Let $M$ and $N$ be $R-$modules, and $$0\longrightarrow M\overset{\iota}{\longrightarrow} N\overset{\pi}{\longrightarrow} N/M\longrightarrow 0$$ an exact sequence. Let $M$ and $N/M$ finitely generated. Show that $N$ is finitely generated.   I know how to prove it, but I wanted to do as following if it's possible. Since $M$ and $N/M$ are finitely generated, then there is a surjection $$\sigma :R^{\oplus s}\longrightarrow M$$ and a surjection $$\tau : R^{\oplus t}\longrightarrow N/M.$$  What I want to show is that there is a surjection $$R^{\oplus s}\oplus R^{\oplus t}\longrightarrow N.$$ If $$\pi_1:R^{\oplus s}\oplus R^{\oplus t}\longrightarrow R^{\oplus s}$$ and $$\pi_2: R^{\oplus s}\oplus R^{\oplus t}\longrightarrow R^{\oplus t}$$ are the projection, we have that $\sigma \circ \pi_1$ and $\tau\circ \pi_2$ are onto, but ow can I do better ?
63931
I know this must be easy. I'm new to modules, so perhaps I must be missing  something important.     Let $M$ be an $R$-module. Show that if there exists a submodule $N$ such that $N$ and $M/N$ are finitely generated, then $M$ is finitely generated.   If the exact sequence $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0 $ (where the second and third arrows are inclusion and projection respectively) splits then $M$ would be isomorphic to the direct sum of $M$ and $M/N$ and the conclusion would follow (I think). But I can't show that it splits.  Thank you.
40012
Say I have a module $M$ over a ring $R$ that is   $1.$ finitely-generated, i.e. $\exists$ finite $X \subset M$ such that $M = \sum_{i=1}^{n} Rx_{i}, x_{i}\in X$ (but $X$ is not necessarily free)  $2.$ free, i.e. $\exists$ a basis $Y \subset M$ a.k.a. $Y$ is free and generates M, (but $Y$ is not necessarily finite).  Now I want to show, that there exists a finite basis for $M$. For vector spaces, this would obviously be trivial, but for some  $R$-Modul, I'm not sure how to proceed.   Furthermore, is every minimal generating subset of $M$ a basis?
112299
Prove or disprove: for any comm. ring $R $,   For three $R $-modules $M $, $N $ and $P $, $M \oplus P \simeq N \oplus P $ implies $M \simeq N $. Let $f:M\to N $ and $g:N\to M $ be two $R $-module homomorphisms such that $gf=1_M $ (where $1_M = \operatorname{id}_M$ means the identity map on $M$). Then, $N\simeq\operatorname{Im}(f)\oplus\ker(g)$.   If $R $ is a PID and the modules are f.g., 1 is true, but I have no idea on general cases.
98611
Let $R$ be a commutative ring with unity and $M$ an $R$-module.   If $M\oplus R^m\cong R^n$ for some integers $m,n \geq 1$ then must $M$ be finitely generated and free?   Can somebody help me with this? Regards.
40525
Let $R$ be a ring and $M$ an $R$-module.  If we suppose $M$ finitely generated then (following Matsumura) if we write $M=Rm_1+\cdots+Rm_n$ we have:  $p\in\operatorname{Supp}M$ if and only if $M_p\neq0$ if and only if there exists an $i$ such that $m_i\neq0$ in $M_p$ if and only if there exists an $i$ such that $\operatorname{Ann}m_i\subset p$ if and only if $\operatorname{ann}M=\bigcap_{i=1}^n\operatorname{Ann}m_i\subset p$. And so $\operatorname{Supp}M=V(\operatorname{Ann}M)$.     If $M$ is not finitely generated where does this proof fail?    It seems to me that if we write $M=\langle m_i\rangle_{i\in I}$ nothing will change.      And if this proof fails could you give me an example of a module such that $\operatorname{Supp}M\neq V(\operatorname{Ann}M)$?
65188
I am currently working through a problem in Rotman:       Let $R$ be a domain and let $Q=$Frac$(R)$.  If $r\in R$ is nonzero and $A$ is an $R$-module for which $rA=0$, prove that for all $n\geq 0$, $\mathrm{Ext}_R^n(Q,A)=0=\mathrm{Tor}_n^R(Q,A)$.   I think I have the Tor part:  Since $Q$ is flat, Tor$_n^R(Q,A)=0$ for all $n\geq 1$.  And Tor$_0^R(Q,A)=Q\otimes_R A=0$ because: for each $\frac{t}{s}\otimes a\in Q\otimes A$, $\frac{t}{s}\otimes a=\frac{tr}{sr}\otimes a=\frac{t}{sr}\otimes ra =0$.  However, I am having trouble with the Ext part.  To do it directly I either need a projective resolution for $Q$ or an injective resolution for $A$, but I am not sure how I would resolve either of these.  Or I considered using a long exact sequence for the short exact sequence $0\rightarrow R\rightarrow Q\rightarrow Q/R\rightarrow 0$, but that didn't seem to get me anywhere either.  I have been stuck on this problem for far too long so any help or hints are greatly appreciated.
49416
Consider the subalgebra $R=k[x,xy,xy^2,xy^3, \ldots] \subset k[x,y]$. How do I prove that $R$ is not finitely generated (over $k$)?    What is the general strategy for proving that an algebra is not finitely generated?
22268
Let $I$ ba a two-sided ideal of a ring $R$ and $M$ be an $R-$ left module. Prove that if $M$ is an $R-$ projective left module then $M/IM$ is an $R/I-$ projective left module.  It is easy to see that $M/IM$ has the structure of $R/I-$module. I need some ideas to prove that $M/IM$ is an $R/I-$ projective left module.  Thank you very much.
79013
Let $R = \mathbb{C}[t]$ be a ring of polynomials in variable $t$ with coefficients in the field of complex numbers $\mathbb{C}$ and let $$N = R[x]/(tx-t).$$  I claim that $N$ is not a flat $R$- module. If we consider the exact sequence $$0 \rightarrow(t) \rightarrow R$$ such that the ideal $(t)$ is viewed as an $R$-module.   We know that the above sequence exact iff the map $(t) \rightarrow R$ is injective. Let us assume that N is flat then this means that $N \otimes  (t) \rightarrow N \otimes R$ has to be injective. How do I go from here?   Thanks.
33643
Let $A$ be a commutative ring and let $F$, $E$ be $A$-modules. Then $E$ is the direct limit of its finitely generated submodules. Serge Lang, in his Algebra, p. 604, remarks that a technique to test whether an element of $F \otimes E$ is zero, is to examine its image in $F \otimes N$, as $N$ varies over the finitely generated submodules of $E$. Could somebody please explain why this is a valid technique? What is the image of an element of $F \otimes E$ in $F \otimes N$?
68559
I'm aware that over a commutative ring, two free finitely generated modules of finite rank are isomorphic if and only if they have the same rank.   I'm trying to understand a curious example of why this does not hold over a noncommutative ring.      Let's take $M=\oplus_{i\in\omega}\mathbb{Z}$. Then let $R=\text{End}(M)$, which is not commutative. Then $R\cong R^2$ as left $R$-modules.   It seems like one could possibly make a basis of $R$ using only endomorphisms which act on the even or odd coordinates of elements of $M$, which might suggest such an isomorphism. Is there a clear proof of why $R\cong R^2$?
59572
This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is bugging me and I don't have any friends also studying this subject (in case it is of any use, I'm an undergraduate).     Let $R$ be a commutative noetherian ring. If $A$, $B$ are finitely generated $R$-modules, then $\operatorname{Hom}_R(A,B)$ is a finitely generated $R$-module.   Here's what I've thought as of yet (maybe the problem is that I haven't been trying to apply the useful theorems...):  First of all, since $R$ is noetherian, it suffices to inject $\operatorname{Hom}_R(A,B)$ into some finitely generated module (in noetherian rings, submodules of f.g. are f.g.).  Now, since $R$ is commutative (or since $R$ is noetherian), a finitely generated $R$-module is a quotient of $R^n$ for some $n$. Then let us write $A\simeq \frac{R^n}{I}$ and $B\simeq \frac{R^m}{J}$. Now we've got:  $\operatorname{Hom}_R(A,B)\simeq \operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right)$.  If I had $\operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right) \hookrightarrow \operatorname{Hom}_R(R^n, R^m)$ (this is what I tend to think is the wrong direction, but a interesting question-rising path nevertheless) then it would be over since $\operatorname{Hom}_R(R^n, R^m)\simeq R^{nm}$ which is finitely generated.  So I ask myself: can I see $\frac{R^n}{I}$ as a submodule of $R^n$? Because if I could, then doing the same thing for $R^m$ and passing it to the hom, then it would be over. Now, this would be true if the sequence $$0\to I\hookrightarrow R^n\to \frac{R^n}{I} \to 0$$ split. But this (of course) doesn't always happen. But if I somehow had the third module to be projective, or the first one to be injective, then it would happen.  But why would, for example, be $\frac{R^n}{I}$ be projective?  And this is where I got stuck. I'm sorry if this is overly detailed, but I remember reading that it is always good, when asking about a textbook question, to add what you've thought up to that moment. Also, even if my thoughts don't lead to the solution, I would like to know if they are correct!
120513
Let $\mathcal{S}$ and $\mathcal{T}$ be two subspaces of $\mathbb{R}^n$, let $P$ be the orthogonal projection of $\mathbb{R}^n$ on $\mathcal{S}$ and let $Q$ be the orthogonal projection of $\mathbb{R}^n$ onto $\mathcal{T}$.    Show that if $P$ and $Q$ commute, then $PQ$ is a projection and $PQ$ is the projection onto $\mathcal{S}\cap \mathcal{T}$.  Is the converse assertion true? Suppose $PQ$ is the orthogonal projection of $\mathbb{R}^n$ onto the intersection $\mathcal{S}\cap \mathcal{T}$. Must $P$ commute with $Q$.   Anybody has advice on how i should start proving this assertion?
28249
I was reading about the construction of a tensor product of any right $R$-module $M$ and left $R$-module $N$ over a ring $R$.  Why is it required that $M$ and $N$ be right and left modules, respectively? How does the construction not work otherwise?
110802
Suppose that $R$ is a ring and $M$ and $N$ are $R$-module. Prove that   a) If $R$ is integral domain, then $Tor(M \bigoplus N) \cong Tor(M) \bigoplus Tor(N)$  b) If $P$ is a submodule of $M$ and $Q$ is a submodule of $N$, then $M \bigoplus N /P \bigoplus Q \cong M/P \bigoplus N/Q$.  For the first part, I kind of confused of the direct sum. The direct sum here means external direct sum or internal direct sum?   For the second part, I define a map $f: M \bigoplus N \rightarrow M/P \bigoplus N/Q$ where $(m,n) \rightarrow (m+P,n+Q)$  Then I manage to show the map is bijective module homomorphism. Just want to confirm with the surjectivity, is it trivial in this case? Like I have $(m+P,n+Q)$, then its preimage is $(m,n)$ right?  EDIT: Part a): Let $(m,n) \in Tor(M) \bigoplus Tor(N)$. Then there exists $q,s \in R$, where $q,s \neq 0$ such that $(qm,sn)=(0,0)$ Note that since $R$ is integral domain and $q,s \neq 0 \Rightarrow qs \neq 0$. Hence, $qs(m,n)=(qsm,qsn)=(s(qm),q(sn))=(0,0)$, which implies that $(m,n) \in Tor(M \bigoplus N)$  Let $(m,n) \in Tor(M \bigoplus N)$. Then there exists $r \in R,r \neq 0$ such that $r(m,n)=(0,0)=(rm,rn) \Rightarrow rm=0,rn=0 \Rightarrow (m,n) \in Tor(M) \bigoplus Tor(N)$. Hence, $Tor(M \bigoplus N)=Tor(M) \bigoplus Tor(N)$
1531
If $R$ is a local ring and $M$ and $N$ are finitely generated $R$-modules such that $M\otimes N=0$ then how does it follow from Nakayama's lemma that either $M=0$ or $N=0$?    This is an exercise in Atiyah and Macdonald. The part I could not show in the hints is $({M{\otimes}_R N)}_{k}=0$ implies $M_{k}{\otimes }_{k} N_{k}=0$, where $k=R/\mathfrak m$ and $\mathfrak m$ is the maximal ideal of $R$.
34218
I'm going to prove following exercise from Bosch's "Algebraic Geometry and Commutative Algebra" (page 461):    I'm struggle with proving that if $m &lt; n$ then the morphism $\varphi: \mathbb{P}^n_R \to \mathbb{P}^m_R$ must be constant.   I tried following:  Assume that we have such non constant $\varphi$. Then we get $m+1$ global sections $s_i \in \Gamma(\mathbb{P}^n, \mathcal{L})$ where $\mathcal{L}:= \varphi^*\mathcal{O}_{P^m}(1)$ where $s_i:= f^{\#}(t_i)$ for canonical $t_i \in \Gamma(\mathbb{P}^m, \mathcal{O}_{P^m}(1))= \oplus_{i=0}^m \mathcal{O}_{P^m}(0) \cdot t_i$.   Since $Pic(\mathbb{P}^n)= \mathbb{Z}$ we conclude $f^*\mathcal{O}_{P^m}(1)= \mathcal{O}_{P^n}(d)$ for a $d \ge 0$ (otherwise it would not have global sections.  Denote $U:= span(s_i \vert i \in \{0,1,...,m\})= &lt;f^{\#}(t_i) \vert i \in \{0,1,...,m\}&gt;_R $ the $R$-module subspace of $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d))$.  Since the $t_i$ generate $\Gamma(\mathbb{P}^m, \mathcal{O}_{P^m}(1))$ then the pullbacks $s_i$ generate $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d))$, therefore $U= \Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d))$.   Since $m&lt;n$ by assumption and $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d)))$ is generated by exactly $n+1$ elements if $d&gt;0$ and otherwise $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(0))=R$ is generated by one element, we conclude $d=0$.  So $\varphi^*\mathcal{O}_{P^m}(1) = \mathcal{O}_{P^n}$ is the structure sheaf and all it's global sections are constants. Therefore also $s_i$.  What I don't understand is why does it already imply that $\varphi$ is constant, so that $\varphi(\mathbb{P}^n)=pt$?  We consider now the construction for $\varphi: X \to P^m$ in detail. As in Hartshorne described it comes by glueing the restrictions $\phi_i: X_{s_i} \to D_+(t_i)$ via ring maps $k[t_0/t_i, ..., t_m/t_i] \to \Gamma(\mathcal{O}_X,X_{s_i}), t_j//t_i \mapsto s_j/s_i$. Since $s_i$ constants we get $X= X_{s_i}$ and therefore all $\varphi_i$ factorise over $Spec(\Gamma(\mathcal{O}_X,X)$. Then also $\varphi$ factorise through this affine scheme.  So that it indeed suffice to show that the composition $X_{s_i}\to Spec(\Gamma(\mathcal{O}_X,X) \to D_+(t_i)$ is already constant. But how?