Let $R$ be a commutative ring with a multiplicative identity. If $a\in R, a \not = 0$, prove that $R/(a)$ is not a free $R$-module.   My initial attempt was to say that R/(a) is not torsion free, thus it cannot be free, but then I realized that the theorem "M is a free module if and only if it is torsion free" only works for finitely generated modules over PIDs.  So I'm pretty lost as to how to prove this.  Any help would be appreciated!
63514
Let $R$ be a commutative ring with $1$ and let $J$ be a proper ideal of $R$ such that $R/J \cong R^n$ as $R$-modules where $n$ is some natural number. Does this imply that $J$ is the trivial ideal?   Basically I am trying to prove/disprove that if $J$ is a proper ideal of $R$ and $R/J$ is free then $J=0$ and above is my work.
105921
Free modules are projective, and projective modules are direct summands of free modules.      Are there examples of projective modules that are not free?    (I know this is not possible for modules of fields.)  Free modules are torsion-free. But is the inverse true?      Are there examples of torsion-free modules that are not free?   Thank you~
26932
The question is motivated by the following multiple-choice problem:  If $R$ is a ring with the property that $r=r^2$ for each $r\in R$, which of the following must be true? I. $r+r=0$ for each $r\in R$. II. $(r+t)^2=r^2+t^2$ for each $r,t\in R$. III. $R$ is commutative.  Here are my questions:  What theorems do I need to solve the problem above? Why is a ring with the property that $r=r^2$ for each $r\in R$ so special? Is there a name for such rings?
94014
It's possible to produce an example of an integral domain $R$ and a free $R$-module $M$ with free submodules $L, L'$ such that $L+L'$ is not free. We can take $R=M=K[x,y]$ , $L=\left&lt;x\right&gt;$ ,  $L'=\left&lt;y\right&gt;$.  If $R$ is a PID and $M$ is free $R$-module, then for every pair of submodules $L, L'$ of $M$ we know that $L+L'$ is free.  My question is the following.      If $R$ is a PID and $M$ is an $R$-module, is it true that $L+L'$ is free whenever $L$ and $L'$ are free submodules of $M$?   Thanks!
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?
119458
Let  $R$  be  a  ring  such  that  the  only  right  ideals  of  $R$  are  $(0)$  and  $R$.  Prove  that  either  $R$  is  a  division  ring  or  that  $R$  is  a  ring  with  a  prime  number  of  elements  in  which  $ab=  0$  for  all  $a,b\in R$.    I  don't  want  the  proof.  I  am  stuck  at  one  point.  Why  does $R$  have  to  be  finite  here?
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 $M$?
96702
This is a question related to :  "...if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is never free. Thus the only commutative rings $R$ for which every finitely generated module is free are those with no nontrivial ideals, and this is one of several equivalent ways to define a field."  Question 1: How do I show that if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is not free? Is this really true? I think the following is a counterexample:  $R = \mathbb Z \times \mathbb Z , I = \langle (1, 0) \rangle $. Define $(a,b) + (c,d) = (a+c, b+d)$ and $(a,b) \cdot (c,d) = (ac, bd)$. Then $R/I = \langle (0,1 ) \rangle$ and the basis of $R/I$ is $\{(0,1)\}$.  Question 2: Can you give me an example of a finitely generated module that doesn't have a basis?  Question 3: What's the relationship between $R$ modules and ideals $I$ of $R$? Are all finitely generated $R$-modules of the form $R/I$? Are there no others?
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?
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!
31515
Let $R$ be a commutative ring. Prove that $R[x]/(x)$ is isomorphic to $R$.  Should I use the first isomorphism theorem for rings?  Can someone give me a hint on how to approach this problem? Also how would I use the first isomorphism theorem to prove this? I know that the theorem says $R /\ker f$ is isomorphic to image of $f$.
81640
Let R be a commutative ring, and let P be a prime ideal of R. Suppose that P has no nontrivial zero divisors in it. Show that R is an integral domain.   My proof:  Take $r,s,a \in R$ with $ar = as$, and $p \in P$. Then  \begin{align*} par &amp;= pas\\ a p r &amp;= a p s &amp;&amp; \text{R is commutative} \end{align*} and since $P$ is an ideal of $R$, then $pr, ps \in P$. Since $P$ is an integral domain, then $a p r = a p s \implies p r = p s$, so $r = s$.  I'm not using the fact that $P$ is a prime ideal which leads me to believe that I messed up a step.
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!
21427
Let $R$ and $R'$ be rings (with 1 but no further assumptions) and $n \in \mathbb{N}$. Does the following implication hold?     If $M_n(R) \simeq M_n(R')$ then $R \simeq R'$.   If the rings are commutative then it follows from considering the centers of $M_n(R)$ and $M_n(R')$. If the rings are division it also holds and even stronger, it holds even if the matrices are not of the same size.  I thought about considering the embeddings of $R$ in $M_n(R)$ as diagonal matrices, but since I do not want to asume anything about the isomorphism I was not able to conclude that these subrings of matrices in $M_n(R)$ and $M_n(R')$ are isomorphic. I hope someone can help me or with a proof, an idea or a counterexample. Thank you.
48645
Let $R$ be a commutative ring and let $S\subseteq R$ be a multiplicative closed subset containing 1.     I need to show that if $S^{-1}R$ is integral over $R$, then $S^{-1}R=R$.   I don't see how to get started.
31304
Let $A$ be a ring. How can I prove that:     $A$ is an Artinian ring $\Leftrightarrow \exists$ a faithful $A$-module which is of finite length.   I know that if a ring has a faithful $A$-module which is Noetherian, then the ring is Noetherian. Can I use this to prove the result above? I need your help to solve this problem. Thank you.
119298
Here $A$, $B$ and $C$ are $R$-modules. Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$? And what if $B$ and $C$ are finitely generated?