# 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 $m_{i,... 13 This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If$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$for ... 12 A proof that every projective module over a PID is free occurs in$\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$be a connected (i.e., without nontrivial idempotents) Noetherian ring. Then every ... 11 I'm denoting your ring with$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 ... 10 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

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 $x$.

9

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 ...

9

If $\phi:\mathbb Z/n\mathbb Z\to \mathbb Z$ is a homomorphism, then $\phi(i)$ must have finite order for all $i\in\mathbb Z/n\mathbb Z$, since $i$ has finite order. But the only element of $\mathbb Z$ with finite order is $0$. So $\phi$ must be trivial, hence $\phi\circ \pi\ne \mathrm{id}$.

8

Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra. This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of ...

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

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, ...

7

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 ...

7

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

7

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 ...

7

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 ...

7

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 $M \... 7 A few remarks, to be expanded below: (1) first is that the proof that$M = \prod_{i=1}^\infty\mathbb{Z}$is not free is elementary, and (2) second is that it might be hard to find simpler examples, at least if "simple" refers to how simple the ring is itself. (1) In fact, here's a proof that I learned from Kaplansky's book "Infinite Abelian Groups": Assume$...

7

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

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 ...

6

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 $\mathbb ... 6 I'll consider the interval$[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 is some elaboration on the wiki entry in George's comment. Suppose$R$is a domain.$R$is called seminormal if whenever$b^2=c^3$in$R$one can find$t \in R$such that$b=t^3, c=t^2$. The relevant thing here is the following fact: R is seminormal if and only if$Pic(R) \cong Pic(R[X])$So if$R$is local and not seminormal then there will ... 6 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$\{i\,:\,f_{i}...

6

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 $R$ ...

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 ...

6

1) I think you made this much harder than it has to be. $\mathbb{Q}$ is a divisible $\mathbb{Z}$-module. This implies that if $f:\mathbb{Q}\to\mathbb{Z}$, then $f(\mathbb{Q})$ is a divisible $\mathbb{Z}$-submodule of $\mathbb{Z}$. How many of those do you know? 2) Hint: PIDs have the nice property that a submodule of a free module is free. With this, what ...

6

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: $$(\ast)\quad\quad 0\to P\to R\frac{\partial}{\partial x}\oplus R\frac{\partial}{\partial y}\oplus R\frac{\partial}{\partial z}\xrightarrow{\alpha := \... 6 Since any projective module is a submodule (even a summand) of a free module, the two formulations of your questions are equivalent. A ring R which has the property that any submodule of a projective R-module 'inherits' projectivity is called hereditary. This is equivalent to saying that the global dimension of R,$$\text{gldim}(R\text{-Mod}) := \...

6

Let $R$ be an integral domain that is not a field. Then $R$ contains a nonzero element that is not invertible. For every nonzero element $r$ that is not invertible, $r$ is not a multiple of $r^2$: if it were, then $r=r^2s$ for some $s\in R$. Since $R$ is an integral domain, $1=rs$, contradicting that $r$ is not invertible. Now, for $I$ any nonzero ...

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