# Tag Info

8

As indicated in the comments, you should saturate $Q$ in $P$ first, i.e. replace it by the preimage of the torsion in $P/Q,$ so that (after changing $Q$ in this way) we get that $P/Q$ is torsion-free. The basic fact you need is that the saturation is again f.g., but this will follows from the fact that the torsion in $Q/P$ is f.g., being a submodule of the ...

6

Here is the answer for finitely generated modules of rank one. Recall that the isomorphism classes of these modules form a group, the Picard group $Pic(R)$, with tensor product as multiplication. Theorem (Traverso, Swan) For a commutative ring $R$ the following are equivalent: a) The reduced ring $R_{red}=R/Nil(R)$ is semi-normal b) The ...

6

I'm denoting your ring with $R$ and your ideal with $I$. We'll just need that $R$ is an integral domain, and that $I$ is a nontrivial ideal. If $R/I$ is a projective $R$ module, then the following exact sequence splits $$0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow 0$$ But integral domains have no proper direct summands, so $I$ would have to be ...

5

I don't know how to reduce to the case you said. However, we can use CRT. We have $M/\mathfrak{m}_iM\cong M_{\mathfrak{m}_i}/\mathfrak{m}_iM_{\mathfrak{m}_i}$ and $R/\mathfrak{m}_1\cdots\mathfrak{m}_n\cong R/\mathfrak{m}_1\times\cdots\times R/\mathfrak{m}_n$. Tensoring this with $M$ we get $M/\prod_i\mathfrak{m}_iM\cong M/\mathfrak{m}_1M\times\cdots\times ... 5 The statement "every element in P can be written as a finite linear combination of some elements of P.", where "some" means a finite set, just says that the module is finitely generated. This has nothing to do with being projective. Take for instance the$\mathbb Z$-module$\mathbb Z/2$. Here every element can be written as a multiple of$[1]$. So$\mathbb ...

5

Thinking of a f.g. projective module as a vector bundle, it seems very likely the answer is no: consider the trivial bundle of rank 2 and two "twisted" subbundles of rank 1 whose intersection is 0-dimensional everywhere except over a closed subset with non-empty interior – so not a vector bundle, in particular. Let's see where this thinking leads. Let $R$ ...

4

Put the fi together to form one giant f from P to R(I), the direct sum of I copies of the ring R. The condition that x = Sum f_i(x) x_i just means that there is some g:R(I)→P such that g(f(x)) = x, namely g((r1,r2,...)) = r1*x1 + r2*x2 + .... In other words, P is a direct summand of the free module R(I).

4

Over a von Neumann regular ring, every right module (and every left module) is flat. Let $V$ be an countable dimensional $F$ vector space, and let $R$ be the ring of endomorphisms of that vector space. It's known that $R$ is a von Neumann regular ring with exactly three ideals. The nontrivial ideal $I$ consists of the endomorphisms with finite dimensional ...

4

A few remarks: 1) You don't need to say "fractional" in your first sentence. By definition a fractional ideal is an $R$-submodule of $K$ of the form $\frac{1}{a} I$ for an integral ideal $I$ and $a \in R^{\bullet}$. As an $R$-module, this is visibly isomorphic to the integral ideal $I$. 2) Yes, I think you're right that this is a genuine hole in the ...

4

No, it isn't. If you take $R=K[x]/(x^2)$, then $R$ is projective over itself as always, but the kernel of the map $R\to R$ given by multiplication by $x$ is $Rx$ which isn't projective. To see this, note that $R$ is local, so the only projective modules are direct sums of copies of $R$, so the dimension (over $K$) of any projective module is divisible by ...

4

Let $R$ be any commutative ring whose projective modules are all free, and let $e\notin \{0,1\}$ be an idempotent of $R$. Then $eR$ and $(1-e)R$ are both projective, hence free of some rank 1 or more, and $eR\oplus(1-e)R=R$, so that we have $R^n\cong R$ as $R$ module for some natural number $n\geq 2$. This is absurd since commutative rings have IBN. This ...

3

I assume that the module structure is induced by the ring map $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \to \mathbb{Q}$, $a \otimes b \mapsto ab$. But this is an isomorphism, so that the module is free of rank $1$. More generally, if $A \to B$ is an epimorphism in the category of commutative rings, then $B \otimes_A B \to B$ is an isomorphism, so that $B$ ...

3

Suppose $R$ is a $k$-algebra, with $k$ a commutative ring. If $M$ is a $k$-module, we can construct the $k$-module $R\otimes_kM\otimes_kR$, which is automatically an $R$-bimodule or, equivalently, an $R^e$-module. If $N$ is an $R$-bimodule, there is a canonical isomorphism $$\hom_{R^e}(R\otimes_kM\otimes_kR, N)\cong\hom_k(M,\bar N)$$ with $\bar N$ the ...

3

Every free module over $R$, that is to say $R,R^n,$ or $R^{\oplus\kappa}$ for any cardinal $\kappa$, is projective. We can't give any other examples in general. Since projective modules are submodules of free modules, in a principal ideal domain every projective is free, since submodules of free modules are direct sums of ideals, and principal ideals in an ...

3

As far as I know, field should not matter, unless your algebra has "weird relations", and in most case, you are pretty safe as long as characteristic of $k$ is not 2. I hope somebody can give some supplement answer here... Back to your specific example, (say if we work over characteristic 0..) the (minimal) projective resolution of $S(1)$ is actually: $$... 3 I believe the (injective) map \bar{a_0} \mapsto a_0 does not work, since it is not a map of \mathbb Q[x,y]-modules. In particular, we have that x\bar{a_0}=0 but xa_0\neq0 for every a_0\in\mathbb Q[x,y]^{\oplus 1}, and a \mathbb Q[x,y]-linear map would require that x\bar{a_0}\mapsto xa_0. In any case, you should try the third method, which ... 3 As Steve D said, you can use the fact that projective modules are always flat. Consider the map \mathbb{Q}[x,y]\to \mathbb{Q}[x,y] defined by multiplying x. This is an injective \mathbb{Q}[x,y]-module map, while tensoring \mathbb{Q} will give an injective map, but it is NOT. So \mathbb{Q} is not flat as \mathbb{Q}[x,y]-module. However, ... 3 The phrasing of your statement a module P is projective iff every element in P can be written as a finite linear combination of some elements of P is pretty bad, because it is not at all evident from that those some elements are fixed: what you mean, but not what you wrote, probably is a module is projective iff there is a set X\subseteq P such ... 3 The statement you're linking to is: A module P is projective if and only if there is a family \{x_{i}\}_{i \in I} \subset P and morphisms f_{i}: P \to R such that for each x \in P we have x = \sum_{i \in I} f_{i}(x) x_{i}. The last statement says three things: In order for the sum to make sense we must have that for all x the set ... 3 Quote from Lam's Serre's problem on projective modules: In 1958, Seshadri showed that Serre's conjecture is true for two variables (i.e. for A = k[x_1,x_2]). In fact, Seshadri proved that f.g. projectives over R[t] are free if R is any commutative PID. Reference: Seshadri, C.S., Triviality of vector bundles over the affine space K^2, Proc. Nat. ... 3 A finitely generated module over a PID (like F[X]) is projective if and only if its free. Certainly any free module is projective. On the other hand, if M is your finitely generated module and M is projective, then for every prime ideal \mathfrak{p} of F[X], M_\mathfrak{p} (the localization of M at \mathfrak{p}) is a finitely generated ... 2 I'm pretty sure that this is incorrect. Indeed, note that since since -6\equiv 2\text{ mod }4 we have that \mathcal{O}_{\mathbb{Q}(\sqrt{-6})}=\mathbb{Z}[\sqrt{-6}] and so \mathbb{Z}[\sqrt{-6}] is a Dedekind domain. It is then a common fact that an integral domain R is a Dedekind domain if and only if every ideal is projective. EDIT: Assuming YACP ... 2 Let \mathcal{R} = [\mathcal{I}, R\text{-}\mathbf{Mod}], let \mathcal{A} = [\mathcal{I}, \mathbf{Ab}], and let \mathcal{S} = [\mathcal{I}, \mathbf{Set}]. First, observe the following: there is an evident forgetful functor \mathrm{Hom}_R (R, -) : \mathcal{R} \to \mathcal{A}, and it has both a left adjoint R \otimes_\mathbb{Z} {-} : \mathcal{A} \to ... 2 If I understood your question correctly then the following provides a counterexample:$$1\to 2\to 3\to 4\to 2$$where the two 2's should be identified. Take A to be the path algebra of this quiver modulo \operatorname{rad}^2 A. If you take the projective dimension of S_1, then P_1 just appears in the beginning as P_0 and then P_2, P_3 and P_4 ... 2 If R is a left regular ring, then the canonical map K_0(R) \to K_0(R[t]) is an isomorphism. This result is due to Grothendieck at least when R is commutative. The general case can be found in the paper "The Whitehead group of a polynomial extension" (Bass, Heller, Swan) or in Rosenberg's book on Algebraic K-Theory. Of course, this does not imply that ... 2 Please see \S 3.5.4 -- "Projective verus free" -- in my commutative algebra notes. In particular, Proposition 27 and the exercise follow it step you through showing that the ideal \langle 3, 1+ \sqrt{-5} \rangle is projective but not free. The same techniques apply to I = \langle 2, 1+ \sqrt{-5} \rangle. (In fact, I view it as a happy accident ... 2 I think it could be easier to work directly with the basic definition of injectivity, i.e. that if A \hookrightarrow B is an embedding, then any A \to J extends to B. Now you are supposed to reduce to the case when A \to J is surjective, and B is projective. You can always just add a copy of J to A and B to get surjectivity. And you can ... 2 On the positive side, if B is finite projective and the sequence splits, then A and C are finite projective as well, because summands of finite projective modules are again finite projective. So, in particular, if B and C are finite projective, then A is too (projectivity of C ensures that the sequence splits). 2 The projectivity of I is not needed. One can suppose I\ne 0. First for all b/c\in I, we have b\in I and  \phi(b)=\phi(c(b/c))=c\phi(b/c), so$$\phi(b/c)=\phi(b)/c.$$Now fix a_0\in I\cap R non-zero. Let x=b/c\in I. Then$$ a_0\phi(x)=\phi(a_0x)=\phi(a_0b/c)=b\phi(a_0/c)=b\phi(a_0)/c. $$So$$\phi(x)=(\phi(a_0)a_0^{-1})x, \quad \forall ...

1

Sorry to show up late to this party, but you were quoting my comment and somehow I missed it. Yes, a condition was overlooked: we must presume that $P$ has constant rank, alternatively, $\operatorname{Spec}(A)$ is connected, or $A$ has no non-trivial idempotents, etc. This result is Serre's Splitting Theorem which states that a projective $A$-module, $P$, ...

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