Proof. For any $k[\mathfrak {S}_n]$ -module W, we have a natural map

$$ \begin{align*} {\mathrm{avg}} \colon W_{\mathfrak{S}_n} \to W^{\mathfrak{S}_n}, \qquad {\mathrm{avg}}(x) = \sum_{g \in \mathfrak{S}_n} gx. \end{align*} $$

This map is an isomorphism if W is free, that is, a sum of regular representations. If M is an $\mathbf {FI}$ -module with an action of $\mathfrak {S}_n$ , then averaging defines a map of $\mathbf {FI}$ -modules

$$ \begin{align*} {\mathrm{avg}} \colon M_{\mathfrak{S}_n} \to M^{\mathfrak{S}_n}. \end{align*} $$

If $M_m$ is a free $k[\mathfrak {S}_n]$ -module for all m then this map is an isomorphism. We apply this with $M=V \otimes \mathbf {P}_n$ . Since $\mathbf {P}_n([m])$ is a free $k[\mathfrak {S}_n]$ -module for all m, we see that $V \otimes \mathbf {P}_n([m])$ is as well, and so the result follows.

Proof. See [Reference Djament7, Theorem A.9]. The result also follows from [Reference Sam and Snowden32] in characteristic 0.

Proof. This was first proved in characteristic 0 in [Reference Sam and Snowden32, Corollary 2.2.6]. The general case follows from [Reference Djament7, Theorem 2].

Proof. This was proved in characteristic 0 in [Reference Sam and Snowden32, Props 2.2.7 and 2.2.8]. In positive characteristic, it is a result of the first author [Reference Nagpal22, §1.2].

Proof. Corollary 2.5 shows that $\varphi $ is surjective. The more precise version of the corollary given in [Reference Nagpal and Snowden25, Theorem 2.7] shows that if V is a simple representation of $\mathfrak {S}_n$ then $[ {\mathrm {soc}}(\overline {\mathbf {P}}(V))]$ is equal to $[\overline {\mathbf {P}}(V)]$ plus terms of the form $[\overline {\mathbf {P}}(W)]$ , where W is a representation of $\mathfrak {S}_m$ with $m<n$ . Thus there is an upper-triangular change of basis between the classes of simples and the classes of induced modules, which completes the proof.

Proof. Since T is exact and kills torsion modules, $T(M) \to T(I^{\bullet })$ is a resolution. Since semi-induced modules are derived saturated, $T(I^i)$ is acyclic for S. Thus this resolution can be used to compute $\mathrm {R} ST(M)$ . Since $S(T(I^i))=I^i$ , claim (a) follows. The remaining claims follow easily.

Proof. By derived adjunction (see §2.10), we have a natural isomorphism

$$ \begin{align*} \mathrm{R} \ {\mathrm{Hom}}(M, \mathrm{R} S(N)) = \mathrm{R} \ {\mathrm{Hom}}(T(M), N) \end{align*} $$

where M is an $\mathbf {FI}$ -module and N is an object of the generic category. Take $M=\mathbf {P}_n$ . The higher $\mathrm {R} {\mathrm {Hom}}$ ’s on the left vanish since M is projective. We thus find

$$ \begin{align*} (\mathrm{R}^i S N)_n = {\mathrm{Ext}}^i(\overline{\mathbf{P}}_n, N). \end{align*} $$

Note that $N=TS(N)$ . Thus the above shows that $\mathrm {R}^i ST(SN)=0$ for all $i>0$ , meaning $SN$ is derived saturated, if and only if (b) holds. The result thus follows from Proposition 2.3.

Proof. Since $\mathbb {V}_n$ is generated by $e_{1,\ldots ,n}$ , it follows that i is injective. Now let $x \in V^{\mathfrak {S}(n)}$ be given. Define a linear map $f \colon \mathbb {V}_n \to V$ by $f(e_{i_1,\ldots ,i_n})=\sigma x$ , where $\sigma \in \mathfrak {S}$ is any element satisfying $\sigma (1)=i_1, \ldots , \sigma (n)=i_n$ . We note that $\sigma $ is well-defined up to right multiplication by $\mathfrak {S}(n)$ , and so f is well-defined since x is $\mathfrak {S}(n)$ -invariant. One easily sees that f is a map of representations. Since $i(f)=x$ , we see that i is surjective. Finally, let $\{x_i\}_{i \in I}$ be a family of generators for V. We have just seen that each $x_i$ is in the image of a map $\mathbb {V}_{n_i} \to V$ for some $n_i$ . We thus have a surjection $\bigoplus _{i \in I} \mathbb {V}_{n_i} \to V$ .

Proof. This result is well-known, but since we do not have a convenient reference we sketch a proof. Let V be a representation of $\mathfrak {S}$ . Then $V^{\mathfrak {S}(n)}$ is a representation of $\mathfrak {S}_n$ , and there is a natural inclusion $V^{\mathfrak {S}(n)} \to V^{\mathfrak {S}(n+1)}$ . One readily verifies that this defines the structure of an $\mathbf {FI}$ -module, and so we have a functor

$$ \begin{align*} \Theta^* \colon {\mathrm{Rep}}(\mathfrak{S}) \to {\mathrm{Mod}}_{\mathbf{FI}}. \end{align*} $$

If M is an $\mathbf {FI}$ -module then giving a map $M \to \Theta ^*(V)$ of $\mathbf {FI}$ -modules is easily seen to be equivalent to giving a map $\Theta (M) \to V$ of $\mathfrak {S}$ -representations, and so $\Theta ^*$ is right adjoint to $\Theta $ . The smoothness condition shows that the natural map $V \to \Theta (\Theta ^*(V))$ is an isomorphism. Finally, it follows from the definitions that $\Theta (M)=0$ if and only if M is a torsion $\mathbf {FI}$ -module. [Reference Gabriel13, III Prop. 5] thus shows that $\overline {\Theta }$ is an equivalence.

Proof. This follows from the corresponding properties of generic $\mathbf {FI}$ -modules discussed in §2.8. See also [Reference Sam and Snowden33, (6.1.6)] for a proof in characteristic 0.

Proof. First consider the tensor product $\mathbb {V}_n \otimes \mathbb {V}_m$ . This is the subspace of $\mathbb {V}^{\otimes (n+m)}$ spanned by tensors

$$ \begin{align*} e_{i_1, \ldots, i_n, j_1, \ldots, j_m} \end{align*} $$

where the i’s are distinct and the j’s are distinct. Now, the span of the vectors where the i’s and j’s are distinct in total is $\mathbb {V}_{n+m}$ . If $i_1=j_1$ and the other indices are distinct, we get a copy of $\mathbb {V}_{n+m-1}$ . Carrying on with this reasoning, we obtain a decomposition

$$ \begin{align*} \mathbb{V}_n \otimes \mathbb{V}_m \cong \bigoplus_{\Gamma} \mathbb{V}_{n+m-\# E(\Gamma)} \end{align*} $$

where the sum is over bipartite matchings $\Gamma $ on $A \amalg B$ , and $E(\Gamma )$ is the edge set of $\Gamma $ . Suppose $\Gamma $ has r edges. Then the $\mathfrak {S}_n \times \mathfrak {S}_m$ orbit of $\Gamma $ contains the standard matching, whose edges are $(1,1), \ldots , (r,r)$ . We thus see that, as a representation of $\mathfrak {S}_n \times \mathfrak {S}_m \times \mathfrak {S}$ , we have

$$ \begin{align*} \mathbb{V}_n \otimes \mathbb{V}_m \cong \bigoplus_{r=0}^{\min(n,m)} {\mathrm{Ind}}_{\mathfrak{S}_r \times \mathfrak{S}_{n-r} \times \mathfrak{S}_{m-r}}^{\mathfrak{S}_n \times \mathfrak{S}_m}(\mathbb{V}_{n+m-r}), \end{align*} $$

where here $\mathfrak {S}_r$ is embedded diagonally in $\mathfrak {S}_n$ and $\mathfrak {S}_m$ .

Now suppose that V is a $k[\mathfrak {S}_n]$ -module and W is a $k[\mathfrak {S}_m]$ -module. Tensor the above with $V \otimes W$ and take invariants under $\mathfrak {S}_n \times \mathfrak {S}_m$ . By Frobenius reciprocity, we have

$$ \begin{align*} \mathbb{P}(V) \otimes \mathbb{P}(W) \cong \bigoplus_{r=0}^{\min(n,m)} (\mathbb{V}_{n+m-r} \otimes V \otimes W)^{\mathfrak{S}_r \times \mathfrak{S}_{n-r} \times \mathfrak{S}_{m-r}}. \end{align*} $$

Here $\mathfrak {S}_r \times \mathfrak {S}_{n-r} \times \mathfrak {S}_{m-r}$ is a Young subgroup of $\mathfrak {S}_{n+m-r}$ , and its action on $\mathbb {V}_{n+m-r}$ is through this inclusion. Another application of Frobenius reciprocity gives

$$ \begin{align*} \mathbb{P}(V) \otimes \mathbb{P}(W) \cong \bigoplus_{r=0}^{\min(n,m)} (\mathbb{V}_{n+m-r} \otimes {\mathrm{Ind}}_{\mathfrak{S}_r \times \mathfrak{S}_{n-r} \times \mathfrak{S}_{m-r}}^{\mathfrak{S}_{n+m-r}}(V \otimes W))^{\mathfrak{S}_{n+m-r}}, \end{align*} $$

which gives the stated result.

Proof. This follows from Proposition 2.11.

Proof. This follows from Corollary 2.5 by applying $\Theta $ . In characteristic 0, it also follows from the results of [Reference Sam and Snowden33].

Proof. It is clear that the class of torsion modules is closed under passing to subquotients and arbitrary direct sums. Now, suppose that

$$ \begin{align*} 0 \to M_1 \to M_2 \to M_3 \to 0 \end{align*} $$

is a short exact sequence of A-modules, with $M_1$ and $M_3$ torsion. We must show that $M_2$ is torsion. Let N be a finitely generated submodule of $M_2$ , and let $\overline {N}$ be its image in $M_3$ . Since $M_3$ is torsion and $\overline {N}$ is a finitely generated submodule, there is a nonzero $\mathfrak {S}$ -ideal $\mathfrak {a}$ of A such that $\mathfrak {a} \overline {N}=0$ . Of course, we can assume $\mathfrak {a}$ is finitely generated. Since $\mathfrak {a} N$ is a finitely generated submodule of the torsion module $M_1$ , there is a nonzero $\mathfrak {S}$ -ideal $\mathfrak {b}$ such that $\mathfrak {b} (\mathfrak {a} N)=0$ . Since A is an $\mathfrak {S}$ -domain, $\mathfrak {a} \mathfrak {b}$ is nonzero. We have thus shown that $M_2$ is torsion, as required.

Proof. Suppose $f \cdot \sigma (g)$ belongs to $\mathfrak {h}_s$ for all $\sigma \in \mathfrak {S}$ , where $f \in R \setminus \mathfrak {h}_s$ and $g \in R$ . We must show $g \in \mathfrak {h}_s$ . Choose $\sigma $ such that f and $\sigma (g)$ contain no common variable. Write $f=\sum _{i=1}^n c_i m_i$ , where the $c_i$ ’s are nonzero scalars and the $m_i$ ’s are distinct monomials. Since $\mathfrak {h}_s$ is a monomial ideal, we have $m_i \sigma (g) \in \mathfrak {h}_s$ for all i. Since $f \not \in \mathfrak {h}_s$ , there is some i such that $m_i \not \in \mathfrak {h}_s$ . Since $m_i \sigma (g) \in \mathfrak {h}_s$ and $m_i$ and $\sigma (g)$ have no common variable, it follows that $\sigma (g) \in \mathfrak {h}_s$ , and so $g \in \mathfrak {h}_s$ , as required.

Now suppose $\mathfrak {p}$ is an $\mathfrak {S}$ -prime containing $\mathfrak {h}_s$ . Thus $x_1^{s+1} \in \mathfrak {p}$ . Let $r \ge 0$ be minimal such that $x_1^{r+1} \in \mathfrak {p}$ . We claim that $\mathfrak {p}=\mathfrak {h}_r$ . Clearly, we have the containment $\mathfrak {h}_r \subset \mathfrak {p}$ . We therefore have equality if $r=0$ . Suppose then that $r \ge 1$ and the containment is strict. Then $\mathfrak {p}$ contains some monomial $x_1^{e_1} \cdots x_n^{e_n}$ where all exponents are $\le r$ , and thus contains $(x_1 \cdots x_n)^r$ . Let n be minimal such that $(x_1 \cdots x_n)^r \in \mathfrak {p}$ . We must have $n \ge 2$ by the minimality of r. We have $(x_1 \cdots x_{n-1})^r \cdot \sigma (x_n^r) \in \mathfrak {p}$ for all $\sigma \in \mathfrak {S}$ . Indeed, if $\sigma (n) \not \in [n-1]$ then this product is simply in the $\mathfrak {S}$ -orbit of $(x_1 \cdots x_n)^r$ , while if $\sigma (n) \in [n-1]$ then some exponent is $>r$ . Thus, since $\mathfrak {p}$ is $\mathfrak {S}$ -prime, we have $(x_1 \cdots x_{n-1})^r \in \mathfrak {p}$ , contradicting the minimality of n. This contradiction proves $\mathfrak {p}=\mathfrak {h}_r$ , as required.

Proof. Let M be a finitely generated torsion $R/\mathfrak {h}_s$ -module. The annihilator $\mathfrak {a}$ of M in R is an $\mathfrak {S}$ -ideal that properly contains $\mathfrak {h}_s$ . The $\mathfrak {S}$ -radical of $\mathfrak {a}$ , denoted $ {\mathrm {rad}}_{\mathfrak {S}}(\mathfrak {a})$ , is the sum of all $\mathfrak {S}$ -ideals $\mathfrak {b}$ such that $\mathfrak {b}^n \subset \mathfrak {a}$ for some n. One easily sees that it is an $\mathfrak {S}$ -ideal. Furthermore, we have

$$ \begin{align*} {\mathrm{rad}}_{\mathfrak{S}}(\mathfrak{a}) = \bigcap_{\mathfrak{a} \subset \mathfrak{q}} \mathfrak{q}, \end{align*} $$

where the intersection is over all $\mathfrak {S}$ -primes $\mathfrak {q}$ containing $\mathfrak {a}$ ; see [Reference Nagpal and Snowden27, Proposition 2.23]. Since $\mathfrak {a}$ properly contains $\mathfrak {h}_s$ , it follows from Proposition 4.1 that $ {\mathrm {rad}}_{\mathfrak {S}}(\mathfrak {a})=\mathfrak {h}_r$ for some $0 \le r < s$ . We thus have $\mathfrak {h}_{s-1} \subset \mathfrak {a}$ . Since R is noetherian, one easily sees that $\mathfrak {a}$ contains a power of its $\mathfrak {S}$ -radical. Thus $\mathfrak {h}_{s-1}^n \subset \mathfrak {a}$ for some n, which completes the proof.

Proof. (a) For $\alpha \in \mathbf {N}^n$ , let $x^{\alpha }=x_1^{\alpha _1} \cdots x_n^{\alpha _n}$ . Then the $x^{\alpha }$ with $0 \le \alpha _i \le s$ form a k-basis of S. Let $\{ y_{\alpha } \}$ be the dual basis of $S^*$ . A simple computation shows that $x^{\alpha } y_{\beta }=y_{\beta -\alpha }$ , which is taken to be 0 if any coordinate is negative. We thus see that $S^*$ is free with basis $y_{(s,\ldots ,s)}$ .

(b) Since S is a finite dimensional k-algebra, the k-linear dual of any finite dimensional projective S-module is an injective S-module. Thus $S \cong S^*$ is injective.

Proof. This follows from the mapping property for $\mathbb {V}_n$ (Proposition 3.2).

Proof. An $\mathfrak {S}(n)$ -invariant element of $\mathcal {Q}_{s,m}$ has the form

$$ \begin{align*} \sum_{i \colon [n] \to [m]} a(i) e_{i(1), \ldots, i(n)} \end{align*} $$

where $a(i)$ is a polynomial in $x_1, \ldots , x_n$ . For i as above, let $b(i)$ be the product of the $x_j^s$ over $j \not \in {\mathrm {im}}(i)$ . Then the above element is annihilated by $x_1, \ldots , x_n$ if and only if $a(i)$ is divisible by $b(i)$ for each i. Of course, we can assume that the variables $x_{i(1)}, \ldots , x_{i(n)}$ do not occur in $a(i)$ . If $a(i)$ is divisible by $b(i)$ and also satisfies this condition then it is simply a k-linear multiple of $b(i)$ . We thus see that the elements $b(i) e_{i(1), \ldots , i(n)}$ form a k-basis of $\mathcal {Q}_{s,m}'$ , to use the notation of Proposition 4.4. The result follows.

Proof. The mapping property for $\mathcal {Q}_{s,n}$ ensures the existence of the map f. One easily sees that f maps a k-basis of $\mathcal {Q}_{s,n}$ to k-linearly independent elements, and so f is injective.

Proof. Let f be given, and put $z=f(e_{1,\ldots ,m})$ . Since z is an $\mathfrak {S}(m)$ -invariant element of $\mathcal {P}_{s,n}$ , we have an expression

$$ \begin{align*} z = \sum_{i \colon [n] \to [m]} a(i) e_{i(1),\ldots,i(n)}, \end{align*} $$

where each $a_{i_1,\ldots ,i_n}$ is a polynomial in $x_1, \ldots , x_m$ . Since z is killed by $x_1, \ldots , x_m$ , it follows that each coefficient $a(i)$ is divisible by $(x_1 \cdots x_m)^s$ , and thus simply a scalar multiple of this monomial. Thus z is contained in $\mathcal {Q}_{s,n}$ . Since $e_{1,\ldots ,m}$ generates $\mathcal {Q}_{s,m}$ , it follows that the image of f is contained in $\mathcal {Q}_{s,n}$ .

Proof. The $\mathfrak {m}$ -adic filtration on S gives a filtration on $\mathcal {P}_s(V)$ where the graded pieces are $\mathcal {Q}_s(\mathfrak {m}^i/\mathfrak {m}^{i+1} \otimes V)$ . Here we have used that the functor $\mathscr {G}$ is exact.

Proof. We have just seen that $\mathcal {P}_{s,n}$ carries a filtration with graded pieces $\mathcal {Q}_s(\mathfrak {m}^i/\mathfrak {m}^{i+1} \otimes k[\mathfrak {S}_n])$ . As a $k[\mathfrak {S}_n]$ -module, $\mathfrak {m}^i/\mathfrak {m}^{i+1} \otimes [\mathfrak {S}_n]$ is isomorphic to a direct sum of $\dim _k(\mathfrak {m}^i/\mathfrak {m}^{i+1})$ copies of $k[\mathfrak {S}_n]$ . Thus we can refine the filtration so that each graded piece is $\mathcal {Q}_s(k[\mathfrak {S}_n])=\mathcal {Q}_{s,n}$ . The length of the refined filtration is $\dim _k(S)=(s+1)^n$ .

Proof. The k-linear dual of S is isomorphic to S as an S-module (Proposition 4.3). Thus if we take the k-linear dual of the minimal free resolution of $S/\mathfrak {m}$ , we obtain a resolution of S-modules

$$ \begin{align*} 0 \to S/\mathfrak{m} \to S^{\oplus a_0} \to S^{\oplus a_1} \to \cdots. \end{align*} $$

Similarly, dualizing a free resolution of $V^*$ as a $k[\mathfrak {S}_n]$ -module, we obtain a resolution of $k[\mathfrak {S}_n]$ -modules

$$ \begin{align*} 0 \to V \to k[\mathfrak{S}_n]^{\oplus b_0} \to k[\mathfrak{S}_n]^{\oplus b_1} \to \cdots. \end{align*} $$

Tensoring these resolutions together and taking the total complex, we obtain a resolution of $\mathfrak {S}_n$ -equivariant S-modules

$$ \begin{align*} 0 \to S/\mathfrak{m} \otimes V \to (S \otimes k[\mathfrak{S}_n])^{\oplus r_0} \to (S \otimes k[\mathfrak{S}_n])^{\oplus r_1} \to \cdots. \end{align*} $$

We regard this as a sequence in $\mathcal {C}$ by letting $\mathfrak {S}(n)$ act trivially. Applying $\mathscr {G}$ yields the result.

Proof. Use notation as in §4.5, with $n=1$ . Thus $S=k[x_1]/(x_1^{s+1})$ . We have a 2-periodic free resolution

Applying the $\mathscr {G}$ functor, we obtain a resolution

where the differentials map $e_i$ to $x_i^s e_i$ or $x_i e_i$ . We can use this resolution to compute $ {\mathrm {Tor}}^{R/\mathfrak {h}_s}$ since the module $\mathcal {P}_{s,1}$ is flat over $R/\mathfrak {h}_s$ . Applying $- \otimes _{R/\mathfrak {h}_s} \mathcal {Q}_{s,1}$ , we obtain the complex

where the differentials map $e_i \otimes m$ to $e_i \otimes x_i^s m$ or $e_i \otimes x_i m$ . Ignoring the $\mathfrak {S}$ action, this complex decomposes as $\bigoplus _{i \ge 1} e_i \otimes C_i$ , where $C_i$ is the complex

Let $\mathcal {Q}_{s,1}^{(i)}$ be the $|R|$ -submodule of $\mathcal {Q}_{s,1}$ generated by $e_i$ . Then

$$ \begin{align*} \mathcal{Q}_{s,1}[x_i^s] = x_i \mathcal{Q}_{s,1} \oplus \mathcal{Q}_{s,1}^{(i)}, \quad \mathcal{Q}_{s,1}[x_i] = x_i^s \mathcal{Q}_{s,1} \oplus \mathcal{Q}_{s,1}^{(i)}. \end{align*} $$

Here $(-)[a]$ denotes the set of elements killed by a. Note that this is indeed a direct sum. We thus see that $\mathrm {H}_r(C_i)=\mathcal {Q}_{1,1}^{(i)}$ for all $r \ge 1$ . Hence

$$ \begin{align*} {\mathrm{Tor}}_r^{R/\mathfrak{h}_s}(\mathcal{Q}_{s,1}, \mathcal{Q}_{s,1}) = \bigoplus_{i \ge 1} e_i \otimes \mathrm{H}_m(C_i) \cong \mathcal{Q}_{s,1} \end{align*} $$

as claimed.

Proof. We first suppose V is the regular representation. Let $M=\Phi _s(\mathbf {P}(V)) = \Phi _s(\mathbf {P}_n)$ . By definition, we have

$$ \begin{align*} M = \bigoplus_{\alpha \in \Delta(s)} \mathbf{P}_n(\Pi_s([\alpha])). \end{align*} $$

Recall that $\mathbf {P}_n(S)=k[ {\mathrm {Hom}}_{\mathbf {FI}}([n], S)]$ . Let $\alpha \in \Delta (s)$ be defined by $\alpha _i=s$ for $i \in [n]$ and $\alpha _i=0$ for $i \not \in [n]$ . Then $\Pi _s([\alpha ])$ is the set of bijections $[n] \to [n]$ , and has a canonical element (the identity). Let $m \in M$ be this canonical element of degree $\alpha $ . Since $\alpha $ is supported on $[n]$ , it follows that m is fixed by $\mathfrak {S}(n)$ . Also, if $\beta $ is a weight with $\beta _i>0$ for some $i \in [n]$ then $\alpha +\beta \not \in \Delta (s)$ , and so $x^{\beta } m=0$ . In particular, m is killed by $x_1, \ldots , x_n$ . It follows that there is a map of R-modules $\mathcal {Q}_{s,n} \to M$ given by $e_{1,\ldots ,n} \mapsto m$ (Proposition 4.4). One easily sees that this induces a bijection on the natural bases of the source and target, and is thus an isomorphism.

We now treat the case of a general representation V. Choose a presentation $F_1 \to F_0 \to V \to 0$ where $F_0$ and $F_1$ are free $k[\mathfrak {S}_n]$ -modules. Consider the diagram

where the left two vertical maps are the isomorphisms from the previous paragraph. One easily sees that the left square commutes. The rows are exact, since $\mathbf {P}$ , $\Phi _s$ , and $\mathcal {Q}_s$ are exact functors. It follows that the right map exists and is an isomorphism.

Proof. First suppose V is the regular representation. Let $M=\Phi _s(\mathbf {T}(V))=\Phi _s(\mathbf {T}_n)$ . By definition, we have

$$ \begin{align*} M = \bigoplus_{\alpha \in \Delta(s)} \mathbf{T}_n(\Pi_s([\alpha])). \end{align*} $$

Recall that $\mathbf {T}_n(S)=k[\text {Bij}([n], S)]$ . Let $\alpha \in \Delta (s)$ be as in the previous proof. Then once again, we have a canonical element $m \in M$ of degree $\alpha $ , which is killed by $x_1, \ldots , x_n$ . If $\beta \in \Delta $ has $\beta _i=s$ for some $i>n$ then $\Pi _s(\alpha +\beta )$ contains $[n] \cup \{i\}$ , and thus is not in bijection with $[n]$ ; hence $M_{\alpha +\beta }=0$ . It follows that m is killed by $x_i^s$ for all $i>n$ . Hence there is a map $\mathcal {Q}_{s-1,n} \to M$ given by $e_{1,\ldots ,n} \mapsto m$ (Proposition 4.4). Again, one checks that it is a bijection on basis vectors. The case of general V follows as in the proof of Lemma 5.4.

Proof. We prove the lemma in five steps. All modules below are finitely generated.

Step 1: M and N are projective. It suffices to treat the case $M=\mathbf {P}_m$ and $N=\mathbf {P}_n$ . We have

$$ \begin{align*} {\mathrm{Hom}}_{\mathbf{FI}}(\mathbf{P}_m, \mathbf{P}_n)=\mathbf{P}_n([m]) = k[{\mathrm{Hom}}_{\mathbf{FI}}([n], [m])]. \end{align*} $$

Similarly,

$$ \begin{align*} {\mathrm{Hom}}_R(\Phi_s(\mathbf{P}_m), \Phi_s(\mathbf{P}_n)) = {\mathrm{Hom}}_R(\mathcal{Q}_{s,m}, \mathcal{Q}_{s,n}) \end{align*} $$

also has a k-basis indexed by injections $[n] \to [m]$ (Corollary 4.5). Since the map in question is injective and the source and target have the same dimension, it is an isomorphism.

Step 2: M is projective and N is induced. Say $N=\mathbf {P}(V)$ , where V is a representation of $\mathfrak {S}_n$ . Choosing a co-presentation of V by free $k[\mathfrak {S}_n]$ -modules and applying $\mathbf {P}$ , we obtain an exact sequence $0 \to N \to P_0 \to P_1$ with $P_0$ and $P_1$ projective. Consider the diagram

The right two vertical maps are isomorphisms by Step 1. It follows that the left vertical map is an isomorphism as well.

Step 3: M is projective and N is semi-induced. Choose an exact sequence

$$ \begin{align*} 0 \to N' \to N \to N" \to 0 \end{align*} $$

with $N'$ induced and $N"$ semi-induced with a shorter filtration length than N. Consider the diagram analogous to the one in Step 2. The left vertical map is an isomorphism by Step 2, and the right vertical map is an isomorphism by induction. Since the top row is exact, as M is projective, it follows that the middle vertical map is an isomorphism.

Step 4: M is projective and N is saturated. Let $0 \to N \to I^0 \to I^1$ be an exact sequence with $I^0$ and $I^1$ semi-induced; this exists since N is saturated (Proposition 2.10). Now proceed as in Step 2, making use of Step 3.

Step 5: the general case. Let $P_1 \to P_0 \to M\to 0$ be a presentation of M, where $P_0$ and $P_1$ are projective $\mathbf {FI}$ -modules. Consider the diagram

The right two vertical maps are isomorphisms by Step 4. It follows that the left vertical map is an isomorphism as well.

Proof. Let N be an $\mathbf {FI}$ -module. Then the map

$$ \begin{align*} {\mathrm{Hom}}_{\mathbf{FI}}(N, M) \to {\mathrm{Hom}}_{\mathbf{FI}}(N, \Phi^*(\Phi_s(M))) = {\mathrm{Hom}}_{R/\mathfrak{h}_s}(\Phi_s(N), \Phi_s(M)) \end{align*} $$

is simply the map induced by $\Phi _s$ , which is an isomorphism by Theorem 5.1. The result thus follows from Yoneda’s lemma.

Proof. Let $i \colon \mathcal {Q}_{s,n} \to \mathcal {P}_{s,n}$ be the usual inclusion. Let M be an $\mathbf {FI}$ -module, and suppose given a map $f \colon \Phi _s(M) \to \mathcal {P}_{s,n}$ . Choose a surjection $P \to M$ where P is a sum of principal projective modules. We then have a surjection $\Phi _s(P) \to \Phi _s(M)$ where $\Phi _s(P)$ is a sum of $\mathcal {Q}_{s,m}$ ’s. The restriction of f to $\Phi _s(P_0)$ maps into $i(\mathcal {Q}_{s,n})$ by Proposition 4.7, and so f does as well, that is, f factors uniquely as $i \circ g$ for some $g \colon \Phi _s(M) \to \mathcal {Q}_{s,n}$ . Now, $\mathcal {Q}_{s,n}=\Phi _s(\mathbf {P}_n)$ . Since $\mathbf {P}_n$ is saturated (Proposition 2.3), it follows from Theorem 5.1 that $g=\Phi _s(g')$ for a unique map $g' \colon M \to \mathbf {P}_n$ . We have thus shown that f factors uniquely as $i \circ \Phi _s(g')$ , and so $\Phi _s^*(P_n)=\mathbf {P}_n$ .

Now, let V be a $k[\mathfrak {S}_n]$ -module, so that $\mathcal {P}_s(V)=(\mathcal {P}_{s,n} \otimes V)^{\mathfrak {S}_n}$ . We have

$$ \begin{align*} \Phi_s^*(\mathcal{P}_s(V)) = (\Phi_s^*(\mathcal{P}_{s,n}) \otimes V)^{\mathfrak{S}_n} = (\mathbf{P}_n \otimes V)^{\mathfrak{S}_n} = \mathbf{P}(V). \end{align*} $$

In the first step, we used that $\Phi _s^*$ is left exact (as it is a right adjoint), and so commutes with $\mathfrak {S}_n$ -invariants; we also used that $\Phi _s^*$ is additive, and so commutes with $-\otimes V$ . In the second step, we used the previous paragraph. This completes the proof.

Proof. By derived adjunction (§2.10), for an $\mathbf {FI}$ -module M and an $R/\mathfrak {h}_s$ -module N, we have

$$ \begin{align*} \mathrm{R} \ {\mathrm{Hom}}_{\mathbf{FI}}(M, \mathrm{R} \Phi_s^*(N)) = \mathrm{R} \ {\mathrm{Hom}}_{R/\mathfrak{h}_s}(\Phi_s(M), N). \end{align*} $$

We apply this with $M=\mathbf {P}_n$ . Since $\mathbf {P}_n$ is projective, the higher $\mathrm {R} {\mathrm {Hom}}$ ’s on the left vanish. We thus find

$$ \begin{align*} (\mathrm{R}^i \Phi_s^*(N))_n = {\mathrm{Ext}}^i_{R/\mathfrak{h}_s}(\mathcal{Q}_{s,n}, N). \end{align*} $$

Taking $N=\mathcal {P}_{s,n}$ yields the stated result.

Proof. First note that the statement holds if $M=\mathcal {Q}_{s,\lambda }$ , as such a module is a summand of $\mathcal {Q}_{s,n}$ with $n=\vert \lambda \vert $ . The general case then follows from dévissage.

Proof. Let X be the set of monomials in $R/\mathfrak {h}_s$ . Then $R/\mathfrak {h}_s$ is identified with the permutation module $k[X]$ . Let $X=\coprod _{i\in I} X$ be the orbit decomposition of $X$ , so that $R/\mathfrak {h}_s$ decomposes as $\bigoplus _{i \in I} k[X]$ . Fix $i \in I$ . Then $X$ contains a unique monomial m of the form $x_1^{e_1} \cdots x_n^{e_n}$ where $e_1 \ge \cdots \ge e_n \ge 1$ . The stabilizer in $\mathfrak {S}$ of m is $G \times \mathfrak {S}(n)$ , where G is a Young subgroup of $\mathfrak {S}_n$ . We thus have

$$ \begin{align*} k[X] \cong {\mathrm{Ind}}_{G \times \mathfrak{S}(n)}^{\mathfrak{S}}(k) \cong \mathbb{P}(k[\mathfrak{S}_n/G]), \end{align*} $$

and so $k[X]$ is a finite length induced module.

Proof. Recall that $\mathscr {F}(\mathcal {P}_{s,d})$ is simply $\mathcal {P}_{s,d}$ regarded as an $\mathfrak {S}_n \times \mathfrak {S}(n)$ equivariant module over S. By definition, $\mathcal {P}_{s,d} = R/\mathfrak {h}_s \otimes \mathbb {V}_d$ . We have $R/\mathfrak {h}_s = S \otimes S'$ . Let $\mathbb {V}_i(n)$ denote the subspace of $\mathbb {V}_i$ where all indices are $\ge n$ ; this is identified with $\mathbb {V}_i$ when $\mathfrak {S}(n)$ is identified with $\mathfrak {S}$ . As a representation of $\mathfrak {S}_n \times \mathfrak {S}(n)$ , we have

$$ \begin{align*} \mathbb{V}_d=\bigoplus_{i=0}^d W_i \otimes \mathbb{V}_i(n) \end{align*} $$

for certain $\mathfrak {S}_n$ -modules $W_i$ . We thus have

$$ \begin{align*} \mathcal{P}_{s,d} \cong \bigoplus_{i=0}^d S \otimes W_i \otimes (S' \otimes \mathbb{V}_i(n)). \end{align*} $$

Since $(S', \mathfrak {S}(n))$ is isomorphic to $(R/\mathfrak {h}_s, \mathfrak {S})$ , Lemma 6.5 shows that $S'$ is a direct sum of induced $\mathfrak {S}(n)$ -modules. Since a tensor product of induced modules is a sum of induced modules (Proposition 3.6), the result follows.

Proof. Let $\mathcal {C}'$ be the category of $\mathfrak {S}_n$ -equivariant modules over S. One easily sees that we have an isomorphism

$$ \begin{align*} \mathrm{R} \ {\mathrm{Hom}}_{\mathcal{C}}(S/\mathfrak{m} \otimes k[\mathfrak{S}_n], S \otimes W \otimes V) \cong \mathrm{R} \ {\mathrm{Hom}}_{\mathcal{C}'}(S/\mathfrak{m} \otimes k[\mathfrak{S}_n], S \otimes W) \otimes_k \mathrm{R} \ {\mathrm{Hom}}_{\mathfrak{S}(n)}(k, V). \end{align*} $$

The second factor is concentrated in degree 0 by Proposition 3.7. We have

$$ \begin{align*} \mathrm{R} \ {\mathrm{Hom}}_{\mathcal{C}'}(S/\mathfrak{m} \otimes k[\mathfrak{S}_n], S \otimes W) \cong \mathrm{R} \ {\mathrm{Hom}}_{\mathcal{C}'}(S/\mathfrak{m}, S \otimes k[\mathfrak{S}_n] \otimes W) \end{align*} $$

since $k[\mathfrak {S}_n]$ is a self-dual representation. Furthermore, $k[\mathfrak {S}_n] \otimes W$ is isomorphic to a direct sum of $\dim (W)$ copies of $k[\mathfrak {S}_n]$ . The result thus follows from the fact that $S \otimes k[\mathfrak {S}_n]$ is injective in $\mathcal {C}'$ . Indeed, if M is an object of $\mathcal {C}'$ then

$$ \begin{align*} {\mathrm{Hom}}_{\mathcal{C}'}(M, S \otimes k[\mathfrak{S}_n]) = {\mathrm{Hom}}_S(M, S), \end{align*} $$

and S is self-injective (Proposition 4.3), so the above functor is exact in M.

Proof of Theorem 6.1

By derived adjunction (§2.10), for $M \in \mathcal {C}$ and $N \in {\mathrm {Mod}}_{R/\mathfrak {h}_s}$ -module we have a natural identification

$$ \begin{align*} {\mathrm{Ext}}^i_{R/\mathfrak{h}_s}(\mathscr{G}(M), N) = {\mathrm{Ext}}^i_{\mathcal{C}}(M, \mathscr{F}(N)). \end{align*} $$

Note that $\mathscr {F}$ is exact, so we do not need to derive it. We apply this with $M=S/\mathfrak {m} \otimes k[\mathfrak {S}_n]$ and $N=\mathcal {P}_{s,d}$ . We find

$$ \begin{align*} {\mathrm{Ext}}^i_{R/\mathfrak{h}_s}(\mathcal{Q}_{s,n}, N) = {\mathrm{Ext}}^i_{\mathcal{C}}(S/\mathfrak{m} \otimes k[\mathfrak{S}_n], \mathscr{F}(\mathcal{P}_{s,d})). \end{align*} $$

This vanishes by the previous two lemmas. Note that since $\mathcal {C}$ is locally noetherian, an arbitrary direct sum of injectives is injective, and so if X is a finite length object then $ {\mathrm {Ext}}^i_{\mathcal {C}}(X, -)$ commutes with arbitrary direct sums (compute with injective resolutions in the second argument).

Proof. By derived adjunction (§2.10), we have

$$ \begin{align*} \mathrm{R} \ {\mathrm{Hom}}_{\mathbf{FI}}(M, \mathrm{R} \Phi_s^*(N)) = \mathrm{R} \ {\mathrm{Hom}}_{R/\mathfrak{h}_s}(\Phi_s(M), N), \end{align*} $$

for an $\mathbf {FI}$ -module M and an $R/\mathfrak {h}_s$ -module N. We apply this with $N=\mathcal {P}_{s,n}$ . By Proposition 5.8 and Corollary 6.2, we have $\mathrm {R} \Phi _s^*(\mathcal {P}_{s,n}) = \mathbf {P}_n$ . We thus find

$$ \begin{align*} {\mathrm{Ext}}^i_{\mathbf{FI}}(M, \mathbf{P}_n) = {\mathrm{Ext}}^i_{R/\mathfrak{h}_s}(\Phi_s(M), \mathcal{P}_{s,n}). \end{align*} $$

Now take $M=\mathbf {T}_m$ . The left side vanishes for all i by Proposition 2.3. Since $\Phi _s(\mathbf {T}_m)=\mathcal {Q}_{s-1,m}$ , the first result follows. The second follows from the first, since $\mathcal {P}_{s-1,m}$ has a finite length filtration with graded pieces $\mathcal {Q}_{s-1,m}$ (Proposition 4.9).

Proof of Theorem 6.8

By dévissage, we reduce to the case $M=\mathcal {Q}_s(V)$ . If $T=\varinjlim T_i$ then

$$ \begin{align*} \mathrm{R} \ {\mathrm{Hom}}_{R/\mathfrak{h}_s}(T, M) = \mathrm{R} \varprojlim \mathrm{R} \ {\mathrm{Hom}}_{R/\mathfrak{h}_s}(T_i, M), \end{align*} $$

and so it suffices to treat the case where T is finitely generated, which we assume. By another dévissage, we reduce to the case where T is killed by $\mathfrak {h}_{s-1}$ (Proposition 4.2). Since T is an $R/\mathfrak {h}_{s-1}$ -module, it admits a resolution $X_{\bullet } \to T$ , where each $X_i$ is a sum of $\mathcal {P}_{s-1,m}$ ’s. By Proposition 4.10, we have a resolution $M \to Y_{\bullet }$ , where each $Y_i$ is a finite sum of $\mathcal {P}_{s,n}$ ’s. Thus the result follows from the previous lemma.

Proof. Typically, this kind of result is proved using a projective resolution of M as an A-module. Since our categories do not have enough projectives, we must modify this approach.

For an $\mathfrak {S}$ -module V, let $ {\mathrm {\underline {Hom}}}(V, -)$ be the right adjoint of the endofunctor $- \otimes V$ of $ {\mathrm {Rep}}(\mathfrak {S})$ . This adjoint exists since $ {\mathrm {Rep}}(\mathfrak {S})$ is a Grothendieck abelian category and $- \otimes V$ commutes with colimits. Thus, if U and W are two other $\mathfrak {S}$ -modules, then we have a natural isomorphism

$$ \begin{align*} {\mathrm{Hom}}_{\mathfrak{S}}(U, {\mathrm{\underline{Hom}}}(V, W)) = {\mathrm{Hom}}_{\mathfrak{S}}(U \otimes V, W). \end{align*} $$

Taking $U=\mathbb {V}_n= {\mathrm {Ind}}_{\mathfrak {S}(n)}^{\mathfrak {S}}(k)$ , one sees that

$$ \begin{align*} {\mathrm{\underline{Hom}}}(V, W)^{\mathfrak{S}(n)} = {\mathrm{Hom}}_{\mathfrak{S}(n)}(V, W). \end{align*} $$

Since $ {\mathrm {\underline {Hom}}}(V, W)$ is a smooth representation, it is the union of its $\mathfrak {S}(n)$ -invariants. Thus we see that $ {\mathrm {\underline {Hom}}}(V, W)$ consists of all linear maps $V \to W$ that commute with some $\mathfrak {S}(n)$ . Taking $n=0$ in the above equation, we have

$$ \begin{align*} {\mathrm{Hom}}_{\mathfrak{S}}(V, W) = \Gamma({\mathrm{\underline{Hom}}}(V, W)), \end{align*} $$

where $\Gamma $ denotes $\mathfrak {S}$ -invariants.