# Tag Info

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

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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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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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