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4

Slup has given you the answer where it is mostly used, though it is not in general true that Trace maps $B$ to $A$, unless you assume something more (typically, one assumes that $A$ is integrally closed). One has standard counterexamples for general cases. For example take a ring $A$ which has a non-free projective module $P$ such that $A\oplus P$ is free. (...


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 $...


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 ...


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

Edit As @Batominovski and @TobiasKildetoft remarked, one should assume characteristic zero here. In positive characteristic $p$ and non-trivial $G$, the statement is indeed not true, at least if one takes the naive definition of characters: For example, the $p$-fold sum of the trivial representation satisfies the assumption but is not of the form $KG^{\oplus ...


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.


1

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 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 $H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i).$


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.


1

Let $M$ be a nonzero finitely generated submodule of $R^+$; you can assume a set of generators is $$ \left\{\frac{a_1}{d},\frac{a_2}{d},\dots,\frac{a_n}{d}\right\} $$ by using a common denominator. The $R$-homomorphism $M\to R$ defined by $x\mapsto dx$ is injective, so $M$ is isomorphic to a nonzero ideal of $R$. (The assumption $M\ne\{0\}$ is of course ...



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