# 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 ...

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 ...

10

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

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 ...

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

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 ...

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 ...

8

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

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

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

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

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}$is not free is elementary, and (2) second is that it might be hard to find simpler examples, at least if "simple" refers to how simple the ring is itself. (1) In fact, here's a proof that I learned from Kaplansky's book "Infinite Abelian Groups": Assume ... 7 Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra. This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of ... 6 Let$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$. This is absurd since commutative rings have IBN. This ... 6 Let$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 ... 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 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 ... 6 Define a category$\mathcal{J}$as follows: The objects are the elements of$s$. The morphisms$s \to t$are the elements$r \in R$such that$r s = t$. Composition and identities are inherited from$R$. Note that$\mathcal{J}$is a filtered category. Define a diagram$F : \mathcal{J} \to \mathbf{Mod} (R)$as follows: For all objects$s$in ... 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 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

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 ...

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.

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

Let $p$ be a prime number. Let $R = \mathbb{Z}/p^2\mathbb{Z}$. Let $M = R/pR$. Since the number of elements of $M$ is $p$, $M$ cannot be free. Hence $M$ cannot be projective.

5

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 ...

5

Put the $f_i$ 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 $g:R^{(I)}\to P$ such that $g(f(x)) = x$, namely $g((r_1,r_2,...)) = r_1x_1 + r_2x_2 +\cdots$. In other words, $P$ is a direct summand of the free module $R^{(I)}$.

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