This section is a continuation of More on Algebra, Section 15.36. Let R be a topological ring and let M be a linearly topologized R-module. When we say “let M_\lambda be a fundamental system of open submodules” we will mean that each M_\lambda is an open submodule and that any neighbourhood of 0 contains one of the M_\lambda . In other words, this means that M_\lambda is a fundamental system of neighbourhoods of 0 in M consisting of submodules. Similarly, if R is a linearly topologized ring, then we say “let I_\lambda be a fundamental system of open ideals” to mean that I_\lambda is a fundamental system of neighbourhoods of 0 in R consisting of ideals.
Example 87.4.1. Let R be a linearly topologized ring and let M be a linearly topologized R-module. Let I_\lambda be a fundamental system of open ideals in R and let M_\mu be a fundamental system of open submodules of M. The continuity of + : M \times M \to M is automatic and the continuity of R \times M \to M signifies
\forall f, x, \mu \ \exists \lambda , \nu ,\ (f + I_\lambda )(x + M_\nu ) \subset fx + M_\mu
Since fM_\nu + I_\lambda M_\nu \subset M_\mu if M_\nu \subset M_\mu we see that the condition is equivalent to
\forall x, \mu \ \exists \lambda \ I_\lambda x \subset M_\mu
However, it need not be the case that given \mu there is a \lambda such that I_\lambda M \subset M_\mu . For example, consider R = k[[t]] with the t-adic topology and M = \bigoplus _{n \in \mathbf{N}} R with fundamental system of open submodules given by
M_ m = \bigoplus \nolimits _{n \in \mathbf{N}} t^{nm}R
Since every x \in M has finitely many nonzero coordinates we see that, given m and x there exists a k such that t^ k x \in M_ m. Thus M is a linearly topologized R-module, but it isn't true that given m there is a k such that t^ kM \subset M_ m. On the other hand, if R \to S is a continuous map of linearly topologized rings, then the corresponding statement does hold, i.e., for every open ideal J \subset S there exists an open ideal I \subset R such that IS \subset J (as the reader can easily deduce from continuity of the map R \to S).
Lemma 87.4.2. Let R be a topological ring. Let M be a linearly topologized R-module and let M_\lambda , \lambda \in \Lambda be a fundamental system of open submodules. Let N \subset M be a submodule. The closure of N is \bigcap _{\lambda \in \Lambda } (N + M_\lambda ).
Proof.
Since each N + M_\lambda is open, it is also closed. Hence the intersection is closed. If x \in M is not in the closure of N, then (x + M_\lambda ) \cap N = 0 for some \lambda . Hence x \not\in N + M_\lambda . This proves the lemma.
\square
Unless otherwise mentioned we endow submodules and quotient modules with the induced topology. Let M be a linearly topologized module over a topological ring R, and let 0 \to N \to M \to Q \to 0 be a short exact sequence of R-modules. If M_\lambda is a fundamental system of open submodules of M, then N \cap M_\lambda is a fundamental system of open submodules of N. If \pi : M \to Q is the quotient map, then \pi (M_\lambda ) is a fundamental system of open submodules of Q. In particular these induced topologies are linear topologies.
Lemma 87.4.3. Let R be a topological ring. Let M be a linearly topologized R-module. Let N \subset M be a submodule. Then
0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge is exact, and
N^\wedge is the closure of the image of N \to M^\wedge .
Proof.
Let M_\lambda , \lambda \in \Lambda be a fundamental system of open submodules. Then N \cap M_\lambda is a fundamental system of open submodules of N and M_\lambda + N/N is a fundamental system of open submodules of M/N. Thus we see that (1) follows from the exactness of the sequences
0 \to N/N \cap M_\lambda \to M/M_\lambda \to M/(M_\lambda + N) \to 0
and the fact that taking limits commutes with limits. The second statement follows from this and the fact that N \to N^\wedge has dense image and that the kernel of M^\wedge \to (M/N)^\wedge is closed.
\square
Lemma 87.4.4. Let R be a topological ring. Let M be a complete, linearly topologized R-module. Let N \subset M be a closed submodule. If M has a countable fundamental system of neighbourhoods of 0, then M/N is complete and the map M \to M/N is open.
Proof.
Let M_ n, n \in \mathbf{N} be a fundamental system of open submodules of M. We may assume M_{n + 1} \subset M_ n for all n. The system (M_ n + N)/N is a fundamental system in M/N. Hence we have to show that M/N = \mathop{\mathrm{lim}}\nolimits M/(M_ n + N). Consider the short exact sequences
0 \to N/N \cap M_ n \to M/M_ n \to M/(M_ n + N) \to 0
Since the transition maps of the system \{ N/N\cap M_ n\} are surjective we see that M = \mathop{\mathrm{lim}}\nolimits M/M_ n (by completeness of M) surjects onto \mathop{\mathrm{lim}}\nolimits M/(M_ n + N) by Algebra, Lemma 10.86.4. As N is closed we see that the kernel of M \to \mathop{\mathrm{lim}}\nolimits M/(M_ n + N) is N (see Lemma 87.4.2). Finally, M \to M/N is open by definition of the quotient topology.
\square
Lemma 87.4.5.reference Let R be a topological ring. Let M be a linearly topologized R-module. Let N \subset M be a submodule. Assume M has a countable fundamental system of neighbourhoods of 0. Then
0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge \to 0 is exact,
N^\wedge is the closure of the image of N \to M^\wedge ,
M^\wedge \to (M/N)^\wedge is open.
Proof.
We have 0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge is exact and statement (2) by Lemma 87.4.3. This produces a canonical map c : M^\wedge /N^\wedge \to (M/N)^\wedge . The module M^\wedge /N^\wedge is complete and M^\wedge \to M^\wedge /N^\wedge is open by Lemma 87.4.4. By the universal property of completion we obtain a canonical map b : (M/N)^\wedge \to M^\wedge /N^\wedge . Then b and c are mutually inverse as they are on a dense subset.
\square
Lemma 87.4.6. Let R be a topological ring. Let M be a topological R-module. Let I \subset R be a finitely generated ideal. Assume M has an open submodule whose topology is I-adic. Then M^\wedge has an open submodule whose topology is I-adic and we have M^\wedge /I^ n M^\wedge = M/I^ nM for all n \geq 1.
Proof.
Let M' \subset M be an open submodule whose topology is I-adic. Then \{ I^ nM'\} _{n \geq 1} is a fundamental system of open submodules of M. Thus M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^ nM' contains (M')^\wedge = \mathop{\mathrm{lim}}\nolimits M'/I^ nM' as an open submodule and the topology on (M')^\wedge is I-adic by Algebra, Lemma 10.96.3. Since I is finitely generated, I^ n is finitely generated, say by f_1, \ldots , f_ r. Observe that the surjection (f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M is continuous and open by our description of the topology on M above. By Lemma 87.4.5 applied to this surjection and to the short exact sequence 0 \to I^ nM \to M \to M/I^ nM \to 0 we conclude that
(f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \longrightarrow M^\wedge
surjects onto the kernel of the surjection M^\wedge \to M/I^ nM. Since f_1, \ldots , f_ r generate I^ n we conclude.
\square
Definition 87.4.7. Let R be a topological ring. Let M and N be linearly topologized R-modules. The tensor product of M and N is the (usual) tensor product M \otimes _ R N endowed with the linear topology defined by declaring
\mathop{\mathrm{Im}}(M_\mu \otimes _ R N + M \otimes _ R N_\nu \longrightarrow M \otimes _ R N)
to be a fundamental system of open submodules, where M_\mu \subset M and N_\nu \subset N run through fundamental systems of open submodules in M and N. The completed tensor product
M \widehat{\otimes }_ R N = \mathop{\mathrm{lim}}\nolimits M \otimes _ R N/(M_\mu \otimes _ R N + M \otimes _ R N_\nu ) = \mathop{\mathrm{lim}}\nolimits M/M_\mu \otimes _ R N/N_\nu
is the completion of the tensor product.
Observe that the topology on R is immaterial for the construction of the tensor product or the completed tensor product. If R \to A and R \to B are continuous maps of linearly topologized rings, then the construction above gives a tensor product A \otimes _ R B and a completed tensor product A \widehat{\otimes }_ R B.
We record here the notions introduced in Remark 87.2.3.
Definition 87.4.8. Let A be a linearly topologized ring.
An element f \in A is called topologically nilpotent if f^ n \to 0 as n \to \infty .
A weak ideal of definition for A is an open ideal I \subset A consisting entirely of topologically nilpotent elements.
We say A is weakly pre-admissible if A has a weak ideal of definition.
We say A is weakly admissible if A is weakly pre-admissible and complete1.
Given a weak ideal of definition I in a linearly topologized ring A and an open ideal J the intersection I \cap J is a weak ideal of definition. Hence if there is one weak ideal of definition, then there is a fundamental system of open ideals consisting of weak ideals of definition. In particular, given a weakly admissible topological ring A then A = \mathop{\mathrm{lim}}\nolimits A/I_\lambda where \{ I_\lambda \} is a fundamental system of weak ideals of definition.
Lemma 87.4.9. Let A be a weakly admissible topological ring. Let I \subset A be a weak ideal of definition. Then (A, I) is a henselian pair.
Proof.
Let A \to A' be an étale ring map and let \sigma : A' \to A/I be an A-algebra map. By More on Algebra, Lemma 15.11.6 it suffices to lift \sigma to an A-algebra map A' \to A. To do this, as A is complete, it suffices to find, for every open ideal J \subset I, a unique A-algebra map A' \to A/J lifting \sigma . Since I is a weak ideal of definition, the ideal I/J is locally nilpotent. We conclude by More on Algebra, Lemma 15.11.2.
\square
Lemma 87.4.10. Let B be a linearly topologized ring. The set of topologically nilpotent elements of B is a closed, radical ideal of B. Let \varphi : A \to B be a continuous map of linearly topologized rings.
If f \in A is topologically nilpotent, then \varphi (f) is topologically nilpotent.
If I \subset A consists of topologically nilpotent elements, then the closure of \varphi (I)B consists of topologically nilpotent elements.
Proof.
Let \mathfrak b \subset B be the set of topologically nilpotent elements. We omit the proof of the fact that \mathfrak b is a radical ideal (good exercise in the definitions). Let g be an element of the closure of \mathfrak b. Our goal is to show that g is topologically nilpotent. Let J \subset B be an open ideal. We have to show g^ e \in J for some e \geq 1. We have g \in \mathfrak b + J by Lemma 87.4.2. Hence g = f + h for some f \in \mathfrak b and h \in J. Pick m \geq 1 such that f^ m \in J. Then g^{m + 1} \in J as desired.
Let \varphi : A \to B be as in the statement of the lemma. Assertion (1) is clear and assertion (2) follows from this and the fact that \mathfrak b is a closed ideal.
\square
Lemma 87.4.11. Let A \to B be a continuous map of linearly topologized rings. Let I \subset A be an ideal. The closure of IB is the kernel of B \to B \widehat{\otimes }_ A A/I.
Proof.
Let J_\mu be a fundamental system of open ideals of B. The closure of IB is \bigcap (IB + J_\lambda ) by Lemma 87.4.2. Let I_\mu be a fundamental system of open ideals in A. Then
B \widehat{\otimes }_ A A/I = \mathop{\mathrm{lim}}\nolimits (B/J_\lambda \otimes _ A A/(I_\mu + I)) = \mathop{\mathrm{lim}}\nolimits B/(J_\lambda + I_\mu B + I B)
Since A \to B is continuous, for every \lambda there is a \mu such that I_\mu B \subset J_\lambda , see discussion in Example 87.4.1. Hence the limit can be written as \mathop{\mathrm{lim}}\nolimits B/(J_\lambda + IB) and the result is clear.
\square
Lemma 87.4.12. Let B \to A and B \to C be continuous homomorphisms of linearly topologized rings.
If A and C are weakly pre-admissible, then A \widehat{\otimes }_ B C is weakly admissible.
If A and C are pre-admissible, then A \widehat{\otimes }_ B C is admissible.
If A and C have a countable fundamental system of open ideals, then A \widehat{\otimes }_ B C has a countable fundamental system of open ideals.
If A and C are pre-adic and have finitely generated ideals of definition, then A \widehat{\otimes }_ B C is adic and has a finitely generated ideal of definition.
If A and C are pre-adic Noetherian rings and B/\mathfrak b \to A/\mathfrak a is of finite type where \mathfrak a \subset A and \mathfrak b \subset B are the ideals of topologically nilpotent elements, then A \widehat{\otimes }_ B C is adic Noetherian.
Proof.
Let I_\lambda \subset A, \lambda \in \Lambda and J_\mu \subset C, \mu \in M be fundamental systems of open ideals, then by definition
A \widehat{\otimes }_ B C = \mathop{\mathrm{lim}}\nolimits _{\lambda , \mu } A/I_\lambda \otimes _ B C/J_\mu
with the limit topology. Thus a fundamental system of open ideals is given by the kernels K_{\lambda , \mu } of the maps A \widehat{\otimes }_ B C \to A/I_\lambda \otimes _ B C/J_\mu . Note that K_{\lambda , \mu } is the closure of the ideal I_\lambda (A \widehat{\otimes }_ B C) + J_\mu (A \widehat{\otimes }_ B C). Finally, we have a ring homomorphism \tau : A \otimes _ B C \to A \widehat{\otimes }_ B C with dense image.
Proof of (1). If I_\lambda and J_\mu consist of topologically nilpotent elements, then so does K_{\lambda , \mu } by Lemma 87.4.10. Hence A \widehat{\otimes }_ B C is weakly admissible by definition.
Proof of (2). Assume for some \lambda _0 and \mu _0 the ideals I = I_{\lambda _0} \subset A and J_{\mu _0} \subset C are ideals of definition. Thus for every \lambda there exists an n such that I^ n \subset I_\lambda . For every \mu there exists an m such that J^ m \subset J_\mu . Then
\left(I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C)\right)^{n + m} \subset I_\lambda (A \widehat{\otimes }_ B C) + J_\mu (A \widehat{\otimes }_ B C)
It follows that the open ideal K = K_{\lambda _0, \mu _0} satisfies K^{n + m} \subset K_{\lambda , \mu }. Hence K is an ideal of definition of A \widehat{\otimes }_ B C and A \widehat{\otimes }_ B C is admissible by definition.
Proof of (3). If \Lambda and M are countable, so is \Lambda \times M.
Proof of (4). Assume \Lambda = \mathbf{N} and M = \mathbf{N} and we have finitely generated ideals I \subset A and J \subset C such that I_ n = I^ n and J_ n = J^ n. Then
I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C)
is a finitely generated ideal and it is easily seen that A \widehat{\otimes }_ B C is the completion of A \otimes _ B C with respect to this ideal. Hence (4) follows from Algebra, Lemma 10.96.3.
Proof of (5). Let \mathfrak c \subset C be the ideal of topologically nilpotent elements. Since A and C are adic Noetherian, we see that \mathfrak a and \mathfrak c are ideals of definition (details omitted). From part (4) we already know that A \widehat{\otimes }_ B C is adic and that \mathfrak a(A \widehat{\otimes }_ B C) + \mathfrak c(A \widehat{\otimes }_ B C) is a finitely generated ideal of definition. Since
A \widehat{\otimes }_ B C / \left(\mathfrak a(A \widehat{\otimes }_ B C) + \mathfrak c(A \widehat{\otimes }_ B C)\right) = A/\mathfrak a \otimes _{B/\mathfrak b} C/\mathfrak c
is Noetherian as a finite type algebra over the Noetherian ring C/\mathfrak c we conclude by Algebra, Lemma 10.97.5.
\square
Comments (2)
Comment #5983 by Dario Weißmann on
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