If $M$ is a Noetherian $R$-module then every submodule of $M$ is a finite intersection of irreducible submodules.   Please show me the way how to get the proof of this statement.
123854
If $M$ is Noetherian, any submodule $N \neq M$ can be written as a finite intersection of intersection-indecomposable ones.   I just don't know how to begin with the proof of this proposition.   Very kind of you for reading or answering.
52999
Let $A$ be a commutative ring, $M$ an $A$-module and $N_1, N_2$ two submodules of $M$.  If we have $M/N_1 \cong M/N_2$, does this imply $N_1 \cong N_2$?  This seems so trivial, but I just don't see a proof... Thanks!
105780
Let $R$ be a ring, $X,Y,Z$ and $T$ four $R-$ modules such that there exists a short exact sequence $$0 \rightarrow X \xrightarrow{f_1} Y \xrightarrow{f_2} Z \xrightarrow{f_3} T \rightarrow 0 $$  Prove the following:   (i)  If $X,Y$ and $T$ are Noetherian modules, then $Z$ is Noetherian.  (ii) If $X,Z$ and $T$ are Noetherian modules, then $Y$ is Noetherian.  I couldn't prove any of them, I assume both parts have similar proofs so I'll write what I've tried to do in (i) and I would appreciate some help with one of the two so I can complete the rest by myself.  In (i), there are alternative ways to show that $Z$ is Noetherian. I chose to show that every ascending chain is stationary.   So let $(Z_k)_{k \in \mathbb N}$ be an ascending chain in $Z$. Then $(f_3(Z_k))_{k \in \mathbb N}$ is an ascending chain in T. The same for $({f_2}^{-1}(Z_k))_{k \in \mathbb N}$ in Y. By hypothesis both of these chains are stationary. Let $N \in \mathbb N$ be such that $f_3(Z_n)=f_3(Z_N)$ for all $n \geq N$ and ${f_2}^{-1}(Z_n)={f_2}^{-1}(Z_N)$ for all $n \geq N$.  Take $x \in Z_n$ for $n \geq N$, I would like to show $x \in Z_N$. We have $f(x) \in f_3(Z_n)=f_3(Z_N)$, so there is $y \in Z_N$ such that $f_3(y)=f_3(x)$. But then $f_3(y-x)=0$, so $x-y \in Ker(f_3)=Im(f_2)$. Here I got stuck, there is $w \in Y$ such that $f_2(w)=x-y$, but in which submodule of $Y$ is $w$? And I don't see where to use that $X$ is also Noetherian. Any help would be greatly appreciated.
101021
As the title suggests, I am asked to prove that, given a Noetherian scheme $(X,\ \mathcal{O}_{X})$ and any open subset $U\subseteq X$, $\Gamma(U,\ \mathcal{O}_{X}):=\mathcal{O}_{X}(U)$ is a Noetherian ring.  Up to now, I have been able to show that the result is true if $U$ is an affine open subset of $X$, i.e. $U\simeq\mathrm{Spec}(A)$ for some ring $A$ (and this is actually true when $X$ is just locally Noetherian). I have also shown that, given $U$ as above, $(U, \mathcal{O}_{X\vert U})$ is a Noetherian scheme as well, which should then allow me to reduce the problem to the case $U=X$. So, when all is said and done, I should try to prove that the ring $\mathcal{O}_{X}(X)$ is Noetherian. However, I can not go any further and I am stuck here.  Any help or suggestion would be grately appreciated.  Thank you.
70246
If $R$ is a Noetherian normal domain, then it is equal to the intersection of its localizations at height one primes.      What is an example of a non-normal domain that is strictly contained in such an intersection?    I'd prefer an example that is a finitely generated $\mathbb{C}$-algebra, and all I want is a candidate suggested. I'd prefer to verify it myself.
54622
Let $I$, $J$ be finitely generated ideals in a ring $A$ (commutative with identity). I know that the intersection need not be finitely generated: can somebody give me an example?   Thanks.
122239
I'm trying to prove that if the ring $R$ is Noetherian then every finitely generated $R$-module is Noetherian.   First of all, it is known that every module is a homomorphic image of a free module, so we can take $R$ as a module over itself, and then we will have a homomorphism $f:R \rightarrow M$, where $M$ is some arbitrary module. Next, according to the well-known theorem, $$\operatorname{Im}(f) \cong \frac{R}{\ker(f)},$$ and  $\frac{R}{\ker(f)}$ is Noetherian because $R$ is Noetherian (we can take finite set of generators for some submodule in $R$ and then its canonical image in $\frac{R}{\ker(f)}$ generates submodule there). Thus, $\operatorname{Im}(f)$ is Noetherian as well, but $\operatorname{Im}(f) \subseteq M$. How then one could say the same about the whole module $M$?
46016
Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes:     Let $R$ be a unitary commutative ring, and $I$ be an ideal, and $N$ is an $R$-module. I'll now show that $$I \otimes_R N \cong IN\,.$$   First, since $1 \in R$, and $I$ is an ideal of $R$, we must have that $I = RI$. So then $$I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.$$  But this must be so wrong, since if this proof is true, I could prove that any $R$-module is flat. So how come this proof is wrong?
24791
My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?  What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then $R^n/N \simeq R^n$ implies $N=0$. It seems like $N\neq 0$ implies $R^n/N$ has nontrivial relations, so $R^n/N$ can't be free.  If $R^n/N \simeq R^n$, we'd have an exact sequence  $0 \rightarrow N \hookrightarrow R^n \twoheadrightarrow R^n/N \simeq R^n \rightarrow 0$  which splits since $R^n$ is free, so $R^n \simeq R^n \oplus N$. Does this imply $N=0$? What if we assume $R$ is commutative, or even local? Maybe Nakayama can come in handy.  I'm interested in noncommutative examples too. Thanks!
66721
Let $aR$ be a nonzero ideal in a PID $R$. Show that $R/aR$ is a ring with only finitely many ideals.  Honestly, I do not know how to start. Appreciate any tips.
99484
Let $R$ be a commutative ring and $M$ an $R$-module such that for every ideal $I \subset R$ the natural map $I \otimes_R M \rightarrow IM$ is an isomorphism.     Why is $M$ flat ?   This result is taken from .
61191
Let $I$ and $J$ be ideals of a Noetherian ring $A$. If $JA_P\subseteq IA_P$ for every $P\in \operatorname{Ass}_A(A/I)$, then $J\subseteq I$.   I'm reading Matsumura's Commutative Ring Theory book on my own. This is exercise $6.4$ in that book, and I fail to prove it. Could you help me with this? Thanks in advance.
54166
I am revising for my Rings and Modules exam and am stuck on the following two questions:  $1.$ Let $M$ be a noetherian module and $ \  f : M \rightarrow M \  $  a surjective homomorphism. Show that $f : M \rightarrow M $ is an isomorphism.  $2$. Show that if a semi-simple module is noetherian then it is artinian.  Both these questions seem like they should be fairly straightforward to prove but I cannot seem to solve them.
15106
Let $M,N$ be $R$-modules with $M$ finitely generated and let $S\subseteq R$ be multiplicatively closed. Then there exists a module isomorphism   $$S^{-1}\text{Hom}_R(M,N) \xrightarrow{\sim} \text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N).$$   As homework, I found a 'proof' of this statement. However,  claims to prove me wrong. I am having trouble in finding the error in my proof. Could someone help me out?  I did prove there exists a map $\phi:S^{-1}\text{Hom}_R(M,N) \to\text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)$ defined by $f/s\mapsto(m/t\mapsto f(m)/st)$, and I am fairly confident that it contains no mistake.     Proof: Let $m_1,\dotsc,m_n$ be the generators for M. Let $f\in \text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)$ be given.   Choose $n_i\in N$ and $s_i\in S$ such that $f(m_i/1)=n_i/s_i$.   Define $\pi_f = s_1\dotsm s_n$. Then   $$\pi f(m/1) = \pi f\left( \sum_{i=1}^n r_i m_i \right) = \pi \sum_{i=1}^n r_i f(m_i) = \pi \sum_{i=1}^n r_i n_i/s_i = \left(\sum_{i=1}\left(\prod_{j=1, j\neq i}^n s_i\right)r_in_i\right)/1,$$   so $\pi f(m/1) \in N$ using the canonical embedding in $S^{-1}N$, so $m\mapsto \pi f(m/1)$ is a map $M\to N$ and is clearly an homomorphism.   If we define the map $\psi:\text{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)\to S^{-1}\text{Hom}_{R}(M,N)$ as $$ f \mapsto (m\mapsto \pi_f f(m)) / \pi_f,$$ then we clearly have a homomorphism if it is well-defined.      Proof: It remains to check that the element we map to does not depend on choice of $\pi_f$.   But if we have both $\pi_f$ and $\pi_f'$ for which $\pi_f f$ and $\pi_f' f$ are homomorphisms to $N$, then   $$(\pi_f f(m))/\pi_f = (\pi_f\pi_f' f(m))/(\pi_f\pi_f') = (\pi_f' f(m))/(\pi_f')$$   for all $m\in M$, so $\psi$ is in fact well defined.   It is now easy to check that $\psi$ and $\phi$ are inverses, which makes them homomorphisms.
123741
Let $L$ be a finitely generated free module over a commutative ring $A$. Set $n=\operatorname{rank} L$. Let $x_1,\dots,x_m$ be generators of $L$. Then $m \ge n$? If $m = n$, then is $x_1,\dots,x_m$ a basis of $L$?
112060
Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free?    What I know about a finitely generated projective module $M$ over a PID $R$ is isomorphic to $R^k\oplus R/(a_1)\oplus\dots\oplus R/(a_n)$, and for the local ring case I don't know how to start.
14188
Let $M$ and $N$ be graded $R$-modules (with $R$ a graded ring). $\varphi:M\rightarrow N$ is a homogeneous homomorphism of degree $i$ if $\varphi(M_n)\subset N_{n+i}$. Denote by $\mathrm{Hom}_i(M,N)$ the group of homogeneous homomorphisms of degree $i$. We define $^*\mathrm{Hom}_R(M,N)=\bigoplus_{i\in\mathbb{Z}}\mathrm{Hom}_i(M,N)$. This is a (graded) $R$-submodule of $\mathrm{Hom}_R(M,N)$.      How can I prove that these two modules are equal if $M$ is finite? And do you know a counterexample if $M$ is not finite?