# Tag Info

10

A proof that every projective module over a PID is free occurs in $\S$ 3.9 of my commutative algebra notes. As Qiaochu Yuan mentions, infinitely generated projective modules long to be free. A generalization of Kaplansky's result is a 1963 theorem of H. Bass: let $R$ be a connected (i.e., without nontrivial idempotents) Noetherian ring. Then every ...

8

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 ...

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'll consider the interval $[0,2\pi]$ for notational simplicity. Consider the matrix $$A = \left( \begin{array}{cc} \sin ^2\tfrac{\theta }{2} & - \sin \tfrac{\theta }{2} \cos \tfrac{\theta }{2} \\ -\sin \tfrac{\theta }{2}\cos \tfrac{\theta }{2} & \cos ^2\tfrac{\theta }{2} \end{array} \right),$$ which defines an $R$-linear ...

6

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 ...

5

This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for ...

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 $\mathbb ... 5 I don't know how to reduce to the case you said. However, we can use CRT. We have$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

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, ...

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 $R$ ...

5

Here are some examples: Localizations or quotients of $R$. More generally any epimorphism of commutative rings with domain $R$. Free $R$-modules of infinite rank. $R \oplus \bigoplus_{i \in I} N$ for any $R$-module $N$ with $N \otimes N = 0$. The finitely generated examples can be classified: Claim: When $M$ is a finitely generated $R$-module with $M ... 5 The truth is (to me) quite surprising: Kaplansky showed that an infinitely generated projective module over any Dedekind domain$D$is free! (The corresponding statement for finitely generated projective modules is equivalent to$D$having trivial class group.) This is referenced, for example, here. 4 As$P$is f. g. we have an exact sequence$0\rightarrow Q\rightarrow A^n\rightarrow P\rightarrow 0$,$Q$denoting the kernel of the map$A^n\rightarrow P$. As$P$is projective, this exact sequence splits,$A^n\cong Q\oplus P$. The exact sequence$A^n\cong Q\oplus P\rightarrow A^n\rightarrow P\rightarrow 0$shows$P$to be f. p. (where$Q\oplus P\rightarrow ...

4

You know the rank of a free module, right? It is the cardinality of a basis (well-defined since $R$ is commutative and $R \neq 0$). This is already well-known from linear algebra ($R$ is a field), where it is called the dimension (but as you see, this is really the same concept). Now it turns out that finitely generated projective $R$-modules are "not far" ...

4

Consider the surjection $R^n \to R$ described by the OP, namely $$(x_1,\ldots,x_n) \mapsto \sum_i r_i x_i.$$ This is a surjection $R^n \to R$. Since $R$ is a free as a module over iself, we may split this surjection. (Concretely, write $1 = \sum_i a_i r_i,$ and define a splitting via $r \mapsto (ra_1,\ldots,r a_n).$) Thus $R^n \cong M \oplus R,$ and so ...

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

Sure. Projective modules $P$ have the property (and actually this is an equivalent characterization) that every epimorphism $F \to P$ splits. Now choose a finite generating system of $P$, this lets you choose $F$ finitely generated free. Of course every direct summand of $F$ is a quotient of $F$ and therefore also finitely generated.

4

As Andrew pointed out, every finitely-generated projective module over a local ring is free (in fact, the hypothesis of being finitely-generated can be dropped - this is a theorem of Kaplansky). Hence, it remains to show that there exists a non-free module. But we have the following characterisation: A commutative ring $R$ is a field if and only if every ...

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

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 ...

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

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 ...

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 ...

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

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

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

I made a similar comment on MO where this question was first posted. Here is an elaboration: Since the circle $S^1$ can be thought of as the unit interval $[0,1]$ with the two endpoints identified, $R$ may be viewed as the ring of all real-valued continuous functions on $S^1$. My hint is to view $M$ as the module of global sections of the Möbius band. ...

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