Suppose that we have a commutative ring $R$ and an exact sequence of graded R-modules $$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$$ It is true that we have an exact sequence of graded R-modules $$0\rightarrow\text{S}^{\bullet}(M')\rightarrow\text{S}^{\bullet}(M)\rightarrow\text{S}^{\bullet}(M'')\rightarrow 0$$ where $\text{S}^{\bullet}(-)$ denotes the symmetric algebra? Thank you for your time.
122076
Let  $$0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$$  be an exact sequence of $A$-modules, where $A$ is a commutative ring with unity.  Then for each natural number $i$ one can look at the induced maps  $$\mathrm{Sym}^i(N) \rightarrow \mathrm{Sym}^i(P)$$  and   $$\mathrm{Sym}^i(M) \rightarrow \mathrm{Sym}^i(N).$$  The first question is, if they are still surjective, resp. injective, i.e. is $\mathrm{Sym}^i$ an exact functor?  The second question is: what are the kernel, resp. cokernel of these maps?
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!
19973
Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?
49646
This is question #66 from        Let $R$ be a ring with a multiplicative identity. If $U$ is an additive subgroup of $R$ such that $ur \in U$ for all    $u \in U$ and for all $r \in R$ , then $U$ is said to be a right ideal of $R$. If $R$ has exactly two right ideals, which of    the following must be true?      I. $R$ is commutative.   II. $R$ is a division ring (that is, all elements except the additive identity have multiplicative inverses).   III. $R$ is infinite.     Here is my reasoning: Because $R$ is a ring, $R$ is also a additive group with some identity element $0$. We have a theorem that says $0r = 0 $ in any ring, so $\{0\}$ is a right ideal of $R$. Also, $R$ is a right ideal of $R$. Now I have found two different right ideals and there mustn't be any more.     Edit: As mentioned in the comments, the example below is not a ring, so it is not applicable to the problem. I could not fix it by taking additive closure because that introduced more than two ideals.  A possible candidate for $R$ could be the set of $2\times2$ matrices $\{0,I,-I,a,-a\}$ where $a = [[^1_0] ,[^0_0]]$. The only right ideals are $R$ and $\{0\}$. This ring satisfies only property I, but the answer key says that II is the correct answer.
60832
It well known that if $R$ is a commutative ring then every ideal of $M_n(R)$ is $M_n (I)$ where $I$ is an ideal of $R$. My question is: Suppose that $I$ and $J$ are ideals of $M_n(R)$, i.e., $I=$M_n(A)$ and $J=$M_n(B)$. When is the statement $M_n(AB)=M_n(A)M_n(B)$ correct?
116946
Suppose that:  $0 \rightarrow M_{1} \rightarrow M \rightarrow M_{2} \rightarrow 0$  Is an exact sequence of $R$-modules (for $R$ commutative integral domain). Show that $\mathrm{rk}(M) = \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})$    I have shown the inequality $\mathrm{rk}(M) \geq \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})$, however for the opposite direction I'm lost. I tried to take $N$ as a linear independent subset and tried to decompose it into $N_{1}$ (the elements in $\mathrm{Ker}(g)$) and $N_{2}$ (the elements not in $\mathrm{Ker}(g)$), but I haven't got far, since $N_{1}$ gives us a linearly independent set in $M_{1}$, however $N_{2}$ can give us a linearly dependent set in $M_{2}$.   We define $\mathrm{rk}(M)$ in this case as the supremum over all cardinalities of the linearly independent subsets of $M$ (where the only finite linear combination equaling $0$ implies trivial coefficients for $r \in R$).  Would this be the proper way to show the result? I think I'm missing something important.
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.
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 $L$?
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!
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 $R\cong R^2$?
24820
I am very interested in studying modules, which I have studied algebra, is basic theory of groups and rings, of Hungerford and Dummit. I was reading the Dummit the module, but I still struggled a bit in the section and the tensor product of exact sequences, as in the book of Atiyah. They know a simple book to begin with?
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!
103834
Let $R$ be a ring, and let $S$ be an $R$-algebra. Let $M$ be an $R$-module. Under what conditions on $R$, $S,$ and $M$ is it true that we have the following isomorphism of $S$-modules? $$S \otimes_R \operatorname{Hom}_R(M, R) \simeq \operatorname{Hom}_S(S \otimes_R M, S)$$  I know there is an obvious map from the left-hand side above to the right-hand side (namely, send a map $\phi \colon M \to R$ to the map $\operatorname{id} \otimes_R\, \phi$), but I do not know what conditions are required for there to be an inverse map. Does this have anything to do with $M$ being torsion-free as an $R$-module?
42711
Let $R$ be a commutative ring with $1 \neq 0$.  It is known that $R^{n} \hookrightarrow R^{m}$ implies $n \leq m$ and $R^{n} \twoheadrightarrow R^{m}$ implies $n \geq m$, and I might use this without mentioning (though I don't think I did). Isomorphisms are $R$-module maps if not mentioned otherwise.  Let $m \geq n$ and consider $R^{n} = R^{n} \times 0 \times \cdots \times 0 \subseteq R^{m}$. We have $\text{Hom}_{R}(R^{n}, R^{m}) \simeq M_{m \times n}(R)$ via $\phi \mapsto M(\phi) = (c_{ij})$ where $\phi(e_{j}) = \sum_{i=1}^{m}c_{ij}e_{i}$. Take any $R$-linearly independent $y_{1}, \cdots, y_{n} \subseteq R^{m}$ and define an $m \times n$ matrix $A = (a_{ij})$ via $y_{j} = b_{1j}e_{1} + \cdots + b_{mj}e_{m}$.  If we understand $e_{j}$ with $1 \leq j \leq n$ as $n \times 1$ column matrix in $R^{n}$ when we multiply an $m \times n$ matrix to it on its left, then $Ae_{j} = y_{j}$ for $1 \leq j \leq n$. Thus, we can understand this matrix as an injection $A : R^{n} \hookrightarrow R^{m}$.  By definition $A$ is an $m \times n$ matrix, but if what I wrote is correct, then $A|_{R^{n}}$ must be an isomorphism $R^{n} \simeq \bigoplus_{j=1}^{n} Ry_{j}$. I am not trying to find an invertible matrix to $A$, but is it okay to think of it as an embedding like how I described?
48095
Consider the ring $\Bbb Z[x_0,\cdots,x_n]$ with the grading defined by the degrees of polynomials, then we have $\mathcal O(\mathrm{Proj}\,\Bbb Z[x_0,\cdots,x_n])=\Bbb Z$.  Let $R$ be a nonzero commutative ring and consider the ring $R[x_0,\cdots,x_n]$ with the grading defined by the degrees of polynomials. What kind of ring $R$ makes $\mathcal O(\mathrm{Proj}\,R[x_0,\cdots,x_n])=R$?
100945
I'm studying recently faithfully flat modules and I'd like to know the following:      Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of nilpotent elements of $R$?
117131
Suppose $R$ is a commutative ring. Show that every ideal of $M_n(R)$ is of the form $M_n(I)$  where $I$ is an ideal of $R$.   I have spent 30 minutes on this question and I still got nowhere. Can anyone give some hints ?