The Stacks project

Given that 𝐾 and 𝑀 are free 𝑅-modules in the short exact sequence 0→𝐾→𝐿→𝑀→ 0, is 𝐿 L necessarily free? Could the structure of 𝑅 influence this conclusion?

Regard Informatika

The family which didn't compile was consisted of intermediate fields generated by $\beta_i$s where $i\in I\setminus I_0$ for various finite $I_0\subseteq I$.

I am failing to see why is the condition (3) of lemma 07P4 satisfied for such intermediate fields $k^p\subseteq k'\subseteq k$. I found that in characteristic zero intersection of two subfields of finite index isn't always of finite index, but I guess in positive characteristic we are suppose to use $p$-basis of $k/k^p$ somehow. Alternatively, I think one can fix a $p$-basis $\{\beta_i\}$ and use the family of intermediate fields $\{k[\{\beta_j:j\in I\setminus J\}]:J\sseq I\text{ finite} \}$, instead of your family $\{k'\}$.

It seems effaceable / erasble delta-functors are not defined in Stack Project, but used in https://stacks.math.columbia.edu/tag/03OS.

End of the first paragraph: ''Thus it is injective as $M'_f \subset M'_K$ as it is free over the domain $R_f$" could probably be replaced by the clearer sentence "Thus it is injective as $M'_f \subset M'_K$ because this inclusion is obtained by tensoring $R \hookrightarrow K$ with the free $R_f$, and hence, flat $R$-module $M'_f$."

Lucas, in the Stacks Project an irreducible topological space is nonempty by definition (see https://stacks.math.columbia.edu/tag/004U#:~:text=Let%20X%20be%20a%20topological,maximal%20irreducible%20subset%20of%20X.). One reason this convention makes sense is that an irreducible topological space can be defined as a topological space which cannot be expressed as a finite union of proper closed subspaces. But the empty space equals the empty union, hence it is not irreducible. A similar line of reasoning explains why the empty topological space is not connected.

Typo: "an projective" is written instead of "a projective" four times in this definition.

It seems that the proof works for not only "first order" but also "finite order" thickenings.

In the last step of the proof of Lemma 57.3.3, how exactly is TR3 used to obtain an element of $\mathrm{Hom}(C, S(A))$ mapping to both $\alpha$ and $\beta$?

Is this proof using that $R$ is noetherian anywhere? I understand in applications it is likely that $R$ is also noetherian.

Here are the omitted details that show that $C^\bullet\to M_{n+1}^\bullet$ is a quasi-isomorphism. (The following is an adaptation of this proof.)

Lemma. Let $\mathcal{A}$ be an abelian category. Let $$\require{AMScd} \begin{CD} A@>f>>B\\ @VgVV@VVhV\\ C@>{\smash{k}}>>D \end{CD}$$ be a cartesian square in $\operatorname{Ch}(\mathcal{A})$. If $k$ is onto and $h$ is a quasi-isomorphism, then $g$ is a quasi-isomorphism.

Proof. Consider the sequence $$0\to A\xrightarrow{(f,g)} B\oplus C\xrightarrow{(-h,k)}D\to 0$$ It is left exact by Homology, Lemma 12.5.11, and it is exact at $D$ for $h$ is onto. It gives rise to a distinguished triangle $$A\xrightarrow{(f,g)} B\oplus C\xrightarrow{(-h,k)}D\to A[1]$$ in $D(A)$. Since $h$ is an iso in $D(A)$, the lemma follows from the following result. $\square$

Lemma. Let $\mathcal{D}$ be a triangulated category. Let $$A\xrightarrow{(f,g)} B\oplus C\xrightarrow{(h,k)}D\to A[1]$$ be a distinguished triangle in $\mathcal{D}$. If $k$ is an isomorphism, then $f$ is an isomorphism.

Proof. Without loss of generality we may assume $k=1_C$, since we have an isomorphism of distinguished triangles

\xymatrix{ A \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r]^{{(f,g)}} & B\oplus C \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r]^{{(h,k)}} & D \ar@{->}[d]^{k^{-1}} \ar@{->}[r] & A[1] \ar@{->}[d]^{\mathrm{id}} \ A \ar@{->}[r]^{{(f,g)}} & B\oplus C \ar@{->}[r]^{{(k^{-1}h,\mathrm{id})}} & C \ar@{->}[r] & A[1] }

But we also have an isomorphism of distinguished triangles

\xymatrix{ A \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r]^{{(f,g)}} & B\oplus C \ar@{->}[d]^{ $$\begin{pmatrix} 1_B&0\\ h&1_C\end{pmatrix}$$ } \ar@{->}[r]^{{(h,\mathrm{id})}} & C \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r] & A[1] \ar@{->}[d]^{\mathrm{id}} \ A \ar@{->}[r]{{(f,0)}} & B\oplus C \ar@{->}[r]{{(0,\mathrm{id})}} & C \ar@{->}[r] & A[1] }

The matrix in the middle is invertible for its inverse is $<script type="math/tex; mode=display">\begin{pmatrix} 1_B&0\\ -h&1_C\end{pmatrix}$ . The right square clearly commutes and the left square commutes as $0=hf+g$, for the top sequence is a complex. Let $T\in\mathcal{D}$. Let $y_T=\operatorname{Hom}_\mathcal{D}(T,-)$. After inspecting the long exact sequence obtained by mapping through $y_T$ the bottom distinguished triangle from the last diagram, we deduce that $y_T(f)$ is an isomorphism. By the Yoneda lemma, $f$ is an isomorphism. $\square$

The converse holds: Let $M^\bullet$ be an acyclic complex. Then the quasi-isomorphism $M^\bullet\to 0$ is mapped through the functor $\operatorname{Tot}(-\otimes_RK^\bullet)$ to a quasi-isomorphism $\operatorname{Tot}(M^\bullet\otimes_RK^\bullet)\to 0$.

Continuation to #9936: The functor $D(\mathcal{A}^{\overline{\omega}})\to D(\mathcal{A})^{\overline{\omega}}$ is actually full [ref]. Also $D(\mathcal{A}^I)\to D(\mathcal{A})^I$ is conservative for any index category $I$ (use Lemma 3 from here).

Ignore #9982, it's wrong.

Two sentences before Lemma 08S6, *solutuion.

One could add the following result to this section.

Let $(A_n)$ be an inverse system of rings and let $\mathbf{M}\subset\mathbf{N}$ be a subset. Define $\textit{Mod}(\mathbf{M}, (A_n))$ to be the category of inverse systems $(M_n)_{n\in\mathbf{M}}$ of abelian groups such that each $M_n$ is given the structure of a $A_n$-module and the transition maps $M_{n'} \to M_{n}$ is $A_{n'}$-linear for each $n',n\in\mathbf{M}$ with $n'\geq n$. This is an abelian category. We have a restriction functor $\textit{Mod}(\mathbf{N}, (A_n))\to\textit{Mod}(\mathbf{M}, (A_n))$, which is exact.

Lemma. Let $\mathbf{M}\subset\mathbf{N}$ be an infinite subset. The commutative diagram of functors

\xymatrix{ Mod(\mathbf{N},(A_n)) \ar@{->}[rd]^{\lim} \ar@{->}[dd] & \ & Mod(A) \ Mod(\mathbf{M},(A_n)) \ar@{->}[ru]_{\lim} & }

promotes to a commutative diagram of right derived functors

\xymatrix{ D(Mod(\mathbf{N},(A_n))) \ar@{->}[rd]^{R\lim} \ar@{->}[dd] & \ & D(Mod(A)) \ D(Mod(\mathbf{M},(A_n))) \ar@{->}[ru]_{R\lim} & }

Proof. Very similar to #9983. $\square$

I propose to add the following result to this section.

For each set $\mathbf{M}\subset\mathbf{N}$, denote $Ab(\mathbf{M})$ to the category $\operatorname{Fun}(\mathbf{M}^{\mathrm{op}},Ab)$, where $\mathbf{M}$ is a poset and hence it may be viewed as a category. We have a restriction functor $Ab(\mathbf{N})\to Ab(\mathbf{M})$, which is exact.

Lemma. Let $\mathbf{M}\subset\mathbf{N}$ be an infinite subset. The commutative diagram of functors

\xymatrix{ Ab(\mathbf{N}) \ar@{->}[rd]^{\lim} \ar@{->}[dd] & \ & Ab \ Ab(\mathbf{M}) \ar@{->}[ru]_{\lim} & }

promotes to a commutative diagram of right derived functors

\xymatrix{ D(Ab(\mathbf{N})) \ar@{->}[rd]^{R\lim} \ar@{->}[dd] & \ & D(Ab) \ D(Ab(\mathbf{M})) \ar@{->}[ru]_{R\lim} & }

Proof. We will apply #9981. Let $(K^\bullet_e)_{e\in\mathbf{N}}\in D(Ab(\mathbf{N}))$ be a complex of inverse systems. By Lemma 15.86.1, there is a quasi-isomorphism $(K^\bullet_e)_{e\in\mathbf{N}}\to (A^\bullet_e)_{e\in\mathbf{N}}$, where $(A^n_e)_{e\in\mathbf{N}}$ is right $\lim$-acyclic for each $n$. We must check that $(A^\bullet_e)_{e\in\mathbf{M}}$ is right $\lim$-acyclic. It turn, by Lemma 15.86.1, it suffices to show that $(A^n_e)_{e\in\mathbf{M}}$ is right $\lim$-acyclic for each $n$.

More generally, we will show: if $(A_e)_{e\in\mathbf{N}}\in Ab(\mathbf{N})$ is right $\displaystyle\lim_{e\in\mathbf{N}}$-acyclic, then $(A_e)_{e\in\mathbf{M}}\in Ab(\mathbf{M})$ is right $\displaystyle\lim_{e\in\mathbf{M}}$-acyclic. By the proof of Lemma 15.86.1, every inverse system of abelian groups has a ML resolution (actually, a resolution by surjective inverse systems). Pick a ML resolution $(A_e)_{e\in\mathbf{N}}\to(B_e^\bullet)_{e\in\mathbf{N}}$. Then, inside $D(\mathcal{A})$, $$\lim_{e\in\mathbf{N}}B_e^\bullet =R\lim_{e\in\mathbf{N}}B_e^\bullet =R\lim_{e\in\mathbf{N}}A_e =\lim_{e\in\mathbf{N}}A_e$$ is concentrated in degree $0$.

Since $(A_e)_{e\in\mathbf{M}}\to(B_e^\bullet)_{e\in\mathbf{M}}$ is also a ML resolution (for $Ab(\mathbf{N})\to Ab(\mathbf{M})$ is exact and preserves ML systems), we get $$\begin{equation} R\lim_{e\in\mathbf{M}}A_e =R\lim_{e\in\mathbf{M}}B_e^\bullet =\lim_{e\in\mathbf{M}}B_e^\bullet =\lim_{e\in\mathbf{N}}B_e^\bullet =\lim_{e\in\mathbf{N}}A_e \end{equation}$$ is concentrated in degree $0$. Hence, by Derived Categories, Lemma 13.16.4, $(A_e)_{e\in\mathbf{M}}$ is right $\displaystyle\lim_{e\in\mathbf{M}}$-acyclic. $\square$

(The hypothesis in #9981 of $RG:D(\mathcal{B})\to D(\mathcal{C})$ being everywhere defined may be dropped since it is already implied by the rest.)

(I missed the title in the reference in #9980, it is J. Lipman, Notes on Derived Functors and Grothendieck Duality.)

This might be added as well. It is Görtz, Wedhorn, Algebraic Geometry II, Remark F.177.

Lemma. Let $\mathcal{A}\xrightarrow{F}\mathcal{B}\xrightarrow{G}\mathcal{C}$ be additive functors between abelian categories. Assume that $F$ is exact, $RG:D(\mathcal{B})\to D(\mathcal{C})$ is everywhere defined, and that every complex $X\in D(\mathcal{A})$ admits a quasi-isomorphism into a complex $A_X$ such that $F(A_X)$ computes $RG$. Then $A_X$ computes $R(G\circ F)$ for every $X\in D(\mathcal{A})$ (hence $R(G\circ F):D(\mathcal{A})\to D(\mathcal{C})$ is everywhere defined) and the morphism $t:R(G\circ F)\to RG\circ RF=RG\circ F$ coming from Lemma 13.14.16 is an isomorphism.

Proof. Just apply #9980. Every complex of $D(\mathcal{A})$ computes $RF$ for $F$ is exact.

I propose to add the following result to this section. It is the translation of J. Lipman, Corollary 2.2.7, to the Stacks Project definitions and results.

Lemma. Let $\mathcal{A}\xrightarrow{F}\mathcal{B}\xrightarrow{G}\mathcal{C}$ be additive functors between abelian categories. Assume that $RG:D(\mathcal{B})\to D(\mathcal{C})$ is everywhere defined and that every complex $X\in D(\mathcal{A})$ admits a quasi-isomorphism into a complex $A_X$ computing $RF$. In particular, $RF:D(\mathcal{A})\to D(\mathcal{B})$ is everywhere defined. The following are equivalent: 1. $F(A_X)$ computes $RG$ for every $X\in D(\mathcal{A})$. 2. $A_X$ computes $R(G\circ F)$ for every $X\in D(\mathcal{A})$ (hence $R(G\circ F):D(\mathcal{A})\to D(\mathcal{C})$ is everywhere defined) and the morphism $t:R(G\circ F)\to RG\circ RF$ coming from Lemma 13.14.16 is an isomorphism.

Proof. Assume 2. Then $R(G\circ F)\to RG\circ RF$ evaluated at $A_X$ becomes $R(G\circ F)(A_X)=G(F(A_X))=G(RF(A_X))\to RG(RF(A_X))$. Thus $RF(A_X)=F(A_X)$ computes $RG$. Conversely, suppose $F(A_X)$ computes $RG$. To show that $A_X$ computes $G\circ F$, we check the hypotheses of Lemma 13.14.15 for this composite and the class $\mathcal{I}=\{A_X\mid X\in D(\mathcal{A})\}$. The first hypothesis is clear. To see the second one, given a quasi-isomorphism $A_X\to A_Y$, $X,Y\in D(\mathcal{A})$, we have that $F(A_X)\to F(A_Y)$ is a quasi-isomorphism since both $A_X$ and $A_Y$ compute $RF$ (Lemma 13.14.4). Hence $GF(A_X)\to GF(A_Y)$ is a quasi-isomorphism since $F(A_X)$ and $F(A_Y)$ compute $RG$ (Lemma 13.14.4 again). Lastly, $t:R(G\circ F)\to RG\circ RF$ is an isomorphism evaluated at each $A_X$. Hence it is globally an isomorphism since each $X\in D(\mathcal{A})$ is isomorphic to $A_X$ in $D(\mathcal{A})$.