Hot answers tagged projective-module
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
How can I find an element $x\not\in\mathfrak mM_{\mathfrak m}$ for every maximal ideal $\mathfrak m$
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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