Tag Info

Hot answers tagged

5

This is true with added generality, that is in the non commutative world. So, assume $\phi\colon R\to S$ is a ring homomorphism between not necessarily commutative rings. If $P_{R}$ is a projective right $R-$module, then, if $P_{R}$ is a direct summand of the free right module $R^{(X)}$ for some set $X$, we get that $$ P_{R}\otimes_{R} S $$ is a direct ...


5

In a category with coproducts and coequalizers, any colimit can be built out of coproducts and coequalizers. This is a nice exercise. In an $\text{Ab}$-enriched category, coequalizers can be replaced with cokernels. This is another nice exercise.


4

If $M$ does not have finite projective dimension: let $R = k[[x,y]]/(xy)$, $M = R/(y) \cong k[[x]]$. Then $x$ is a zerodivisor in $R$, but acts as a nonzerodivisor on $M$. If $R$ is not local: let $R = k \times k$, $k$ a field, $M = k \times 0 \subseteq R$, $x = (1,0) \in R$.


4

It (usually) doesn't hold when $M$ is free of infinite rank. But when $M$ is f.g. projective, then $M$ is a direct summand of a f.g. free module, and you are done: Show that the class of $M$s which satisfy the claim is closed under finite direct sums, direct summands, and contains $\Delta$. It follows that the class contains all f.g. projective modules. You ...


3

Since $f$ is finite, $f_*$ is exact. (This is more or less exercise III.8.2 in Hartshorne -- namely $R^1f_* = 0$. More generally, this property is true for affine morphisms.) Taking (co)homology of a complex commutes with exact functors since cohomology is computed using kernels and cokernels (which exact functors preserve).


3

Things get clearer if one expands the sequences to the right: $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} & \ra{} & E_q(C) & \ra{} & ...


3

Tunococ's comment provides half the answer: Flatness of $S$ implies that $\{P_i\otimes_RS\}$ is a resolution of $M\otimes_RS$. The other half is to observe that this resolution consists of projective $S$-modules. Indeed, each $P_i$ is a projective $R$-module, hence a summand of a free $R$-module. That direct sum decomposition is preserved by tensoring with ...


3

This follows from the naturality of the connecting homomorphism of a short exact sequence of complexes (homological convention) : if $$ \begin{matrix} 0\longrightarrow &K_{\bullet}&\longrightarrow &L_{\bullet}&\longrightarrow &M_{\bullet}&\longrightarrow 0\\ ...


2

Dualizing $G \overset{\varphi}\to F \overset{\pi}\to M \to 0$ we get $0\to M^*\overset{\pi^*}\to F^* \overset{\varphi^*}\to G^* \to D(M) \to 0$. Now take $K\to H\overset{p}\to M^*\to 0$ a finite presentation of $M^*$. Thus we get a part of a free resolution of $D(M)$, that is, $$K\to H\to F^* \overset{\varphi^*}\to G^* \to D(M) \to 0,$$ and therefore we can ...


2

The injective envelope is $\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}$ We define $$[k]\mapsto \frac{k}{n} +\mathbb{Z}$$ we must show that this is an essential extension and that $\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}$ is injective (for groups =divisible) First a typical element of $\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}$ is $\frac{a}{n^l}+\mathbb{Z}$ with $n\not | a$. ...


2

Do you know Shapiro's Lemma? Your module is the induced module of the subgroup $S_{n-1}$ of $S_n$ acting on the trivial module, so by Shapiro's Lemma $H^n(S_n,{\mathbb F}_p^n) \cong H^n(S_{n-1},{\mathbb F}_p)$ for all $n \ge 1$. For $n=1$, $H^1(S_{n-1},{\mathbb F}_p) = {\rm Hom}(S_{n-1},{\mathbb F}_p)$ which, for $n \ge 3$, has order $2$ when $p=2$ and $1$ ...


2

You will find this kind of result in Blakers, A. "Some relations between homology and homotopy groups". Ann. of Math. (2) 49 (1948) 428--461. I am pretty sure it is in Massey's book on Singular Homology, from a cubical viewpoint. Proposition 14.7.1 of Nonabelian Algebraic Topology gives a deformation of the singular cubical complex of a space onto that ...


1

Pick an $A_m$-module $M$, view it as an $A$-module and find a projective resolution. Tensor it with $A_m$: the resulting complex is an $A_m$-projective resolution of $M$. It follows immediately that the global dimension of $A$ bounds that of its localizations.


1

A presheaf is a sheaf iff it is mono and conjunctive (maybe there is other terminology) For mono: Say that $f \in i^{-1}F(X)$ and if $r_x(f) \in i^{-1}F(\{x\})$ and $r_y(f) \in i^{-1}F(\{y\})$ are both zero then $f$ must be zero. this means that the map given by restrictions $$i^{-1}F(X) \rightarrow i^{-1}F(\{x\})\times i^{-1}F(\{y\})$$ is injective. For ...


1

If $C$ is any symmetric monoidal category, then the tensor product of commutative algebras in $C$ is their coproduct. The tensor product of algebras in $C$, not necessarily commutative, is the "commutative coproduct": We have to add to the universal property that the test morphisms commute with each other. Now apply this to $C=$ chain complexes.


1

Well, to show that $f_!$ takes acyclic complexes of soft sheaves to acyclic complexes, we need to split up an acyclic complex of soft sheaves $$0 \to \mathcal F_1 \xrightarrow{\phi_1} \mathcal F_2 \xrightarrow{\phi_2} \mathcal F_3 \xrightarrow{\phi_3} \mathcal F_4 \xrightarrow{\phi_4} \dots$$ into short exact sequences, namely $$0 \to \mathcal F_1 ...


1

First, a general element of $S\otimes G$ is of the form $\sum \alpha_i\otimes g_i$. But that is not the problem. The boundary of this is $\sum (\partial \alpha_i)\otimes g_i$ and you are assuming that its a cycle so it is zero in $S\otimes G$, however this does not imply that it is zero in $B\otimes G$. Here is an example of this phenomena, which arises ...


1

If $M$ is any $kG$-module, then $kG \bigotimes_{k} M$ is a free $kG$-module, and the kG-module homomorphism $$ m \mapsto \sum_{g \in G} g \otimes g^{-1} m$$ is injective. If $M$ is injective, then since free $kG$-modules are injective, $M$ is now a direct summand of a free $kG$-module and is therefore projective.


1

To try to give a complete answer I use and ideal $I \subseteq R$ and ideal $\mathfrak{m}$ in our case $I=xR$ and assume $I \subseteq \mathfrak{m}$, since $x \in \mathfrak{m}$ I believe. Then we claim $\operatorname{Tor}(R/I,R/\mathfrak{m})=I/I\mathfrak{m}$ consider the free resoultion $$0\rightarrow I\rightarrow R\rightarrow R/I \rightarrow 0$$ and this ...


1

You're right. I think what he means to ask for is a natural isomorphism of complexes $$\operatorname{Tot}^{\prod}\operatorname{Hom_{Ab}}(\operatorname{Tot}^{\oplus}(P\otimes Q),I) \cong \operatorname{Tot}^{\prod}\operatorname{Hom_R}(P,\operatorname{Tot}^{\prod}\operatorname{Hom_{Ab}(Q,I)}).$$


1

Use the short exact sequence $0\to R\to F\to F/R\to 0$ and consider the long exact sequence you get when applying $-\otimes_R B$. Use the fact that $F$ is flat to get an exact sequence $0\to \mathrm{Tor}_1(F/R,B)\to B\to F\otimes_R B$.



Only top voted, non community-wiki answers of a minimum length are eligible