For questions related to projective modules, their structures, and properties.
2
votes
1answer
44 views
Prove that if $P$ and $Q$ are projective and finitely generated $R$-modules then $\operatorname{Hom}_{R}(P,Q)$ is projective and finitely generated.
Suppose $R$ is a commutative ring and $P$ and $Q$ are projective and finitly generated $R$-modules. Prove that $\operatorname{Hom}_{R}(P,Q)$ is projective and finitely generated.
suppose I proved ...
3
votes
1answer
75 views
A problem about an $R$-module that is both injective and projective.
Let $R$ be a domain that is not a field, and let $M$ be an $R$-module that is both injective and projective. Prove that $M= \left \{ 0 \right \}$.
This is exercise 7.52 of Rotman's Advanced ...
1
vote
2answers
31 views
Direct sum of projective module
Let $\left\{ P_{i}\right\} _{i\in I}$ be a family of $R$-module. I know that if each $P_{i}$ is projective then $\oplus_{i\in I}P_{i}$ is projective. Is the converse true, i.e if $\oplus_{i\in ...
0
votes
1answer
38 views
If $A \cong A^*$, is every projective module also injective?
Suppose $A$ is a finite-dimensional algebra over $k$. Assume further that $A \cong A^* = \text{Hom}(A,k)$ as $A$-modules.
My question is: is every finite dimensional projective module over $A$ ...
2
votes
1answer
40 views
Why are projective modules cohomologically trivial?
Let $G$ be a finite group, $H\subset G$ a subgroup, $k$ a commutative ring, $M$ a $kG$-module, $n\in\mathbb{Z}$, and $\hat{H}\,^n(H,M)$ the $n$th Tate cohomology group as defined in this question, ...
0
votes
0answers
37 views
trace of projection independent of direct sum of modules
Let $P$ be a finitely generated projective $R$-module. Suppose $P\oplus Q\cong R^n$. Show that the trace of projection map of $R^n$ onto $P$ does not depend on $Q$.
3
votes
1answer
64 views
Module has finitely generated projective resolution
Let $M$ be a finitely generated module (over a local noetherian ring $(R,\mathfrak m))$ such that the projective dimension of $M$ is finite $(pd\ M=n<\infty)$. We know that
i) There is a free ...
1
vote
1answer
50 views
Projective Modules, Annihilators, and Idempotents
Let $Ra$ be the left ideal of a ring $R$ generated by an element $a \in R$. Show that $Ra$ is a projective left $R$-module if and only if the left annihilator of a, $\{r \in R | ra = 0\}$ is of the ...
2
votes
1answer
42 views
Free modules are projective
Free modules are projective.
Let $M$ be a free module. We have the diagram below (where the second row is exact)
Since $M$ is free, it has a basis (call it $X$). For every $x_i \in X$, there ...
3
votes
0answers
33 views
Projective Module as a Direct Sum of Left Ideals
I wonder if the following statement is true:
Every projective $R$-module is a direct sum of projective left ideals of $R$.
Most examples of non-free projective modules I have seen are all left ...
1
vote
1answer
62 views
Prove that $Hom_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$ and show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module.
1) Prove that $Hom_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$
2) Show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module.
1) We know that $\Bbb{Q}$ is an injective $\Bbb{Z}$-module.
This implies ...
0
votes
1answer
62 views
Localization at a monic polynomial: Horrocks' theorem
A theorem of Horrocks states that if $P$ is a projective $R[x]$-module ($R$ a local ring) such that $P_S$ is a free $R[x]_S$-module (where $S\subset R[x]$ is the set of all monic polynomials in ...
1
vote
1answer
63 views
Finitely Presented Modules and Exact Sequences
I came across this passage in Rotman's Advanced Modern Algebra:
If $M$ is finitely presented, there is a short exact sequence
$\begin{equation*}0 \rightarrow K \rightarrow F \rightarrow M ...
8
votes
1answer
74 views
A direct product of projective modules which is not projective
I am looking for an elementary example of a family $\{M_\alpha\}_\alpha$ of projective $R$-modules whose direct product is not projective. The simplest example that I know is the $\Bbb{Z}$-modules, ...
2
votes
0answers
38 views
Projective modules are stably isomorphic if localization at a monic polynomial is isomorphic
Let $R$ be a ring and let $P,Q$ be finitely generated projective $R[x]$-modules. Let further $f\in R[x]$ be a monic polynomial. If $P_f\cong Q_f$ (localization on $f$) then $P$ and $Q$ are stably ...
4
votes
1answer
92 views
direct sum of projective modules
I find the following question: Is the direct sum of two projective and not free $R$-modules projective/free? What if $R$ has got no zero divisors?
The answer to the first question is very simple, ...
2
votes
2answers
92 views
Is a finitely generated projective module a direct summand of a *finitely generated* free module?
Let $R$ be a (not necessarily commutative) ring and $P$ a finitely generated projective $R$-module. Then there is an $R$-module $N$ such that $P \oplus N$ is free.
Can $N$ always be chosen such that ...
0
votes
1answer
44 views
Non-trivial stably free, projective right ideals
I am working on Stably free, projective right ideals (1985) by J.T. Stafford. Trying to get through Theorem 1.2 on page 66 [4].
From the beginning of the paper we are supposed to know: taking two ...
2
votes
2answers
99 views
Example of a simple module which does not occur in the regular module?
Let $K$ be a field and $A$ be a $K$-algebra.
I know, if $A$ is artinain algebra, then by Krull-Schmidt Theorem $A$ , as a left regular module, can be written as a direct sum of indecomposable ...
0
votes
1answer
46 views
When are free modules extended
I am looking for help to understand the following:
Let $R$ be a commutative ring and $P$ a projective $R[x]$-module. If $P_{\mathfrak m}$ (localization at $R-{\mathfrak m}$, for $\mathfrak m$ a ...
3
votes
1answer
98 views
Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$
The question I have is on how to calculate Projective Resolution of $\mathbb{Q}$ over $\mathbb{Z}$. I have found that $\mathbb{Q}$ is flat, but that it is not projective. I mention this as I wonder ...
1
vote
0answers
43 views
Rejects and injectives
Let $A$ be any ring and consider modules on the left.
It is well known that the trace $Tr(M,A)$ is a two-sided ideal of $A$. If $A$ is a unitary ring then:
$Tr(P,A)P=P$, for $P$ projective;
...
1
vote
1answer
80 views
Projective cover of a simple module and indecomposable factor modules
Let $P$ a non-zero projective module. Prove that $P$ is the projective cover of a simple module if and only if every non-zero factor module of $P$ is indecomposable.
Thank you.
5
votes
1answer
135 views
Modules which are isomorphic to their tensor product.
Suppose that we have a commutative ring $R$. I am interested in finding the (finitely generated and projective, if you want) $R$-modules $M,$ such that $M\cong M\otimes_R M$ as $R$-modules.
I know ...
7
votes
1answer
103 views
Question on Projective Dimensions
$\require{AMScd}$I have a question regarding a claim in A first course of homological algebra by Northcott. I think it's very easy, since the author didn't provide a proof, and just kind of claimed ...
8
votes
2answers
148 views
Finitely generated flat modules that are not projective
Over left noetherian rings and over semiperfect rings, every finitely generated flat module is projective. What are some examples of finitely generated flat modules that are not projective?
Compare ...
1
vote
2answers
99 views
Questions about epimorphisms and projectives in functor categories
Suppose $I$ is a small category, $R$ is a ring and $_R\mathrm{Mod}$ is the category of left $R$-modules. How do I show that the category $[I,~_R\mathrm{Mod}]$ of all functors from $I$ to ...
2
votes
2answers
35 views
Partial cycles in projective resolutions of square-free algebra
Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand?
I suspect not, but have not ...
7
votes
1answer
86 views
Specific projective dimension of a module over bound quiver
Suppose $K$ is an algebraically closed field, and $A$ is the algebra presented by the quiver
$$\require{AMScd}
\begin{CD}
1 @>>> 2\\
@V{}VV @V{}VV \\
3 @>>> 4 @>>> 5
...
8
votes
2answers
153 views
How does this step in the proof of the structure theorem for f.g. modules over a Dedekind domain work?
I am trying to show that every finitely generated projective module $P$ over a Dedekind domain $D$ is a direct sum of (fractional) ideals. May's notes on Dedekind domains claim the result can be ...
4
votes
2answers
155 views
Injective and Projective module
Ok, this problem is driving me nuts. At first, I thought I did it. But when reading another textbook (having a similar proposition, they (the problem in my textbook, and the proposition in the other ...
0
votes
1answer
28 views
A module which is not singular
Suppose that M is a projective R-module and that it's simple, isomorphic to $R/I$ where $I$ is a maximal left ideal of $R$ such that $I$ is not a direct summand of $R$. How to get to a contradiction?
2
votes
2answers
51 views
Exact sequences and finitely generated projective modules
Let $R$ be a ring and $A,B,C$ be $R$-modules in an exact sequence
$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0.$$
Submodules or quotients of finitely generated projective modules are not ...
2
votes
1answer
98 views
Resolutions of bimodules as $R^e$-modules.
Let $k$ be a commutative ring, let $R$ be a $k$-algebra, a $R$-Bimodule $M$ over $R$ is a $k$-module with two actions of $R$ on $M$, on the left and on the right, the classical example of this being ...
0
votes
2answers
95 views
Projectivity of $\mathbb Q$ over $\mathbb Q\otimes_{\mathbb Z}\mathbb Q$
Consider $\mathbb Q\otimes \mathbb Q$, where $\mathbb Q$ is considered as $\mathbb Z$-algebra and consider $\mathbb Q$ as a right $\mathbb Q\otimes\mathbb Q$ module. Then is it true that $\mathbb Q$ ...
2
votes
3answers
180 views
Showing an ideal is a projective module via a split exact sequence
Let $R=\mathbb{Z}[\sqrt{-6}]$ and $I=(2,\sqrt{-6})$ the ideal generated by $2$ and $\sqrt{-6}$.
I want to show that $I$ is a projective $R$-module by producing a short exact sequence that splits, ...
3
votes
1answer
110 views
vector bundles on the affine line over a PID
Let $R$ be a PID. Is every finitely generated projective $R[T]$-module free? In other words, is every vector bundle on $\mathbb{A}^1_R$ trivial?
For $R=k[X]$ this is true by the Theorem of ...
0
votes
0answers
32 views
What is the relationship between various groups
Does the projective linear group of a given dimension share a representation with a group of automorphisms?
3
votes
0answers
71 views
Separability of finitely generated projectives over commutative ring
A class $\mathcal{C}$ of $R$-modules is called
-separative if $A \oplus A \simeq A \oplus B \simeq B \oplus B$ implies $A \simeq B$ for each $A,B \in \mathcal{C}$
-cancelative if $A \oplus C \simeq ...
2
votes
1answer
68 views
1
vote
1answer
36 views
Cancellation law of morphisms?
The title is probably misleading, but I wasn't quite sure how to boil my question down to one line.
My problem is this:
Assume I have two projective $R$-modules, $P$ and $Q$, and epimorphisms ...
1
vote
0answers
22 views
For a finite group $G$ and field $k$ of char$=p$, if $P,P'$ are projective $k[G]$-modules with $[P]=[P']$, is it true that $P=P'$?
That is -- is it true that if projective $k[G]$-modules have same composition factors then they are isomorphic?
This is easy to see for $\text{char}(k)=0$, or if $G$ is a composition of a $p$-group ...
2
votes
1answer
65 views
The indecomposable projective $\mathbb{F}_pG$-module with $UJ/J\cong \mathbb{F}_p$
Let:
$G$ be a finite group;
$p$ be prime;
$J$ be the Jacobson radical of $\mathbb{F}_pG$.
A paper I'm trying to read mentions the following object:
The indecomposable projective ...
1
vote
1answer
168 views
Exactness of short exact sequences
Let $A$ be a ring and let $P$ be a projective $A$-module. Then, the exactness of the sequence:
$$0\longrightarrow M_1 \overset{f}{\longrightarrow}M_2\overset{g}{\longrightarrow}M_3\longrightarrow 0 ...
8
votes
1answer
125 views
Projective Modules over the Ring of Trigonometric Functions
Let $ R = \mathbb{R}[ \cos x, \sin x] $ and consider the ideal $ \langle 1 - \cos x, \sin x\rangle $. Is this ideal a projective module over $R$ ?
2
votes
1answer
56 views
$R$-linear maps of projective fractional ideals
Let $R$ be an integral domain with field of fractions $K$, and let $I$ be a fractional ideal. If $I$ is projective then every $R$-linear maps $I\to R$ is multiplication by an element of $K$.
The ...
1
vote
1answer
165 views
Projective module over a ring
If $R$ is domain, as a projective module always exist over R. But how to produce such a module over $R$.
1
vote
0answers
230 views
Is the module quotient of projective modules projective?
Let $R$ be a commutative ring, let $M$, $N$ and $P$ be $R$-modules, and let $N' \subseteq N$ and $P' \subseteq P$ be submodules. Let $\mu:M\times N \to P$ be a surjective bilinear map. Define the ...
3
votes
2answers
223 views
Are projective modules over an artinian ring free?
Quoting a comment to this question:
By a theorem of Serre, if $R$ is a commutative artinian ring, every projective module [over $R$] is free. (The theorem states that for any commutative ...
2
votes
0answers
71 views
Projective dimension of simple module
Let $R$ be a commutative ring and $M$ a simple $R$-module. Then $\mathfrak{m}=Ann(M)$ is a maximal ideal of $R$. Then it is known that
$$
\mathrm{pdim}_{R}(M)=\mathrm{pdim}_{R_{\mathfrak{m}}}(M),
$$
...
