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3

This is a very general question and can not be answered in a few words. So, may be let me describe one aspect. If $R$ is a commutative ring and $M$ an $R$-module, giving an $R$-algebra homomorphism $A\to \operatorname{End}_R M$, where $A$ is an $R$-algebra makes $M$ into an $A$-module. In the example you write above, $R=\mathbb{Z}$. Now, let me look at a ...


3

Ok Slup, here goes. Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $...


2

I will prove that given a family $\{M_j\}_{j \in J}$ of $R$-modules then $Hom(\oplus M_{j},N) \cong \prod Hom(M_{j},N)$ for any given $R$-module. Then your question follows by setting $M_{j}=R$ for all $\thinspace$ $j \in J$. Assume that $i_{j} : M_{j} \rightarrow \oplus M_{j}$ is the canonical inclusion. Now define the above homomorphism $\phi: Hom(\oplus ...


2

Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$. Example. (i) If $A$ is local Noetherian then $(S)$ holds. (ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $...


2

Finite dimensional algebras have integrals, and using one you can show selfinjectivity at once —in fact, they are Frobenius. See for example the Lectures on Hopf algebras by Schneider, which Google will find for you, or Susan Montgomery's book in Hopf algebras.


1

For a vector space, having finite dimension is the same as being finitely generated. One direction is obvious: if the vector space has a finite basis it clearly is finitely generated. Suppose $V$ is finitely generated. From any finite spanning set $S$, one can extract a minimal spanning set $B$, in the sense that $B\subseteq S$ is a spanning set and no ...


1

Yes, of course, but it's not that useful. Suppose you have two ring homomorphisms $f\colon A\to E(M)$ and $g\colon A\to E(N)$. Suppose you also have a group homomorphism $h\colon M\to N$ is a module. For each $a\in A$, you have $f(a)\colon M\to M$ and $g(a)\colon N\to N$; so you can form the square $$\require{AMScd} \begin{CD} M @>f(a)>> M \\ @VhVV ...


1

By the Serre-Swan theorem, at least if $M$ is closed, taking smooth sections defines an equivalence of categories between smooth vector bundles over $M$ and finitely generated projective modules over $C^{\infty}(M)$. This is more or less equivalent to the claim that every smooth vector bundle is a subbundle of a trivial vector bundle.


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It should be true for any PID. See the book by Lam, Serre's Conjecture


1

HINT: Do you see how to generate $1$ using $2$ and $3$? That is, do you see how to write $1$ as a $\mathbb{Z}$-linear combination of $2$ and $3$? If so, do you see why this means that $\{2, 3\}$ spans?


1

You distribute the $n$ and the $x$ by using your very own definition: $$(x+y)(n+1) = (x+y)n + (x+y)$$ for non-negative $n$, and in the second case $$x(n+m+1) = x(n+m) + x$$ for $n+m$ non-negative. Use induction. For the negative case, use induction there too, together with your definition of what $xn$ means for negative $n$.


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The annihilator of $0\in M$ is $R$, which is always going to be essential. The singular submodule always has at least that.


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You can use an injective resolution for $N$: let $E$ be injective and $0\to N\to E\to E/N\to 0$ be exact. Then the long exact sequence $$\DeclareMathOperator{\E}{Ext}\DeclareMathOperator{\H}{Hom} 0\to \H_R(\bigoplus_{i\in I}M_i,N)\to \H_R(\bigoplus_{i\in I}M_i,E)\to \H_R(\bigoplus_{i\in I}M_i,E/N)\to\\ \E_R^1(\bigoplus_{i\in I}M_i,N)\to \E_R^1(\bigoplus_{i\...


1

The proof is correct. If $(C_i)_{i\in I}$ is any family of cochain complexes, then by writing out the definitions you immediately see $H^n(\prod C_i)=\prod H^n(C_i)$. For if $D$ denotes the differential on $\prod C_i$, $d_i$ the differential on $C_i$, then $H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i).$



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