free finite-generation form [F5], and no map $k\to \mathbb{C}$ or basis of an infinite-dimensional rational space is used.