Proof. [Reference KoziolKoz] and [Reference Ollivier and SchneiderOS18] Proposition 3.25 and its proof.

Proof. Clearly, ${\mathrm {Mod}}_k^I(G)$ is closed under the formation of arbitrary direct sums. Let $0 \rightarrow V_0 \rightarrow V_1 \rightarrow V_2 \rightarrow 0$ be an exact sequence in ${\mathrm {Mod}}_k(G)$ such that $V_1$ lies in ${\mathrm {Mod}}_k^I(G)$ . Obviously, then also $V_2$ lies in ${\mathrm {Mod}}_k^I(G)$ . For $V_0$ , we consider the commutative diagram

The upper horizontal row is exact by the left exactness of the functor $(-)^I$ and the fact that $k[G/I]$ is flat as a (right) H-module. By the category equivalence in Theorem 2.1.1, the middle and right perpendicular arrows are isomorphisms. Hence, the left one is an isomorphism as well. This shows that $V_0$ lies in ${\mathrm {Mod}}_k^I(G)$ .

Proof. It suffices to consider the case where V is generated by a single vector v. We can then find an open normal subgroup $N \subseteq J$ such that N fixes v. Then V is a quotient of the $\mathcal {G}$ -representation $k[\mathcal {G}/N]$ . Hence, it actually suffices to consider the case $V = k[\mathcal {G}/N]$ . In V, we then have the J-subrepresentation $k[J/N]$ . Let $\mathfrak {m}$ denote the augmentation ideal of the ring $k[J/N]$ . Since $J/N$ is a finite p-group, we find an integer $\ell \geq 0$ such that $\mathfrak {m}^\ell = 0$ . Observe that J acts trivially on each subquotient $\mathfrak {m}^i / \mathfrak {m}^{i+1}$ . We now define $V_i \subseteq V$ as the $\mathcal {G}$ -subrepresentation generated by $\mathfrak {m}^{\ell - i}$ . The subquotient $V_i /V_{i-1}$ then is generated by the image of $\mathfrak {m}^{\ell - i} / \mathfrak {m}^{(\ell - i)+1}$ , which is contained in $(V_i/V_{i-1})^J$ .

Proof. By Lemma 1(iv) [Reference SchneiderSch], ${\mathrm {Mod}}_k(G)$ is a Grothendieck category. We have to show that any finitely generated G-representation V in ${\mathrm {Mod}}_k(G)$ is noetherian. Lemma 2.1.3 reduces us to the case that V is generated by finitely many I-fixed vectors. According to Lemma 2.1.2, any increasing chain of G-subrepresentations of V already lies in ${\mathrm {Mod}}_k^I(G)$ . By Theorem 2.1.1, it therefore corresponds to a chain in the finitely generated H-module $V^I$ . Hence, it must become stationary.

Proof. a) is obvious. b) holds since I is pro-p. c) is a consequence of the functor being right adjoint to a (left) exact functor. For d), let $V = \varinjlim _j V_j$ be a filtered colimit in ${\mathrm {Mod}}_k(G)$ . Since filtered colimits are exact in the Grothendieck category ${\mathrm {Mod}}_k(G)$ , we have the inclusion $\varinjlim _j V_j(I) \subseteq V(I)$ . But $V^I = \varinjlim _j V_j^I$ . Hence, this inclusion is an equality.

Proof. By our assumption, E is $\mathcal {C}_\alpha $ -closed and contains a nonzero subobject S lying in $\mathcal {C}_{\alpha + 1}$ (cf. [Reference GabrielGab] p. 383). Since E is an essential extension of any of its nonzero subobjects (cf. [Reference GabrielGab] Proposition 11 on p. 361) it follows from the stability of $\mathcal {C}_{\alpha + 1}$ that E lies in $\mathcal {C}_{\alpha + 1}$ .

Proof. The computation of $Z(H)$ in [Reference Ollivier and SchneiderOS18] §3.2.4 (together with [Reference McConnell and RobsonMCR] Corollary 6.4.8) shows that $Z(H)$ has Krull dimension one. By [Reference Ollivier and SchneiderOS18] Corollary 3.4, the centre $Z(H)$ contains a polynomial ring $k[\zeta ]$ over which H is a finitely generated free module. Using [Reference McConnell and RobsonMCR] Corollary 6.5.3, it then follows that H has Krull dimension one as well. Hence, ${\mathrm {Mod}}_k^I(G) \simeq {\mathrm {Mod}}(H)$ has Krull dimension one. By the proof of Lemma 2.1.2, the subcategory ${\mathrm {Mod}}_k^I(G)$ is closed under the passage to subobjects in ${\mathrm {Mod}}_k(G)$ . Therefore, the Krull dimensions of an object in ${\mathrm {Mod}}_k^I(G)$ when viewed in ${\mathrm {Mod}}_k^I(G)$ or in ${\mathrm {Mod}}_k(G)$ , coincide.

Since the subcategory ${\mathrm {Mod}}_k(G)_1$ is closed under extensions, it follows from Lemma 2.1.3 that every finitely generated representation lies in ${\mathrm {Mod}}_k(G)_1$ . The assertion about ${\mathrm {Mod}}_k(G)$ then is immediate from the fact that ${\mathrm {Mod}}_k(G)_1$ also is closed under inductive limits.

Proof. In [Reference StenströmSte] VI§1, the functor $r_0$ is called a left exact idempotent preradical of $\mathcal {C}$ and the asserted properties of the $r_j$ can be found in the Exercises 1 and 2 of that Chapter VI.

Proof. a) follows from [Reference StenströmSte] Propositions VI.1.2 and VI.1.7. For b) we observe that for any homomorphism $f : V_0 \rightarrow V_1$ in $\mathcal {C}$ we have $r_j(f) = f|r_j(V_0)$ . c), d) and f) hold by construction. e) We proceed by induction with respect to j. For $j = 0$ , the claim is part of the Definition 3.1.1. Now, assume that the claim holds for some j. Let $V = \varinjlim _i V_i$ be a filtered colimit in $\mathcal {C}$ . We then have

$$ \begin{align*} r_{j+1}(V)/r_j(V) & = r_0(V/r_j(V)) = r_0(\varinjlim_i V_i / r_j(\varinjlim_i V_i)) = r_0(\varinjlim_i V_i / \varinjlim_i r_j(V_i)) \\ & = r_0(\varinjlim_i (V_i / r_j(V_i))) = \varinjlim_i r_0(V_i / r_j(V_i)) = \varinjlim_i (r_{j+1}(V_i) / r_j(V_i)) \\ & = \varinjlim_i r_{j+1}(V_i) / \varinjlim_i r_j(V_i) = (\varinjlim_i r_{j+1}(V_i)) / r_j(V) \ , \end{align*} $$

where we use for Equations $4$ and $7$ that filtered colimits in $\mathcal {C}$ are exact. It follows that $r_{j+1}(V) = \varinjlim _i r_{j+1}(V_i)$ .

Proof. Since $r_j$ is left exact by Lemma 3.1.2.a, we have the commutative exact diagram

which exhibits $Y/r_j(Y)$ as a quotient object of Z. Since $\mathcal {A}$ is closed under quotient objects it follows that $r_0(Y/r_j(Y)) = Y/r_j(Y)$ . On the other hand, by construction we have $r_{j+1}(Y)/r_j(Y) = r_0(Y/r_j(Y))$ . It follows that $r_{j+1}(Y) = Y$ .

Proof. Write V as a filtered union of noetherian subobjects. Using Lemma 3.1.2.c, we see that it suffices to prove the assertion for a noetherian V. In this case, however, it follows from Definition 3.1.1.c and an iterated application of Lemma 3.1.4 that $V = r_j(V)$ for sufficiently large j.

Proof. By [Reference Bazzoni and PositselskiBP] Lemma 3.4, the category $\mathcal {A}_j$ is Grothendieck. That it is locally noetherian is clear.

Proof. This is immediate from $\mathcal {A}_j$ being closed under the passage to subobjects in $\mathcal {C}$ .

Proof. We have the adjunction ${\mathrm {Hom}}_{\mathcal {A}}(A, r(V)) = {\mathrm {Hom}}_{\mathcal {C}}(A,V)$ . Since r, being right adjoint to a (left) exact functor, preserves injective objects, this equation extends to the asserted composed functor spectral sequence.

Proof. a) Let $U \subseteq V$ be a nonzero subobject. Since r preserves nonzero objects, we then have $0 \neq r(U) \subseteq r(V)$ and hence $U \cap r(V) \neq 0$ .

b) We obviously have $V = r(V) \subseteq r(E_{\mathcal {C}}(V)) \subseteq E_{\mathcal {C}}(V)$ . Since r, as a right adjoint of a (left) exact functor, preserves injective objects, both terms in the essential extension $V = r(V) \subseteq r(E_{\mathcal {C}}(V))$ are injective in $\mathcal {A}$ . Hence, we must have equality.

c) Suppose that $r(V) = U_1 \oplus U_2$ with $U_i \neq 0$ (and necessarily lying in $\mathcal {A}$ ). Using [Reference StenströmSte] Proposition V.2.6, we see that V, being an injective hull of $r(V)$ by a), is isomorphic to the direct sum of injective hulls of $U_1$ and $U_2$ and hence is decomposable. On the other hand, if $V = V_1 \oplus V_2$ with $V_i \neq 0$ , then $r(V) = r(V_1) \oplus r(V_2)$ with $r(V_i) \neq 0$ .

Proof. [Reference HerzogHer] Theorem 2.8.

Proof. Each of the categories $\mathcal {A}$ and $\mathcal {L}$ is closed under subobjects, quotient objects and arbitrary direct sums. Therefore, the same holds for $\mathcal {A} \cap \mathcal {L}$ . It remains to consider extensions. Let $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$ be a short exact sequence in $\mathcal {A}$ such that X and Z lie in $\mathcal {L}$ . Then Y must lie in $\mathcal {L}$ and hence in $\mathcal {A} \cap \mathcal {L}$ .

Proof. We obviously have $\mathcal {A}^{noeth} = \mathcal {A} \cap \mathcal {C}^{noeth}$ . Hence, Proposition 3.4.1 reduces the asserted bijectivity to the bijectivity of the map

$$ \begin{align*} \text{collection of all Serre}\, & \xrightarrow{\;\simeq\;} \text{collection of all Serre} \\ \text{subcategories of}\ {\mathcal{C}}^{\text{noeth}} & \qquad \text{subcategories of}\ {\mathcal{A}}^{\text{noeth}} \\ & {\mathcal{S}} \longmapsto {\mathcal{A}} \cap {\mathcal{S}}. \end{align*} $$

But it follows from Definition 3.1.1.c that any such $\mathcal {S}$ is the smallest Serre subcategory of $\mathcal {C}^{noeth}$ which contains $\mathcal {A} \cap \mathcal {S}$ . On the other hand, let $\mathcal {T}$ be a Serre subcategory of $\mathcal {A}^{noeth}$ . We define $\mathcal {S}$ to be the full subcategory of $\mathcal {C}^{noeth}$ whose objects V have a finite filtration $0 = V_0 \subseteq V_1 \subseteq \ldots \subseteq V_m = V$ with $V_j/V_{j-1}$ in $\mathcal {T}$ for any $1 \leq j \leq m$ . It is straightforward to check that $\mathcal {S}$ is a Serre subcategory of $\mathcal {C}^{noeth}$ and that $\mathcal {A} \cap \mathcal {S} = \mathcal {T}$ .

Proof. Using [Reference HerzogHer] Proposition 3.2 and Corollary 3.5, we see that Definition 3.1.1.c implies that the sets

$$ \begin{align*} O_{\mathcal{C}}(U) & := \{[E] \in \mathbf{Sp}(\mathcal{C}) : {\mathrm{Hom}}_{\mathcal{C}}(U,E) \neq 0\} , \\ \text{resp.}\ O_{\mathcal{A}}(U) & := \{[r(E)] \in \mathbf{Sp}(\mathcal{A}) : {\mathrm{Hom}}_{\mathcal{A}}(U,r(E)) \neq 0\}, \end{align*} $$

for $U \in {\mathrm {ob}}(\mathcal {A})$ , form a base for the Ziegler-open subsets of $\mathbf {Sp}(\mathcal {C})$ , resp. $\mathbf {Sp}(\mathcal {A})$ . But, for such U, we have ${\mathrm {Hom}}_{\mathcal {C}}(U,E) = {\mathrm {Hom}}_{\mathcal {A}}(U,r(E))$ since r is right adjoint to the inclusion functor.

Proof. For the equivalence of a. and b. we consider more generally an arbitrary object of $\mathcal {D}$ , and we let $\langle C \rangle $ denote the smallest localising subcategory containing C. Let E be any injective object in $\mathcal {D}$ . If ${\mathrm {Hom}}_{\mathcal {D}}(-,E)$ vanishes on $\langle C \rangle $ , then obviously ${\mathrm {Hom}}_{\mathcal {D}}(C,E) = 0$ . But the injectivity of E easily implies that the converse holds as well. It follows that

$$ \begin{align*} A(\langle C \rangle) = \{ [E] \in \mathbf{Sp}(\mathcal{D}) : {\mathrm{Hom}}_{\mathcal{D}}(C,E) = 0\}. \end{align*} $$

For the equivalence of b. and c., see [Reference HerzogHer] Corollary 3.9.

Proof. We argue similarly as in [Reference GolanGol] Proposition 11.3.

$a) \Longrightarrow b)$ : Let E be an indecomposable injective object in $\mathcal {D}$ , and let $t_{\mathcal {L}}(E)$ denote the largest subobject of E contained in $\mathcal {L}$ . Suppose that $t_{\mathcal {L}}(E) \neq 0$ . Then E is an injective hull of $t (E)$ and hence, by stability, is contained in $\mathcal {L}$ .

$b) \Longrightarrow a)$ : We write an injective hull $E(V)$ of an object V lying in $\mathcal {L}$ as a direct sum $E(V) = \oplus _{i \in I} E_i$ of indecomposable injective objects $E_i$ . Since $E_i \cap V \neq 0$ lies in $\mathcal {L}$ for any $i \in I$ , we see that all $E_i$ and hence $E(V)$ lie in $\mathcal {L}$ .

Proof. $a) \Longrightarrow b)$ : Let $[E'] \in A(\mathcal {L})$ such that ${\mathrm {Hom}}_{\mathcal {D}}(E,E') \neq 0$ . Then E does not lie in $\mathcal {L}$ . Since $\mathcal {L}$ is stable, Lemma 3.5.1 applies and tells us that E does not have any nonzero subobject lying in $\mathcal {L}$ . Hence, $[E] \in A(\mathcal {L}) = A$ .

$b) \Longrightarrow a)$ : Let $\mathcal {L}$ be the localising subcategory of $\mathcal {D}$ cogenerated by the $E'$ for $[E'] \in A$ . This means that $\mathcal {L}$ is the full subcategory of those objects V in $\mathcal {D}$ which satisfy ${\mathrm {Hom}}_{\mathcal {D}}(V,E') = 0$ for any $[E'] \in A$ . It is immediate that $A \subseteq A(\mathcal {L})$ . Consider any $[E] \in A(\mathcal {L})$ . Then E cannot lie in $\mathcal {L}$ . Hence, there must exist an $[E'] \in A$ such that ${\mathrm {Hom}}_{\mathcal {D}}(E,E') \neq 0$ . It follows from (b) that $[E] \in A$ . This shows that $A = A(\mathcal {L})$ . To establish that $\mathcal {L}$ is stable, we use Lemma 3.5.1. We have just seen that the E which do not lie in $\mathcal {L}$ must have $[E] \in A(\mathcal {L})$ . By the very definition of $A(\mathcal {L})$ , such E do not have a nonzero subobject lying in $\mathcal {L}$ .

Proof. It suffices to consider the map $\mathbf {Sp}(\mathcal {C}) \rightarrow \mathbf {Sp}(\mathcal {A})$ sending $[E]$ to $[r(E)]$ . Let $S \subseteq \mathbf {Sp}(\mathcal {C})$ be any stable subset and $[r(E')]$ be any point in the image of S. Now, suppose that we have $[U] \in \mathbf {Sp}(\mathcal {A})$ such that ${\mathrm {Hom}}_{\mathcal {A}}(U,r(E')) \neq 0$ . Then ${\mathrm {Hom}}_{\mathcal {C}}(U,E') \neq 0$ and hence ${\mathrm {Hom}}_{\mathcal {C}}(E_{\mathcal {C}}(U),E') \neq 0$ . Using Lemma 3.5.3, we deduce that $[E (U)] \in S$ . But $U \cong r(E_{\mathcal {C}}(U))$ . It follows that $[U]$ lies in the image of S. Using again Lemma 3.5.3, we conclude that the image of S is stable.

Proof. Let V be any object in $\mathcal {L}$ . Since $\mathcal {L}$ is stable an injective hull $E(V)$ of V in $\mathcal {C}$ lies already in $\mathcal {L}$ . Hence, we find an injective resolution $V \xrightarrow {\simeq } I^\bullet $ of V in $\mathcal {C}$ all of whose terms lie in $\mathcal {L}$ . The values $R^j r(V)$ of the right derived functors in question are the cohomology objects of the complex $r(I^\bullet )$ . On the other hand, Definition 3.1.1.b easily implies that r maps $\mathcal {L}$ to $\mathcal {A} \cap \mathcal {L}$ . It follows that the complex $r(I^\bullet )$ and then also its cohomology objects lie in $\mathcal {A} \cap \mathcal {L}$ .

Proof. a) By Proposition 3.2.1, we have the convergent Grothendieck spectral sequence

$$ \begin{align*} {\mathrm{Ext}}^i_{\mathcal{A}}(A, R^jr(V)) \Rightarrow {\mathrm{Ext}}^{i+j}_{\mathcal{C}}(A,V) . \end{align*} $$

The exact sequence of low degree terms is

$$ \begin{align*} 0 \rightarrow {\mathrm{Ext}}^1_{\mathcal{A}}(A, r(V)) \rightarrow {\mathrm{Ext}}^1_{\mathcal{C}}(A,V) \rightarrow {\mathrm{Hom}}_{\mathcal{A}}(A, R^1r(V)) \rightarrow {\mathrm{Ext}}^2_{\mathcal{A}}(A, r(V)) . \end{align*} $$

Since V lies in $\mathcal {A}$ , we have $r(V) = V$ . Moreover, using Equation (5) for the second equality, we obtain

$$ \begin{align*} {\mathrm{Hom}}_{\mathcal{A}}(A, R^1r(V)) & = {\mathrm{Hom}}_H({\mathrm{Hom}}_{\mathcal{A}}(P,A),{\mathrm{Hom}}_{\mathcal{A}}(P, R^1 r(V))) \\ & = {\mathrm{Hom}}_H({\mathrm{Hom}} (P,A),{\mathrm{Ext}}_{\mathcal{C}}^1(P,V)) . \end{align*} $$

b) Since V is injective, we have ${\mathrm {Ext}}^1_{\mathcal {A}}(A,V) = {\mathrm {Ext}}^2_{\mathcal {A}}(A,V) = 0$ .

Proof. We first note that we have $r(V) \neq 0$ by Definition 3.1.1.b.

Let $E_{\mathcal {C}}(V)$ be an injective hull of V in $\mathcal {C}$ , and let $E = r_1(E_{\mathcal {C}}(V))$ . Then $r(E) = r(E_{\mathcal {C}}(V))$ is an injective object in $\mathcal {A}$ , and $r(V) \subseteq r(E)$ . In fact, we have the inclusions

By Lemma 3.3.1.a, the extension $r(V) \subseteq V$ is essential. Hence, the extension $r(V) \subseteq r(E)$ is essential as well. Since $\mathcal {A} \cap \mathcal {L}$ is stable in $\mathcal {A}$ and since $r(V)$ lies $\mathcal {A} \cap \mathcal {L}$ by assumption b), we see that $r(E)$ lies in $\mathcal {A} \cap \mathcal {L}$ .

Now, because $r_1$ is left exact, $V = r_1(V) \subseteq r_1(E_{\mathcal {C}}(V)) = E$ . Next, $(r(E) + V)/r(E) \cong V / (V \cap r(E)) = V / r(V)$ because r is left exact. Since $r(V)$ is the largest subobject of V contained in $\mathcal {L}$ , we see that $V/r(V)$ is $\mathcal {L}$ -torsion-free. Also, $V = r_1(V)$ means that $V / r(V)$ lies in $\mathcal {A}$ . We deduce that $(r(E) + V)/r(E) \cong V / r(V)$ lies in $\mathcal {A}$ and is $\mathcal {L}$ -torsion-free. Hence, we need to show that $V \subseteq r(E)$ .

Consider any intermediate object $r(E) \subseteq V^{\prime \prime } \subseteq V' := r(E) + V \subseteq E$ in $\mathcal {C}$ . We obtain extension classes $e' = [0 \rightarrow r(E) \rightarrow V' \rightarrow V'/r(E) \rightarrow 0] \in {\mathrm {Ext}}_{\mathcal {C}}^1(V'/r(E),r(E))$ and analogously $e^{\prime \prime } \in {\mathrm {Ext}}_{\mathcal {C}}^1(V^{\prime \prime }/r(E),r(E))$ . Note that $V'/r(E)$ , $V^{\prime \prime }/r(E)$ and $r(E)$ are objects in $\mathcal {A}$ with $V'/r(E)$ and $V^{\prime \prime }/r(E)$ being $\mathcal {A} \cap \mathcal {L}$ -torsion-free and $r(E)$ being injective in $\mathcal {A}$ . Hence, Lemma 3.5.6 applies. We define the H-modules $N^{\prime \prime } := {\mathrm {Hom}}_{\mathcal {A}}(P,V^{\prime \prime }/r(E)) \subseteq N' := {\mathrm {Hom}}_{\mathcal {A}}(P,V'/r(E))$ and obtain the commutative diagram

with horizontal isomorphisms. Obviously, the left perpendicular arrow maps $e'$ to $e^{\prime \prime }$ . Since $r(E)$ lies in $\mathcal {A} \cap \mathcal {L}$ the H-module ${\mathrm {Ext}}^1_{\mathcal {C}}(P, r(E))$ lies in $\mathcal {L}^H$ by our assumptions. We have the H-linear map $f' := \varphi '(e') : N' \rightarrow {\mathrm {Ext}}^1_{\mathcal {C}}(P, r(E))$ . Suppose now that $N' \neq 0$ , and choose a nonzero element $v \in N'$ . Let $J := {\mathrm {Ann}}_H(f'(v)) \subseteq H$ be the annihilator left ideal. Then $H/J \cong Hf'(v)$ lie in $\mathcal {L}^H$ . We now consider specifically the H-submodule $N^{\prime \prime } := Jv \subseteq N'$ . By the category equivalence, there is a unique $V^{\prime \prime }$ as above such that $N^{\prime \prime } = {\mathrm {Hom}}_{\mathcal {A}}(P,V^{\prime \prime }/r(E))$ . We first check that $N^{\prime \prime } \neq 0$ . Otherwise, we would have $J \subseteq {\mathrm {Ann}}_H(v)$ so that $H/J$ surjects onto $Hv \subseteq N'$ , which implies that $Hv$ lies in $\mathcal {L}^H$ . On the other hand the modules $Hv \subseteq N'$ are $\mathcal {L}^H$ -torsion-free, which is a contradiction. Hence, $N^{\prime \prime } \neq 0$ . Now,

$$ \begin{align*} {\mathrm{res}}(\varphi'(e'))(N^{\prime\prime}) = {\mathrm{res}}(f')(N^{\prime\prime}) = f'(N^{\prime\prime}) = f'(Jv) = J f'(v) = 0 . \end{align*} $$

The commutativity of the above diagram then implies that the extension class $e^{\prime \prime } = 0$ , that is, the short exact sequence $0 \rightarrow r(E) \rightarrow V^{\prime \prime } \rightarrow V^{\prime \prime }/r(E) \rightarrow 0$ in $\mathcal {C}$ splits. But then because $V^{\prime \prime }/r(E) \neq 0$ , we see $r(E)$ is not essential in E, which contradicts Lemma 3.3.1.a. It follows that $N'$ must be zero, that is, that $r(E) + V = r(E)$ .

Proof. a) $\Rightarrow $ b). This is Lemma 3.5.5.

b) $\Rightarrow $ c). This follows from Equation (5) and the definition of $\mathcal {L}^H$ .

c) $\Rightarrow $ d) $\Rightarrow $ e). These are trivial.

e) $\Rightarrow $ a). Let E be an indecomposable injective object in $\mathcal {C}$ , and let Y be its largest subobject in $\mathcal {L}$ . We suppose that $Y \neq 0$ and will aim to show that $Y = E$ . Let $E_n := r_n(E)$ for $n \geq 0$ , and note that $E = \bigcup \limits _{n=0}^\infty E_n$ by Lemma 3.1.5. Correspondingly, $Y = \bigcup \limits _{n=0}^\infty Y_n$ with $Y_n := r_n(Y)$ , and note that $Y_n = Y \cap V_n$ for all $n \geq 0$ because each $r_n$ is left exact.

Since Y is the largest subobject of E contained in $\mathcal {L}$ , $E/Y$ is $\mathcal {L}$ -torsion free. Hence, for any $n \geq 0$ , $(E_n + Y)/Y$ is also $\mathcal {L}$ -torsion free. But $(E_n + Y)/Y \cong E_n / (E_n \cap Y) = E_n / Y_n$ . Hence, for all $n \geq 0$ , $E_n / Y_n$ is $\mathcal {L}$ -torsion-free, and $Y_n$ is the largest subobject of $E_n$ contained in $\mathcal {L}$ .

With $Y \neq 0$ , also $Y_0 = r(Y)$ is nonzero. Since E is an indecomposable injective in $\mathcal {C}$ , it follows that $Y_0$ is essential in E.

Suppose for a contradiction that $Y \neq E$ , and let n be minimal such that $Y_n \neq E_n$ . Now, $Y_0 \subseteq E_0$ is an essential extension in $\mathcal {A}$ with $Y_0$ lying in $\mathcal {A} \cap \mathcal {L}$ ; since $\mathcal {A} \cap \mathcal {L}$ is stable in $\mathcal {A}$ , we see that $Y_0 = E_0$ . Hence, $n \geq 1$ , and the minimality of n implies that $Y_{n-1} = E_{n-1}$ so that $E_{n-1} \subseteq Y_n$ .

Now, consider the short exact sequence

$$ \begin{align*} 0 \rightarrow Y_n / E_{n-1} \rightarrow E_n / E_{n-1} \rightarrow E_n / Y_n \rightarrow 0 \end{align*} $$

in $\mathcal {A}$ . Note that $Y_n / E_{n-1}$ lies in $\mathcal {L}$ and that $E_n / Y_n$ is $\mathcal {L}$ -torsion-free and nonzero. If $Y_n / E_{n-1}$ is also nonzero, then because $\mathcal {A} \cap \mathcal {L}$ is stable in $\mathcal {A}$ , this extension is not essential. This gives us some subobject W of $E_n$ containing $E_{n-1}$ such that $W / E_{n-1}$ is isomorphic to a nonzero subobject of $E_n/Y_n$ ; thus, $W / E_{n-1}$ is also nonzero and $\mathcal {L}$ -torsion-free. If $Y_n = E_{n-1}$ , then we can take $W = E_n$ to obtain an object with the same properties.

Write $E_{-1} := 0$ and $X := W/E_{n-2} \subseteq E_n / E_{n-2}$ , and consider the short exact sequence

$$ \begin{align*} 0 \rightarrow E_{n-1} / E_{n-2} \rightarrow X \rightarrow W / E_{n-1} \rightarrow 0 . \end{align*} $$

Since the outer terms in this short exact sequence lie in $\mathcal {A}$ , we see that $r_1(X) = X$ . Also, $r(X) \subseteq r(E_n / E_{n-2}) = E_{n-1}/E_{n-2} = r( E_{n-1}/E_{n-2} ) \subseteq r(X)$ , where the left equality uses Lemma 3.1.2.b, shows that

$$ \begin{align*} r(X) = E_{n-1} / E_{n-2} \subsetneqq X . \end{align*} $$

Since $W / E_{n-1}$ is $\mathcal {L}$ -torsion-free and $E_{n-1} / E_{n-2} = Y_{n-1} / E_{n-2}$ is a subquotient of Y, we see that $r(X)$ is the largest subobject of X contained in $\mathcal {L}$ . Applying now Proposition 3.5.7 with X leads to a contradiction.

Proof. a) [Reference GabrielGab] Proposition 12 on p. 431. b) The equality is [Reference McConnell and RobsonMCR] Corollary 10.1.10, and the stability then follows from a). c) Because of b) this follows from Remark 2.2.1.

Proof. (Recall the notation $A(\mathcal {L})$ from $\S $ 3.4.) Using Corollary 3.5.2, we compute

$$ \begin{align*} A(\varrho^{-1}(\mathcal{L})) & = \{\mathfrak{p} \not\in {\mathrm{Spec}}(H) : E_{\mathfrak{p}} \in {\mathrm{ob}}(\varrho^{-1}(\mathcal{L})) \} \\ & = \{\mathfrak{p} \in {\mathrm{Spec}}(H) : H/\mathfrak{p} \not\in {\mathrm{ob}}(\varrho^{-1}(\mathcal{L})) \} \\ & = \{\mathfrak{p} \in {\mathrm{Spec}}(H) : \varrho(H/\mathfrak{p}) \not\in {\mathrm{ob}}(\mathcal{L}) \} \\ & \supseteq \{\mathfrak{p} \in {\mathrm{Spec}}(H) : Z(H)/\varphi(\mathfrak{p}) \not\in {\mathrm{ob}}(\mathcal{L}) \} \\ & = \varphi^{-1}(A(\mathcal{L})) . \end{align*} $$

For the reverse inclusion, we suppose that $Z(H)/\varphi (\mathfrak {p})$ lies in $\mathcal {L}$ . Being a finitely generated $Z(H)/\varphi (\mathfrak {p})$ -module $\varrho (H / \varphi (\mathfrak {p}) H)$ also must lie in ${\mathrm {ob}}(\mathcal {L})$ and hence $H / \varphi (\mathfrak {p}) H \in {\mathrm {ob}}(\varrho ^{-1}(\mathcal {L}))$ . It follows that $H/\mathfrak {p}$ , as a quotient of $H / \varphi (\mathfrak {p}) H$ , is contained in ${\mathrm {ob}}(\varrho ^{-1}(\mathcal {L}))$ .

Proof. a) Let L be a uniform left ideal in $H/\mathfrak {p}$ so that $E_{\mathfrak {p}}$ is the injective hull $E(L)$ of L.

Suppose first that $z \notin \mathfrak {p}$ . Multiplication by z is an injective H-linear map $\ell _z : H/\mathfrak {p} \to H/\mathfrak {p}$ : If $x + \mathfrak {p} \in H/\mathfrak {p}$ is such that $z(x+\mathfrak {p}) = \mathfrak {p}$ , then $(HzH)(HxH) \subseteq \mathfrak {p}$ since z is central in H, so $x \in \mathfrak {p}$ since $\mathfrak {p}$ is prime and $z \notin \mathfrak {p}$ . Since L is an H-submodule of $H/\mathfrak {p}$ , $\ell _z : L \to L$ is also injective. Hence, it extends to an injective H-linear map $\ell _z : E(L) \to E(L)$ by the injectivity of $E(L)$ . The image of $\ell _z$ must admit a complement in $E(L)$ since it is itself injective. Since $E(L) \cong E_{\mathfrak {p}}$ is indecomposable, $\ell _z : E(L) \to E(L)$ is an isomorphism, as claimed.

Now, suppose that $z \in \mathfrak {p}$ and let $x \in E_{\mathfrak {p}}$ be nonzero. Since L is essential in $E_{\mathfrak {p}}$ , we see that $Hx \cap L$ is essential in $Hx$ . Since H is noetherian and z is central, the ideal $zH$ of H satisfies the left Artin–Rees property by [Reference McConnell and RobsonMCR] Proposition 4.2.6. Since $z \in \mathfrak {p}$ kills L, it also kills $Hx \cap L$ . So $z^n$ kills $Hx$ for some $n \geq 1$ by the implication (i) $\Rightarrow $ (iii) of [Reference McConnell and RobsonMCR] Theorem 4.2.2.

b) Let $\phi : E_{\mathfrak {p}} \rightarrow E_{\mathfrak {q}}$ be a nonzero map. Suppose that there is a $z \in (\mathfrak {p} \cap Z(H)) \setminus \mathfrak {q}$ . Then, by a), z acts invertibly on $E_{\mathfrak {q}}$ but locally nilpotently on $E_{\mathfrak {p}}$ . Consider any $x \in E_{\mathfrak {p}}$ such that $\phi (x) \neq 0$ . We find a $t \geq 1$ such that $z^t x = 0$ . Hence, $z^t \phi (x) = \phi (z^t x) = \phi (0) = 0$ , which is a contradiction.

Proof. Since $\mathcal {L}_{\varphi ^{-1}(W)}$ is localising we may assume that M is finitely generated. If $v_1, \ldots , v_r$ are generators of the H-module M, then

$$ \begin{align*} V({\mathrm{Ann}}_{Z(H)}(M)) = V\left(\bigcap_{i=1}^r {\mathrm{Ann}}_{Z(H)}(v_i)\right) = \bigcup_{i=1}^r V({\mathrm{Ann}}_{Z(H)}(v_i)) . \end{align*} $$

This shows the equivalence of b) and c).

For the equivalence of a) and b), we first note that, because H is noetherian, M contains an essential submodule N of the form $N = N_1 \oplus \cdots \oplus N_r$ , where each $N_i$ is a uniform H-module ([Reference Goodearl and WarfieldGW] Corollary 5.18 and Proposition 5.15). Hence, $E(M) = E(N) \cong \bigoplus _{i=1}^n E(N_i)$ . By [Reference Goodearl and WarfieldGW] Lemma 5.26, for each $i = 1,\cdots , r$ there is a unique prime ideal $\mathfrak {p}_i$ of H which is equal to the annihilator of some nonzero H-submodule $N_i'$ of $N_i$ , and which contains the annihilators of all nonzero H-submodules of $N_i$ . This $\mathfrak {p}_i$ is the assassinator of $N_i$ in the sense of [Reference Goodearl and WarfieldGW] Definition on p. 102.

Since H is assumed to be a finitely generated module over its centre, it is an FBN-ring by [Reference Goodearl and WarfieldGW] Proposition 9.1(a). Then we can apply [Reference Goodearl and WarfieldGW] Proposition 9.14 to see that the uniform injective H-module $E(N_i)$ is isomorphic to the H-injective hull $E(L_i)$ of any uniform left ideal $L_i$ of $H/\mathfrak {p}_i$ . In other words, ([Reference Goodearl and WarfieldGW] Lemma 5.14), $E(N_i) \cong E_{\mathfrak {p}_i}$ and

$$ \begin{align*} E(M) \cong E(N) \cong \bigoplus_{i=1}^r E(N_i) \cong \bigoplus_{i=1}^r E_{\mathfrak{p}_i} . \end{align*} $$

We identify M with its image inside $E(M)$ . Note that M is not contained in the direct sum $\bigoplus _{i \neq j} E_{\mathfrak {p}_i}$ for any j because otherwise we would have $M \cap E_{\mathfrak {p}_j} = 0$ , which contradicts the fact that M is essential in $E(M)$ . So, ${\mathrm {Hom}}_H(M, E_{\mathfrak {p}_j})$ is nonzero for any j.

a) $\Rightarrow $ b). Suppose that M lies in $\mathcal {L}_{\varphi ^{-1}(W)}$ . From ${\mathrm {Hom}}_H(M,E_{\mathfrak {p}_j}) \neq 0$ for any j, we deduce that $\mathfrak {p}_j \notin \varphi ^{-1}(W)$ for all j. Now, consider any nonzero $v \in M$ and any $\mathfrak {p} \in \varphi ^{-1}(V({\mathrm {Ann}}_{Z(H)}(v)))$ . Because $\mathfrak {p}_i \cap Z(H)$ acts locally nilpotently on $E_{\mathfrak {p}_i}$ by Lemma 3.6.3.a, we see that

$$ \begin{align*} ((\mathfrak{p}_1 \cap Z(H)) (\mathfrak{p}_2 \cap Z(H)) \cdots (\mathfrak{p}_r \cap Z(H))^k \cdot v = 0 \end{align*} $$

for some $k> 0$ . This implies that $\mathfrak {p}_j \cap Z(H) \subseteq \mathfrak {p} \cap Z(H)$ for some j. Since W is generalization-closed and $\mathfrak {p}_j \cap Z(H) \notin W$ , we see that $\mathfrak {p} \cap Z(H) \notin W$ . Hence, $V({\mathrm {Ann}}_{Z(H)}(v)) \cap W = \emptyset $ .

b) $\Rightarrow $ a). Suppose that ${\mathrm {Hom}}_H(M, E_{\mathfrak {q}}) \neq 0$ for some $\mathfrak {q} \in \varphi ^{-1}(W)$ . Then ${\mathrm {Hom}}_H(E_{\mathfrak {p}_i}, E_{\mathfrak {q}}) \neq 0$ for some $1 \leq i \leq r$ and so $\mathfrak {p}_i \cap Z(H) \subseteq \mathfrak {q} \cap Z(H)$ by Lemma 3.6.3.b. Hence, $\mathfrak {p}_i \cap Z(H) \in W$ since W is generalization-closed.

Choose a nonzero element $v \in N_i'$ ; since $\mathfrak {p}_i$ kills $N_i' \supseteq Hv$ , we have $\mathfrak {p}_i \subseteq {\mathrm {Ann}}_H(Hv)$ ; on the other hand, $\mathfrak {p}_i$ contains ${\mathrm {Ann}}_H(Hv)$ as we saw above and hence $\mathfrak {p}_i = {\mathrm {Ann}}_H(Hv)$ . Since $Z(H)$ is central in H, we have ${\mathrm {Ann}}_{Z(H)}(v) = {\mathrm {Ann}}_{Z(H)}(Hv)$ . This ideal of $Z(H)$ is equal to ${\mathrm {Ann}}_H(Hv) \cap Z(H) = \mathfrak {p}_i \cap Z(H)$ . Thus, ${\mathrm {Ann}}_{Z(H)}(v) = \mathfrak {p}_i \cap Z(H)$ lies in W and therefore $V({\mathrm {Ann}}_{Z(H)}(v)) \cap W \neq \emptyset $ .

Proof. In the Grothendieck category $\mathcal {C}$ filtered colimits are exact. Moreover, because $\mathcal {C}$ is locally noetherian and M is a noetherian object in $\mathcal {C}$ , [Reference GabrielGab] Corollary 1 on p. 358 tells us that:

• – the full subcategory of injective objects in $\mathcal {C}$ is closed under filtered colimits;

• – the functor $F := {\mathrm {Hom}}_{\mathcal {C}}(M,-)$ from $\mathcal {C}$ to the category of abelian groups commutes with filtered colimits.

In this situation, [Reference Kashiwara and SchapiraKS] Proposition 15.3.3 implies that the derived functors $R^jF = {\mathrm {Ext}}^j_{\mathcal {C}}(M,-)$ of F commute with filtered colimits.

Proof. By Lemma 3.7.2, the functors ${\mathrm {Ext}}^j_{\mathcal {C}}(P,-)$ , for any $j \geq 0$ , commute with arbitrary direct sums. It follows that $\mathcal {J}$ annihilates ${\mathrm {Ext}}_{\mathcal {C}}^*(P,F)$ for any F which is a possibly infinite direct sums of copies of P. Since P is a generatorFootnote ² of $\mathcal {A}$ , using [Reference StenströmSte] p. 93 Proposition 6.2 repeatedly we find an exact sequence in $\mathcal {A}$ of the form

$$ \begin{align*} \ldots \longrightarrow F_n \longrightarrow \ldots \longrightarrow F_0 \longrightarrow A \longrightarrow 0, \end{align*} $$

where each $F_n$ is a direct sum of copies of P. We break it up into short exact sequences

$$ \begin{align*} & 0 \rightarrow Y_0 \rightarrow F_0 \rightarrow A \rightarrow 0 \\ & 0 \rightarrow Y_1 \rightarrow F_1 \rightarrow Y_0 \rightarrow 0 \\ & \cdots \\ & 0 \rightarrow Y_n \rightarrow F_n \rightarrow Y_{n-1} \rightarrow 0 \\ & \cdots \end{align*} $$

Applying ${\mathrm {Ext}}^*_{\mathcal {C}}(P,-)$ we obtain exact sequences

$$ \begin{align*} & {\mathrm{Ext}}^i_{\mathcal{C}}(P,F_0) \rightarrow {\mathrm{Ext}}^i_{\mathcal{C}}(P,A) \xrightarrow{\delta} {\mathrm{Ext}}^{i+1}_{\mathcal{C}}(P,Y_0) \\ & {\mathrm{Ext}}^{i+1}_{\mathcal{C}}(P,F_1) \rightarrow {\mathrm{Ext}}^{i+1}_{\mathcal{C}}(P,Y_0) \xrightarrow{\delta} {\mathrm{Ext}}^{i+2}_{\mathcal{C}}(P,Y_1) \\ & \cdots \\ & {\mathrm{Ext}}^{i+n}_{\mathcal{C}}(P,F_n) \rightarrow {\mathrm{Ext}}^{i+n}_{\mathcal{C}}(P,Y_{n-1}) \xrightarrow{\delta} {\mathrm{Ext}}^{i+n+1}_{\mathcal{C}}(P,Y_n) \ , \\ & \cdots \end{align*} $$

where $\delta $ denotes the connecting homomorphisms. These are sequences of $Z\otimes Z$ -bimodules. The first term in each sequence is annihilated by $\mathcal {J}$ . Now, choose $n := d - i$ . Then ${\mathrm {Ext}}^{i+n+1}_{\mathcal {C}}(P,Y_n) = 0$ . The assertion now follows by downward induction along the above sequences.

Proof of Theorem 3.7.1.

Since $\mathcal {L}_W^H = \varrho ^{-1}(\mathcal {L}_W^Z)$ , it follows from Lemma 3.6.1 that $\mathcal {L}_W^H$ is stable in ${\mathrm {Mod}}(H)$ . Hence, $\mathcal {L}_W^{\mathcal {A}} = \mathcal {L}_W^{\mathcal {C}} \cap \mathcal {A}$ is stable in $\mathcal {A}$ . Therefore, by Theorem 3.5.8, it is enough to show that the functor ${\mathrm {Ext}}^1_{\mathcal {C}}(P,-)$ maps $\mathcal {L}_W^{\mathcal {A}}$ to $\mathcal {L}_W^H$ . By Lemma 3.7.2, the functor ${\mathrm {Ext}}^1_{\mathcal {C}}(P,-)$ commutes with filtered colimits. Hence, it suffices to show that for any noetherian object A in $\mathcal {L}_W^{\mathcal {C}} \cap \mathcal {A}$ , the H-module $M := {\mathrm {Ext}}^1_{\mathcal {C}}(P,A)$ lies in $\mathcal {L}_W^H$ . Since the functor ${\mathrm {Ext}}^1_{\mathcal {C}}(P,-)$ is also half-exact, we may assume further, using a prime series for ${\mathrm {Hom}}_{\mathcal {A}}(P,A)$ as in [Reference Goodearl and WarfieldGW] Proposition 3.13, that the annihilator of any nonzero H-submodule of ${\mathrm {Hom}}_{\mathcal {A}}(P,A)$ is a prime ideal $\mathfrak {p}$ of H. Then necessarily $\mathfrak {p} = {\mathrm {Ann}}_H({\mathrm {Hom}}_{\mathcal {A}}(P,A))$ , and we write $\mathfrak {p}_Z := \mathfrak {p} \cap Z \in \Xi $ .

By Lemma 3.7.3, the $Z \otimes Z$ -module $M = {\mathrm {Ext}}^1_{\mathcal {C}}(P,A)$ is killed by $\mathcal {J}^{d+1}$ . Letting $\mathcal {R}_d := {\mathrm {Spec}}((Z \otimes Z) / \mathcal {J}^{d+1})$ , we can then regard M as an $\mathcal {O}(\mathcal {R}_d) = (Z \otimes Z) / \mathcal {J}^{d+1}$ -module in a natural way. Next, consider the commutative diagram

(8)

where i is the closed embedding, and the $f_j$ are, similarly as before, the restrictions of the two projection maps.

Let $S := \Xi \setminus W$ ; then our assumption $\pi _2(\pi _1^{-1}(W)) \subseteq W$ is equivalent to $\pi _1^{-1}(W) \subseteq \pi _2^{-1}(W)$ , which is the same as $\pi _2^{-1}(S) \subseteq \pi _1^{-1}(S)$ . Since $\pi _j = f_j \circ i$ , this is equivalent to $i^{-1} f_2^{-1}(S) \subseteq i^{-1} f_1^{-1}(S)$ . However, i is a bijection because $\mathcal {J}$ is nilpotent modulo $\mathcal {J}^{d+1}$ , so this is in turn equivalent to $f_2^{-1}(S) \subseteq f_1^{-1}(S)$ .

Now, $\mathfrak {p}$ is the annihilator of any nonzero H-submodule of ${\mathrm {Hom}}_{\mathcal {A}}(P,A)$ , so $\mathfrak {p}_Z = \mathfrak {p} \cap Z = {\mathrm {Ann}}_Z(v)$ for all nonzero $v \in {\mathrm {Hom}}_{\mathcal {A}}(P,A)$ . Since ${\mathrm {Hom}}_{\mathcal {A}}(P,A)$ lies in $\mathcal {L}_W^H$ , this means that $V(\mathfrak {p}_Z) \cap W = \emptyset $ by Proposition 3.6.4, and hence $V(\mathfrak {p}_Z) \subseteq S = \Xi \setminus W$ .

By assumption, $\mathcal {O}(\mathcal {R}_d) = (Z \otimes Z) / \mathcal {J}^{d+1}$ is noetherian. Let $\overline {i}_2 : Z \to (Z \otimes Z) / \mathcal {J}^{d+1}$ , sending z to $1 \otimes z + \mathcal {J}^{d+1}$ , be the ring homomorphism which induces $f_2$ . Let $Q_1,\cdots , Q_k \subseteq \mathcal {O}(\mathcal {R}_d)$ be the minimal prime ideals lying above $\overline {i}_2(\mathfrak {p}_Z) \mathcal {O}(\mathcal {R}_d)$ ; then $Q_1^{a_1}\cdots Q_k^{a_k} \subseteq \overline {i}_2(\mathfrak {p}_Z) \mathcal {O}(\mathcal {R}_d)$ for some positive integers $a_1,\cdots , a_k$ . We have $f_2(Q_j) = (\overline {i}_2)^{-1}(Q_j) \in V(\mathfrak {p}_Z)$ because $Q_j$ contains $\overline {i}_2(\mathfrak {p}_Z)$ . Therefore, $Q_j \in f_2^{-1}(V(\mathfrak {p}_Z)) \subseteq f_2^{-1}(S) \subseteq f_1^{-1}(S)$ . Thus, we see that $\mathfrak {q}_j := f_1(Q_j)$ lies in S for all $j = 1,\cdots , k$ . Hence, $V(\mathfrak {q}_j) \subseteq S$ for all j because S is specialization-closed.

With ${\mathrm {Hom}}_{\mathcal {A}}(P,A)$ also A is killed by $\mathfrak {p}_Z$ . Hence, $M = {\mathrm {Ext}}^1_{\mathcal {C}}(P, A)$ is killed by $Z \otimes \mathfrak {p}_Z$ as well as $\mathcal {J}^{d+1}$ . We see that M is killed by $\overline {i}_2(\mathfrak {p}_Z) \mathcal {O}(\mathcal {R}_d) \supseteq Q_1^{a_1}\cdots Q_k^{a_k}$ . Hence, $\mathfrak {q}_1^{a_1} \cdots \mathfrak {q}_k^{a_k} \cdot v = 0$ for any $v \in M$ , so $V({\mathrm {Ann}}_Z(v)) \subseteq V(\mathfrak {q}_1) \cup \cdots \cup V(\mathfrak {q}_k) \subseteq S$ . Therefore, the H-module M (where H acts on M through its action on P) lies in $\mathcal {L}^H_W$ by Proposition 3.6.4 as required.

Proof. Put $W = \mathbf {q}^{-1}(U)$ . Then

$$\begin{align*}\pi_2(\pi_1^{-1}(W)) = \pi_2(\pi_1^{-1} \mathbf{q}^{-1}(U)) = \pi_2(\pi_2^{-1}\mathbf{q}^{-1}(U)) \subseteq \mathbf{q}^{-1}(U) = W.\end{align*}$$

Now, apply Theorem 3.7.1.

Proof. An induction on the number of generators of M together with [Reference McConnell and RobsonMCR] Lemma 6.2.4 shows the first equality, namely $\kappa (M) = \sup _{v \in M} \kappa (Av)$ .

Let I be an ideal of A. Since A is noetherian, [Reference McConnell and RobsonMCR] Lemma 6.2.4 implies that $\kappa (A/I) = \kappa (A / I^m)$ for all $m \geq 0$ . Since $(\sqrt {I})^m \subseteq I$ for sufficiently large m, we have $\kappa (A/I) \geq \kappa (A/\sqrt {I}) = \kappa (A/(\sqrt {I})^m) \geq \kappa (A/I)$ , so in fact we have the equality $\kappa (A/I) = \kappa (A/\sqrt {I})$ . Since $\sqrt {I}$ is equal to the intersection of the finitely many minimal primes $\mathfrak {p}_1, \cdots , \mathfrak {p}_n$ containing it, there is a natural A-linear embedding $A / \sqrt {I} \hookrightarrow \oplus _{i=1}^n A / \mathfrak {p}_i$ . So

$$\begin{align*}\kappa(A/I) = \kappa(A/\sqrt{I}) \leq \sup_{1\leq i \leq n} \kappa(A/\mathfrak{p}_i) \leq \sup_{\mathfrak{p} \in V(I)} \kappa(A/\mathfrak{p}) \leq \kappa(A/I).\end{align*}$$

Therefore, for any $v \in M$ , $\kappa (Av) = \kappa (A/{\mathrm {Ann}}_A(v)) = \sup \limits _{\mathfrak {p} \in V({\mathrm {ann}}(v))} \kappa (A/\mathfrak {p})$ .

Proof. By Gabriel’s definition of Krull dimension found at [Reference GabrielGab] p. 382, we see that $\kappa (V) = \sup _i \kappa (V_i)$ if we write $V = \varinjlim V_i$ as a filtered colimit of its noetherian subobjects $V_i$ . Therefore, it is enough to show that for noetherian $V \in \mathcal {C}$ , we have $V \in \mathcal {L}^{\mathcal {C}}_{W_\alpha }$ if and only if $\kappa (V) \leq \alpha $ . Since $\mathcal {C}$ is a thickening of $\mathcal {A}$ , we may further assume that $V \in \mathcal {A}$ . Hence, we are reduced to showing that for finitely generated H-modules M, we have

$$\begin{align*}M \in \mathcal{L}^H_{W_\alpha} \quad \Leftrightarrow \quad \kappa(M) \leq \alpha.\end{align*}$$

However, since H is a finitely generated Z-module, $\kappa (M) = \kappa (\rho (M))$ by [Reference McConnell and RobsonMCR] Corollary 10.1.10, whereas $M \in \mathcal {L}^H_{W_\alpha }$ if and only if $\rho (M) \in \mathcal {L}^Z_{W_\alpha }$ by equation (7). Next, $\rho (M) \in \mathcal {L}^Z_{W_\alpha }$ if and only if for all $v \in M$ and all $\mathfrak {p} \in {\mathrm {Spec}}(Z)$ containing ${\mathrm {Ann}}_Z(v)$ , we have $\kappa (Z/\mathfrak {p}) \leq \alpha $ . As $\rho (M)$ is a finitely generated Z-module, this is equivalent to $\kappa (\rho (M)) \leq \alpha $ by Lemma 3.7.6.

Proof. By assumption, the k-algebra $Z \otimes _k Z$ is also finitely generated, so it is noetherian by Hilbert’s basis theorem. By Theorem 3.7.1 and Remark 3.7.5, it is enough to show that

$$\begin{align*}{\mathrm{pr}}_2( {\mathrm{pr}}_1^{-1}(W_\alpha)) \subseteq W_\alpha.\end{align*}$$

Let $\mathcal {S}_\alpha := \Xi \backslash W_\alpha $ be the complement of $W_\alpha $ in $\Xi $ ; then it is enough to show that

$$\begin{align*}{\mathrm{pr}}_1({\mathrm{pr}}_2^{-1}(\mathcal{S}_\alpha)) \subseteq \mathcal{S}_\alpha.\end{align*}$$

Let $P \in {\mathrm {pr}}_2^{-1}(\mathcal {S}_\alpha )$ , and set $\mathfrak {p}_i := {\mathrm {pr}}_i(P)$ for $i=1,2$ . Then $\mathfrak {p}_2 \in \mathcal {S}_\alpha $ , so $\kappa (Z/\mathfrak {p}_2) \leq \alpha $ . The morphisms ${\mathrm {pr}}_1$ and ${\mathrm {pr}}_2$ induce the following diagram of residue fields:

$$\begin{align*}k(\mathfrak{p}_1) \hookrightarrow k(P) \hookleftarrow k(\mathfrak{p}_2).\end{align*}$$

Since ${\mathrm {pr}}_2 : \mathcal {R} \to \Xi $ is quasi-finite, $k(P)$ is a finite-dimensional $k(\mathfrak {p}_2)$ -vector space. Hence,

$$\begin{align*}{\mathrm{tr}}.\deg_k(k(\mathfrak{p}_1)) \leq {\mathrm{tr}}.\deg_k(k(P)) = {\mathrm{tr}}.\deg_k(k(\mathfrak{p}_2)).\end{align*}$$

Since Z is a finitely generated k-algebra, we can apply a version of the Noether normalisation theorem – see [Reference MatsumuraMat] Theorem 5.6 – to deduce that

$$\begin{align*}\kappa(Z / \mathfrak{p}_i) = {\mathrm{tr}}.\deg_k(Z/\mathfrak{p}_i) = {\mathrm{tr}}.\deg_k(k(\mathfrak{p}_i)) \quad\text{for}\quad i=1,2.\end{align*}$$

Hence, $\kappa (Z/\mathfrak {p}_1) \leq \kappa (Z/\mathfrak {p}_2) \leq \alpha $ , which means that $\mathfrak {p}_1 = {\mathrm {pr}}_1(P) \in \mathcal {S}_\alpha $ .