Does there exist a nonzero $R$-module $M$ so that $M \cong M\oplus M$, where here $R$ is any ring with unity?  I'm not sure if this is true; I don't think it's true but I can't prove that. Clearly $M$ must be infinite for this to be true.
66341
I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0  The setup: $V$ is a $k$-vector space and let $R = \mathrm{End}_{k}(V)$.   Prove that $\mathrm{End}_{k}(V\oplus V) \cong R^4$ as an $R$-module Prove that $R$ doesn't satisfy the IBN property if $V = k^{\oplus \mathbb N}$.   For the first, I used to the fact that $V \oplus V$ is both the product and coproduct (in $k$-Vect) of $V$ with itself to get the isomorphism. What I just realized is I only showed that the two are isomorphic as groups not $R$-modules. So what would be the $R$-module structure on $\mathrm{End}_{k}(V \oplus V)$?   For the second, I used the fact that $V = k^{\oplus \mathbb N}$ implies $V \cong V \oplus V$ which in turn implies $R = \mathrm{End}_{k}(V) \cong \mathrm{End}_{k}(V \oplus V)$. Again, I just realized that I only showed the latter two are isomorphic as groups.   It may be obvious (and maybe why my professor let it pass?) but I can't come up with a good $R$-module structure that makes the two group isomorphisms $R$-linear.  Edit:  Explicitly, these are the isomorphisms I'm dealing with. Let $\pi_j, i_j$ be the natural projection/inclusion maps of the $j$-th factor resp. and $\psi: k^{\oplus \mathbb N} \oplus k^{\oplus \mathbb N} \to k^{\oplus \mathbb N}$ the isomorphism given by $\psi(e_i, 0)=e_{2i-1}$ and $\psi(0, e_i)=e_{2i}$.  Then the first isomorphism $\mathrm{End}_k(V \oplus V)\to R^4$ is given by $\varphi \mapsto (\pi_1\varphi i_1,\pi_2\varphi i_1,\pi_1\varphi i_2,\pi_2\varphi i_2)$  The second isomorphism $R \to \mathrm{End}_k(V \oplus V)$ is given by $\alpha \mapsto \psi^{-1} \alpha \psi$  The composition doesn't seem to be $R$-linear if I use the obvious $R$-module structure on $R$ and $R^4$.
112299
Prove or disprove: for any comm. ring $R $,   For three $R $-modules $M $, $N $ and $P $, $M \oplus P \simeq N \oplus P $ implies $M \simeq N $. Let $f:M\to N $ and $g:N\to M $ be two $R $-module homomorphisms such that $gf=1_M $ (where $1_M = \operatorname{id}_M$ means the identity map on $M$). Then, $N\simeq\operatorname{Im}(f)\oplus\ker(g)$.   If $R $ is a PID and the modules are f.g., 1 is true, but I have no idea on general cases.
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.
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)?
55624
I have a homework question as follows:  Let $M$ be an $R$-module with submodule $K$. A complement to $K$ in $M$ is a submodule $L$ of $M$ such that $M = K \bigoplus L$. An $R$-module $M$ is semisimple if every submodule of $M$ has a complement.      For each of the following rings $\mathbb{Z}, \mathbb{C}[t]$, and $\mathbb{C}[\mathbb{Z}]$ find a module which is not semisimple.    For the integers, I think I found one: $K = 2\mathbb{Z}$, which is a $\mathbb{Z}$-module, but the odd integers are not a module, so $K$ does not have a complement. Is this a correct example?  For the second one, I thought perhaps $\mathbb{R}[x]$ would work, but I'm not sure that this is an $R$-module for the ring $\mathbb{C}[t]$, and for the third one, I am unsure of where to start.   Any hints/examples would be appreciated.
14129
Let $R=\prod_{n\in\mathbb{N}}\mathbb{Z}$ and $M$ be $R$ regarded as $R$-module is usual way.  The submodule $N=\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ is it finitely generated?  Thanks
92027
Question: Given a ring without unity and with no zero-divisors, is it possible that there are idempotents other than zero?   Def: $a$ is idempotent if $a^2 = a$.   Originally the problem was to show that $1$ and $0$ are the only idempotents in a ring with unity and no zero-divisors, but I wonder what happens if we remove the unity condition.   I am trying to find a ring with idempotents not equal to $0$ or $1$. So far my biggest struggle has been coming up with examples of rings with the given properties.   Does anyone have any hints? How should I attack this problem?
13732
I am solving an exercise from Dummit, Foote. Let $R$ be a ring with $1$. Then the following are equivalent:   Every $R$-module is projective. Every $R$-module is injective.   Proof:  (1) implies that for any R-mod $A$ we have $Ext^i(-,A)$ is a zero functor. But this means that for any R-mod $Ext^i(L,A) = 0$ for all $A$ so $L$ is injective. I show $(2) \Rightarrow (1)$ exactly the same way.  It is quite easy so I am afraid I missed something. Is my proof correct?
77908
Let $R$ be the subring of $M_2(\mathbb{R})$ defined by $$R=\left\{\begin{pmatrix}a&amp;b\\0&amp;d\end{pmatrix}\mid a\in\mathbb{Q},b,d\in\mathbb{R}\right\}$$  Show that $R$ is right-Artinian but not left-Artinian.  The "obvious" approach to show it is not left-Artinian is to find an $A\in R$ such that $$RA\supset R(A^2)\supset R(A^3)\supset\dots$$  is an infinite decreasing chain of proper left submodules of $R$ (as a module over itself). However, after spending a lot of time in this endeavor I've come to the conclusion that no such element $A$ exists (I can post a proof if anybody is interested but it'd be long and not worth the time for me otherwise). Does anybody have an idea for another approach I can take to this? I have no idea what else submodules of $R$ would look like.
22268
Let $I$ ba a two-sided ideal of a ring $R$ and $M$ be an $R-$ left module. Prove that if $M$ is an $R-$ projective left module then $M/IM$ is an $R/I-$ projective left module.  It is easy to see that $M/IM$ has the structure of $R/I-$module. I need some ideas to prove that $M/IM$ is an $R/I-$ projective left module.  Thank you very much.
31414
Let $A$ be a ring and consider the free modules $A^{\oplus n}$, $A^{\oplus k}$, with $n,k\in \mathbb{N}$. Can $A^{\oplus n}$ be isomorphic to $A^{\oplus k}$ if $k\neq n$?   Thanks in advance for the help.
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.
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.
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.
16154
Question. Let $R$ be a ring, $\mathfrak{p}$ a prime, $M$ a finitely-generated $R$-module, and $N$ any $R$-module. Is the natural map $$\textrm{Hom}_R(M, N)_\mathfrak{p} \to \textrm{Hom}_{R_\mathfrak{p}}(M_\mathfrak{p}, N_\mathfrak{p})$$ an isomorphism (of $R_\mathfrak{p}$-modules)?  I can prove this in the case when $M$ is finitely presented: indeed, let $S$ be any flat $R$-algebra, and let $$R^m \to R^n \to M \to 0$$ be a right-exact sequence; tensoring with $S$ gives another right-exact sequence $$S^m \to S^n \to M \mathbin{\otimes_R} S \to 0$$ and applying hom functors, we get left-exact sequences $$0 \to \textrm{Hom}_R(M, N) \to N^n \to N^m$$ $$0 \to \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S) \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ and $S$ is flat, so tensoring the first sequence yields $$0 \to \textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ but extending the sequences by $0$ to the left, and putting in vertical maps between the last two, we conclude that $\textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \cong \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S)$ by the five lemma (modulo checking commutativity of diagrams). This is essentially the proof Eisenbud gives [Commutative Algebra, Prop. 2.10].  The trouble with extending it to a proof for finitely-generated modules is that we have to replace $R^m$ with a potentially arbitrary submodule $K$ of $R^n$, and that may not be free or even finitely-generated without some additional assumptions on $R$. I can't see an abstract nonsense proof of the claim, but I admit I haven't tried a bare-hands proof. However, is the claim even true?
11499
I am looking for rings that are integral domains but not factorization domains, that is, rings in which it is not possible to express a nonzero nonunit element as a product of irreducible elements.    Do you know any example?
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.