# Tag Info

15

$\def\id{\operatorname{id}}$Suppose $M\otimes N$ is isomorphic to $R^n$. Pick a basis $\{x_1,\dots,x_n\}$ of $M\otimes N$, with $x_i=\sum_{j=1}^{r_i}m_{i,j}\otimes n_{i,j}$ for each $i\in\{1,\dots,n\}$. Let $r=r_1+\cdots+r_n$, let $\{e_{i,j}:1\leq i\leq n, 1\leq j\leq r_i\}$ be a basis of $R^r$, and consider the map $f:R^r\to M$ which maps $e_{i,j}$ to ...

11

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 ...

9

Let $M, N$ be $R-$modules. Then the following holds. If $M$ and $N$ is flat, then so is $M\otimes_{R}N$: see related question here. If $M$ and $N$ are projective, then so is $M\otimes_{R} N$. Indeed, writing $M\oplus M'=F,\ N\oplus N'=F'$, for free $R-$modules $F,\ F'$, one has that $$F'':=F\otimes_{R}F'$$ is free (tensor product of free modules) and ...

9

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 ...

7

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 ...

7

A projective module over a domain has no nonzero torsion element, since it is a submodule of a free module. But every element of your module is a torsion element: it is killed by $x$.

7

Since $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}$ is a field, every module over it is projective.

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

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, ...

6

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 ...

6

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.

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

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 $M ... 6 Let$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$. This is absurd since commutative rings have IBN. This ... 5 Thinking of a f.g. projective module as a vector bundle, it seems very likely the answer is no: consider the trivial bundle of rank 2 and two "twisted" subbundles of rank 1 whose intersection is 0-dimensional everywhere except over a closed subset with non-empty interior – so not a vector bundle, in particular. Let's see where this thinking leads. Let$R$... 5 For the case$R$is a local ring it's a corollary of Nakayama's lemma. As the notation in the above link, suppose$M$is a finite generated projective module over$R$, then, first pick a minimal number of generators, i.e.,$M=Rm_1+\cdots +Rm_k$, and$k$is the minimal number with this property, so we get a decomposition $$R^k=M\oplus N,$$ then, we are ... 5 This has been answered on MathOverflow here. 5 A prominent counter-example is the following: Take$R := {\mathbb R}[x,y,z]/(x^2+y^2+z^2-1)$, the ring of real-valued polynomial functions on the$2$-sphere, and consider the following short exact sequence:$$(\ast)\quad\quad 0\to P\to R\frac{\partial}{\partial x}\oplus R\frac{\partial}{\partial y}\oplus R\frac{\partial}{\partial z}\xrightarrow{\alpha := ... 5 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 ... 5 I don't know of any other meaning of a projective ideal other than the one suggested by Boris Novikov, i.e. an ideal of a ring$R$that is also projective as an$R$-module. I want to emphasize that such an ideal$I$need NOT be a direct summand of$R$(Boris never implied that condition to be necessary - only sufficient!) as well as give more examples. The ... 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

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

Take any product $R_1 \times R_2$, a projective $R_1$-module $M_1$ and a non-projective $R_2$-module $M_2$. Then $M_1 \times M_2$ is a counterexample.

4

Hint: $P\oplus P'=R^{(\Lambda)}$ (a free module) for some set $\Lambda$ and some module $P'$.

4

Well, take $P=0$, and you have a counterexample

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 ...

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