Let $R$ be a ring, and $M$ an $R$-leftmodule. Let $\operatorname{Ann}_R(M)$ be the annihilator of M, meaning that $r m = 0 \space\space\space\space  \forall r \in \operatorname{Ann}_R(M), m \in M$.  Let $I \subseteq \operatorname{Ann}_R(M)$ be a two-sided ideal. Show that M is naturally an $R/I$-module.  Thanks in advance! I'm not that used to annihilators, so any help would be appreciated.
33590
Suppose that $I$ is a two-sided ideal in the ring $R$, and that $M$ is a module over the quotient ring $R/I$. Why can we naturally regard $M$ as a $R$-module that is annihilated by $I$? Conversely, suppose that $M$ is a $R$-module annihilated by $I$. Why is $M$ naturally a module over $R/I$?
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
38705
Let $R$ a commutative ring and $M$ a $R$-module. Prove there is a bijection between the set of $R[x]$-module structures on $M$ and $\mathrm{End}_{R-\mathsf{Mod}}(M)$.  I really don't even know how to start with this question.
46851
The following is exercise number 10 in Hungerford's Algebra, page 190:  Let $R$ be a ring with no zero divisors such that for all $r, s \in R$ there exist $a, b \in R$, not both zero, such that $ar + bs = 0$.  a) If $R = K \oplus L$ (module direct sum) then $K = 0$ or $L = 0$  b) If $R$ has an identity, then $R$ has the invariant dimension property.  I have proved part a), but I'm having difficulty with b). I realize that it suffices to show that $R^m \cong R^n$ implies $m = n$, but I can't seem to use part a) to conclude this.
4257
What are the left and right ideals of matrix ring? How about the two sided ideals?
88321
Let $R$ be a (not necessarily commutative) ring and suppose $R$ has a maximal left ideal $I$ consisting of nilpotent elements, that is for $x\in R$ there exists some $n\in\mathbb{Z}_{\geq 0}$ such that $x^{m}=1$. I want to show that in this case, $I$ is the unique maximal left ideal.  To attempt to prove this, assume $J$ is some other maximal left ideal of $R$. Since $x^{m}=0$, we have $x^{m}\in J$. This is where I get stuck.  In the commutative case we can note that $I$ is necessarily a prime ideal, so either $x\in J$ or $x^{m-1}\in J$. Iterating this process shows that $x\in J$, then $I=J$ by maximality.  In the non-commutative case I can't seem to proceed. This is probably really easy but I'd appreciate a hint.
92335
Let $R$ be a PID, let $B$ be a torsion $R$-module, and let $p$ be a prime in $R$. Show that if $pb = 0$ for some nonzero $b \in B$, then $\operatorname{Ann}(B)$ is a subset of $(p)$.  Well we can let $\operatorname{Ann}(B) = (t)$ for some $t \in R$ since it is an ideal in $R$. How to make use of the fact that $B$ is torsion (i.e. there is r in R such that rm=0 for each m in M) to prove the desired result?
14194
Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is commutative)?   Note that $R$ is not Noetherian ring. Is there any counterexample (for any cases)?
1899
How to prove the following?  $R$ is an associative ring with identity. $R$ contains element $r$. The element is not invertible on the right and is not a left divisor of zero. Then the ring $R$ cannot be Artinian on the right.
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.
42525
Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$.     Must the kernel of $f$ be finitely generated?   This is true by definition if the generating set $m_1,\ldots,m_n$ is the one coming from the finite presentation of $M$.
96939
I am stuck on proving a left ideal $I$ of a ring $R$ is a direct summand of $R$ if and only if $I = Rr$ with $r^2 = r$. Could you help me with that? Any help will be very appreciated. Thanks!
61481
Let $R$ be a commutative ring with $1$. We know that the Krull dimension of $R$ is by definition the length of the longest chain of prime ideals of $R$.   Now if $M$ is a  $R$-module, the  is by definition $\dim(M):=\dim(R/\mathrm{Ann}_R(M))$. Since every ideal $I$ of $R$ is also a $R$-module, the Krull dimension of $I$ is $\dim(I)=\dim(R/\mathrm{Ann}_R(I))$.  However, in the literature, the Krull dimension of an ideal is $\dim(I):=\dim(R/I)$.   Are the two definitions equivalent?
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?
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
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!
50854
$$Rad(I)=\{a \in R | \exists n \in \mathbb{N} \text{ such that } a^n \in I\}$$   R is a commutative ring, I is an ideal.   To show that $Rad(I)$ is an ideal of $I$, we have to show that for $a,b \in Rad(I)$ it stands that $a-b \in Rad(I)$ and for $r\in R$ it stands that $a\cdot r\in Rad(I)$, right??   When $a,b \in Rad(I)$ we have that $a,b \in R$ and $a^n, b^m \in I$.   $a-b \in R$.   But how can we show that $(a-b)^x \in I, x \in \mathbb{N}$ ??