If $M$ is finitely generated left module over a left noetherian ring $R$, then $M$ is a noetherian module.  I need to show that an arbitrary ascending chain of submodules of $M$ must satisfy ACC, or equivalently, every submodule of $M$ is finitely generated.   Since $R$ is a left noetherian ring, every left ideal in $R$ is finitely generated. I am trying to find a 1-1 correspondence between left ideals of $R$ and submodules of $M$.  Is it a correct way, or how can I show the assertion by a different way? 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 $M$?
107961
Can a finitely generated module $M$ over a commutative ring have $\operatorname{Ann}(x) \neq 0$ for all $x \in M$ while $\operatorname{Ann}(M) = 0$?  It's not difficult to show that there is no such module if the ring is a integral domain. For general, I guess the answer is yes. But I failed to find a desired example.
121070
I want to understand why the following is true:     Let $S \subseteq R$ be commutative rings with $1$ and assume that $R$ is finitely generated as an $S$-module by at most $k$ elements. For every maximal ideal $M$ of $S$ there are at most $k$ maximal ideals of $R$ lying over $M$.   So let $T$ be a maximal ideal of $R$ lying over $M$ then $T \cap S = M$. Now it can be shown that $R/T$ is a finitely generated $S/M$-module. Hence $R/T$ is Artinian and thus it has finitely many maximal ideals. From here I have two questions:  1) Why from this follows that there can be only finitely many $T$ lying over $M$? I suspect they are using some correspondence which I don't see.  2) Why exactly at most $k$ ?  Can you please explain?
122471
please first read these then answer my question  Let $D$ be a division ring and r, n be two positive integer numbers such that $0 ≤ r ≤ n$. Then $H_r(D)$ denotes the left ideal of $M_n(D)$ containing all matrices whose $j$th column is zero, for every $r &lt; j ≤ n$.     Theorem A: Let $D$ be a division ring and $n$ be a positive integer number.   Then for every left ideal $I$ of $M_n(D)$, there exist an invertible matrix $P$ and integer $r$,   $0 ≤ r ≤ n$, such that $I = PH_r(D)P^{−1}$.   Now from theorem A how I can conclude that each left ideal of $M_2(D)$ is minimal? I'm totally confused with this
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.
52508
This is an unproven proposition I've come across in multiple places.   Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian.  Why is this? I thought about taking some chain of submodules $$ S^{-1}M_1\subset S^{-1}M_2\subset\cdots $$ and pulling back to a chain  $$ M_1\subset M_2\subset\cdots $$ of submodules of $A$ which must eventually stablize. Is there more to it than this? I kind of wary of assuming all submodules of $S^{-1}A$ have form $S^{-1}M$ for $M\leq A$.
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.
97142
Please give me a hint for this problem.  Let $R$ be a ring, and $I$ a right ideal of $R$. Show if $M$ is a left $R$-module, then $$ f  \   \colon (R/I) \otimes_{R} M \to  M/IM$$  defined by $ f((r+I) \otimes m) = rm + IM$ is an isomorphism of commutative groups.   And show that if $L$ is a left ideal, then  $$ (R/I) \otimes _{R} (R/L) \cong R/(I+L)$$ as commutative groups.
76227
Exercise 2.1 in Matsumura's Commutative Ring Theory reads as follows: "Let $A$ be a commutative ring and $I$ an ideal that is finitely generated and $I=I^2$. Then $I$ is generated by an idempotent."   In trying to solve it, i first followed a constructive approach, where e.g. for the case of two generators i tried to construct an idempotent generator. However, it seemed difficult. Then i realized that i could apply Nakayama's lemma to the $A$-module $I$ and the existence of the idempotent generator follows.   My question is: How could one go about finding this idempotent generator? Is there a systematic way?
17502
Consider the subring $R=k[x,xy,xy^2,\ldots]$ of $k[x,y]$. I want to prove that $R$ is not noetherian.   An ascending chain of ideals is the following: $$(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$$ It is intuitively clearly to me that this is an ascending chain of ideals. But how do I prove it rigorously that $$xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})$$ or that this chain of ideals can never stabillize?
71011
I've just learned the definitions of Artinian and Noetherian module and I'm now trying to think of examples. Can you tell me if the following example is correct:  An example of a $\mathbb Z$-module $M$ that is not Noetherian: Let $G_{1/2}$ be the additive subgroup of $\mathbb Q$ generated by $\frac12$. Then $G_{1/2} \subset G_{1/4} \subset G_{1/8} \subset \dotsb$ is a chain with no upper bound  hence $M = G_{1/2}$ as a $\mathbb Z$-module is not Noetherian.  But $M$ is Artinian: $G_{1/2^n}$ are the only subgroups of $G_{1/2}$. So every decreasing chain of submodules $G_i$ is bounded from below by  $G_{1/2^{\min i}}$.  Edit In Atiyah-MacDonald they give the following example:     Let $G$ be the subgroups of $\mathbb{Q}/\mathbb{Z}$ consisting of all elements whose order is a power of $p$, where $p$ is a fixed prime.   Then $G$ has exactly one subgroup $G_n$ of order $p^n$ for each $n \geq 0$, and $G_0 \subset G_1 \subset \dotsb \subset G_n \subset \dotsb$ (strict inclusions) so that $G$ does not satisfy the a.c.c.   On the other hand the only proper subgroups of $G$ are the $G_n$, so that $G$ does satisfy d.c.c.      (Original images  and .)   Does one have to take the quotient $\mathbb{Q}/\mathbb{Z}$?
36865
How to prove a commutative, with unit, Noetherian ring $A$ only has finitely many minimal prime ideals via the following step?   I have proved:     Step. All radical ideals of Noetherian ring $A$ can be expressed as an intersection of finitely many prime ideals.   How to continue?
116478
I proved part (i) of the following:  Proposition 6.3. Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then  i) $M$ is Noetherian $\iff$ $M'$ and $M'"$ are Noetherian;  ii) $M$ is Artinian $\iff$ $M'$ and $M''$ are Artinian  Proof. We shall prove i); the proof of ii) is similar.  $\implies$: An ascending chain of submodules of $M'$ (or $M''$) gives rise to a chain in $M$, hence is stationary.   Can you tell me if my proof of $(i)\Longleftarrow$ is correct? Here goes:  Let $M^\prime$ and $M^{\prime \prime}$ be Noetherian. Let $L_n$ be an ascending chain of submodules in $M$. Then $\alpha^{-1}(L_n)$ is an ascending chain of submodules in $M^\prime$. Hence $\alpha^{-1}(L_n)$ is stationary, that is, $\alpha^{-1}(L_n) = \alpha^{-1}(L_{n+1})$ for $n$ large enough. Hence $L_n = L_{n+1}$ for $n$ large enough since $\alpha$ is injective hence $L_n$ is stationary.   The proof given in Atiyah-Macdonald is the following but I don't understand why they need both maps, $\alpha$ and $\beta$:  $\Longleftarrow$: Let $(L_n)_{n \ge 1}$ be an ascending chain of submodules of $M$; then $(\alpha^{-1}(L_n))$ is a chain in $M'$, and $(\beta(L_n))$ is a chain in $M''$. For large enough $n$ both these chains are stationary, and it follows that the chain $(L_n)$ is stationary. $\Box$
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!
73365
Let $A$ be a commutative Noetherian ring and let $I$ and $J$ be ideals of $A.$  Suppose that $I\subseteq J$ and that $A/I \cong A/J$ as rings.  I want to prove that $I=J.$    Observations so far:  1) If we drop the hypothesis $I\subseteq J,$ then the result is false:  $$\mathbb{Q}[X,Y]/(X)\cong \mathbb{Q}[X,Y]/(Y) \text{ but } (X)\neq (Y).$$  2) If $A/I\cong A/J$ as $A$-modules, then $I=J$:  $$I=\mathrm{Ann}_A(A/I)=\mathrm{Ann}_A(A/J)=J.$$  I'm now a bit stuck as to how to proceed. Any hints would be most helpful!  Many thanks :)
106320
So I'm stuck with this problem.  Let $A$ be a nonzero commutative ring (with unit). I have several questions that are really close to each other.  1) Let $M$ be a finitely generated module over $A$. Can we have $\{x_n, n \in \mathbb{N}\} \subset M$ a linear independent family in $M$ ?  2) Let $M$ be a module over $A$ generated by $f_1, \ldots, f_n \in M$. Can we have $x_1,\ldots, x_{n+1} \in M$ a linear independent family ? (Obviously a negative answer to the latter implies a negative answer to the former.)   3) Let $N$ be a free module over $A$, and $M$ a module such that there exists an exact sequence $0 \to N \to M$ and another exact sequence $N \to M \to 0$. Is $M$ a free module as well (is $M \simeq N$)? If not, can we add $N$ of finite rank to get a positive answer ?  4) Let $M$ be finitely generated by $f_1, \ldots, f_n$, and $x_1, \ldots, x_n$ be a linear independent family in $M$. Is $M$ free?  Thank you for your help!
117346
Prove that if $R$ is a commutative ring in which all the prime ideals are finitely generated, then $R$ is Noetherian.   Here is what I been told to do: Suppose that $R$ is not Noetherian, and use Zorn’s lemma to obtain a maximal element $I$ in the collection of all ideals of $R$ that are not finitely generated.  Then use the following proposition: Let $I$ be an ideal of a commutative ring $R$, and let $r ∈ R$. If the ideals $I +rR$ and {$s ∈ R : sr ∈ I$} are finitely generated, then $I$ is a finitely generated ideal.  Can anyone help please. thanks a lot.