The Stacks project

Suggested alternative proof: Use Lemmas 13.31.6 and 13.14.14.

One could add to the statement:

(4) Any quasi-isomorphism $I^\bullet\to N^\bullet$ is a homotopy equivalence.

Proof of the equivalence. (2)$\Rightarrow$(4). Precomposition by $I^\bullet\to N^\bullet$ induces a bijection $\operatorname{Hom}_{K(\mathcal{A})}(N^\bullet,I^\bullet)\to\operatorname{Hom}_{K(\mathcal{A})}(I^\bullet,I^\bullet)$. This implies that $I^\bullet\to N^\bullet$ has a unique left inverse in $K(\mathcal{A})$. In a preadditive category, this implies invertibility.

(4)$\Rightarrow$(3). A morphism $M^\bullet\to I^\bullet$ in $D(\mathcal{A})$ is of the form $s^{-1}f$, where $f:M^\bullet\to N^\bullet$ and $s:I^\bullet\to N^\bullet$ is a quasi-isomorphism. Hence $\operatorname{Hom}_{K(\mathcal{A})}(M^\bullet,I^\bullet)\to\operatorname{Hom}_{D(\mathcal{A})}(M^\bullet,I^\bullet)$ is onto. It is also injective: if $f:M^\bullet\to I^\bullet$ is a map in $K(\mathcal{A})$ that becomes zero in $D(\mathcal{A})$, then by Homology, Comment #9465 there is a quasi-isomorphism $s:I^\bullet\to N^\bullet$ with $s\circ f=0$ in $K(\mathcal{A})$. Thus $f=0$ in $K(\mathcal{A})$. $\square$

Suggested slogan: A bounded below complex of injectives is K-injective.

For future reference, and in case anyone wants to chime in. To this section, I would add the following result:

Lemma. Let $\mathcal{C}$ be a preadditive category. Let $S$ be a LMS in $\mathcal{C}$. Let $f:X\to Y$ be a morphism in $\mathcal{C}$. Then $Q(f)=0$ if and only if there exists $s:Y\to Z$ in $S$ with $s\circ f=0$.

Proof. This is a particular case of Categories, Lemma 4.27.6. $\square$

This was originally proposed in #8372.

For anyone that might care: from Lemma 13.31.2 we get that $$\label{1}\tag{1} \operatorname{Hom}_{K(\mathcal{A})}(N^\bullet,I^\bullet)\to\operatorname{Hom}_{K(\mathcal{A})}(I^\bullet,I^\bullet)$$ is bijective. In particular, $I^\bullet\to N^\bullet$ has a left inverse. This alone already implies that $I^\bullet\to I^\bullet$ is a final object of $I^\bullet/\operatorname{Qis}(\mathcal{A})$.

Even though one does not need it in the proof, actually $I^\bullet\to N^\bullet$ is invertible in $K(\mathcal{A})$. Bijectivity of \eqref{1} means that $I^\bullet\to N^\bullet$ has a unique left inverse, and this implies existence of an inverse in a preadditive category. The dual result is proven here.

Sorry, forget about

You need to assume $\mathcal{B}$ has countable direct sums and products.

Suggestion: it would be easier to rememeber the statement if $(2)$ was written with the word "unit" at the same place as in $(1)$ i.e. if it was written:

"(2) $x$ is prime if and only if the image of $x$ in $S^{−1}A$ is a prime element or a unit in $S^{−1}A$."

The final statement should probably be: "Moreover, then $A$ is a UFD if and only if every nonzero nonunit element of $A$ has a factorization into irreducibles and $S^{−1}A$ is a UFD."

In (c) in the proof, $A_f$ should be $A_{f_i}$ and $M_{i,f_i}$ is flat over $A_{f_i}$.

@#9456 Hints: use 43.23.3, use that $T \cap r^{-1}(r(Z)) = T \cap r^{-1}(r(T) \cap r(Z))$, and use that the maps on the $r_j \circ \ldots \circ r_{N - 1} \circ r_N$ for $j \geq n + 1$ restrict to finite maps on the $T_i$.

Lemma 0B2V: Irreducibility of polynomials corresponds to integralness of closed subschemes. Therefore, I think irreducibility in the proof concerning schemes should be replaced with integralness.

How does the induction for (4) in the proof of [0B0E] work? I couldn't figure it out. In the reference of [Roberts], it is done by an analysis of the dimension in the Grassmannian variety, instead of by induction.

Thanks!

1. In the proof of Lemma[0B0E], the sentence "For any irreducible component of Ti∩Z contained in Ei we have the desired dimension bound" should be changed to "For any irreducible component of Ti∩Zj contained in Ei we have the desired dimension bound.
2. On the second to last line of the proof of Lemma[0B1U], the word "transversally" should be changed to "properly".

If one wishes, it is faster to conclude part (4) from part (5) via lemma 29.37.7

Lemma 0B2V: I think here maybe its better to explain this element is geometrically irreducible, and reduce to an algebraically closed field so image could be considered at closed points, which are decent linear morphisms.

Lemma 0B2T: $p$ did not exist in the conclusion of the statement.

Typo in the proof: says "universall" instead of "universally"

I think it could be interesting to add the following result to this section:

Lemma. Let $M$ be a finite module over a Noetherian ring $R$. Then $$\sqrt{\operatorname{Ann}M}=\bigcap\operatorname{Supp}M=\bigcap\operatorname{Ass}M.$$

Proof. Since $M$ is finite, then $V(\operatorname{Ann}M)=\operatorname{Supp}M$, see Algebra, Lemma 10.40.5. Thus, by Algebra, Lemma 10.17.2, point 7, we have $\sqrt{\operatorname{Ann}M}=\bigcap V(\operatorname{Ann}M)=\bigcap\operatorname{Supp} M$. Suppose we have shown that for $S\in\{\operatorname{Ass}M,\operatorname{Supp}M\}$, every prime $\mathfrak{p}\in S$ contains a minimal prime of $S$. Then it will follow $\bigcap\operatorname{Supp} M=\bigcap\operatorname{Ass}M$ from Lemma 10.63.6. On the one hand, the assertion is true for $S=\operatorname{Ass}M$ by Lemma 10.63.5. On the other hand, each point in $S=\operatorname{Supp}M$ is contained in some irreducible component of $S$ by Topology, Lemma 5.9.2 ($S$ is Noetherian since $\operatorname{Spec}R$ is). Since $\operatorname{Supp}M=V(\operatorname{Ann}M)$ is sober (use Topology, Lemma 5.8.7 and that $\operatorname{Spec}R$ is sober), this means that every point in $\operatorname{Supp}M$ generalizes (in the sense of Topology, Definition 5.19.1) to a minimal prime of $\operatorname{Supp}M$, as we wanted to show. $\square$

In the second paragraph, first sentence, I think we should replace “which do not have finitely many irreducible components” by “which cannot be written as a finite union of irreducible closed subsets” (and in “by construction $Z'=\bigcup Z'_i$ and $Z''=\bigcup Z''_j$ are finite unions of their irreducible components,” change components by subsets). Otherwise, unless we appeal to Lemma 5.8.3, point 3 (which requires Zorn's lemma), we cannot write $Z'$ and $Z''$ as a union of irreducible closed subsets.

In the statement, point 2, one could add “and it is the union of these.”

At the end of second paragraph, the last sentence could be replaced by “since $A$ is empty, $X$ is a finite union of some irreducible subsets $X_i$; removing redundant $X_i$'s, we win by Lemma 5.8.4.”