I am trying to solve part (c) of the following Representation Theory question:  Let $D$ be a division ring and let $n$ be a positive integer. For $ 1 \leq l \leq n $ let $$C_l= \{A = (a_{ij}) \in {M_n}{(D)} : a_{ij} = 0 \text{ for } j \neq l \} .$$  (a) Show that each $C_l$ is a simple submodule of ${M_n}{(D)}$.  (b) Deduce that ${M_n}{(D)}$ is a semisimple ring.  (c) Also deduce that all simple modules of ${M_n}{(D)}$ are isomorphic to $D^n$.  Now, I have already managed to show (a) and (b), but I am struggling with (c) and specifically with defining a homomorphism. Any help with (c) would be a great benefit, thanks.
12199
I've a basic question.      $R$ is a simple left artinian ring. I want to show that all simple left $R$-modules are isomorphic.    A simple $R$-module $M$ is isomorphic to $R/J$ where $J$ is a maximal left ideal of $R$. Does this imply that $M\cong Ra$ for some $a\in R$, $a\neq 0$?   Many thanks.
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!
6706
Question:      Find the number of non trivial ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$. ($f$ is not necessarily unitary, i.e., $f(1)$ need not be $1$.)    Suppose $f$ is a ring homomorphism from $\mathbb Z_{12}$ to $\mathbb Z_{28}$.  Consider $f$ as a additive group homomorphism.  Let $k= |\ker f|$ and  $ t = |\operatorname{im}(f)|$. Then $k\mid 12$ and $t\mid 24$ and $kt=12$, by first isomorphism  theorem of groups.  There are two possibilities $k=3$, $t=4$   and $k=6$, $t=2$.  For the first case $f$ should map $1$ to an element of the subgroup generated by $7$ as there is a unique subgroup of $\mathbb Z_{28}$ of order $4$ generated by $7$. For the second case $1$ has to map to $14$, for the same reasoning.  So there are at most two ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$. Question is how to check the possible maps which are ring homomorphisms.  Thanks.
120354
I hope to show that $$\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$$ for $(a,b)=1$.  I made an effort to find a homomorphism but I failed.  Can you give a hint?
20705
In general, the group $\mathbb Z_{mn}$ under addition $\bmod mn$ is isomorphic to the direct product $\mathbb Z_m \times \mathbb Z_n$ if and only if $\gcd(m,n)=1.$ This is easy to see since $\mathbb Z_{mn}$ is generated by the element $(1,1)$, and if $\gcd(m,n)=1$, then this will only be zero if it is multiplied by itself $mn$ times (resulting in the cyclic group).   However I'm interested in constructing an isomorphism between the two groups as an explicit expression. I've been playing around a bit with some examples, one of which is the function $\phi : \mathbb Z_2 \times \mathbb Z_3 \to \mathbb Z_6$ where $\phi(a,b) = 3a + 4b \pmod 6$, which gives us the desired isomorphism. I'm interested here in the relationship between the integers $2$, $3$, and the coefficients $3, 4$ of $a$ and $b$. After playing around in Wolfram|Mathematica, I managed to obtain the following isomorphisms $\phi_{m,n} : \mathbb Z_m \times \mathbb Z_n \to \mathbb Z_{mn}$ for coprime $m,n$ in $\{1, 2, \dots, 10\}$.   First of all, the following results seem to hold in general.  $$\phi_{1, n}(a, b) = a + b \qquad \text{for all $n$}$$  $$\phi_{m, 1}(a, b) = a + b \qquad \text{for all $m$}$$  $$\phi_{2, n}(a, b) = na + (n + 1)b \qquad \text{for all $n$}$$ $$\phi_{m, n}(a, b) = K a + L b \iff \phi_{n,m}(a, b) = L a + K b \qquad \text{for all $m, n$}$$  Beyond this point, I could not spot a general pattern. The isomorphisms I obtained are:  $$\phi_{3, 4}(a, b) = 4a + 9b$$ $$\phi_{3, 5}(a, b) = 10a + 6b$$ $$\phi_{3, 7}(a, b) = 7a + 15b$$ $$\phi_{3, 8}(a, b) = 16a + 9b$$ $$\phi_{3, 10}(a, b) = 10a + 21b$$ $$\phi_{4, 5}(a, b) = 5a + 16b$$ $$\phi_{4, 7}(a, b) = 21a + 8b$$ $$\phi_{4, 9}(a, b) = 9a + 28b$$ $$\phi_{5, 6}(a, b) = 6a + 25b$$ $$\phi_{5, 7}(a, b) = 21a + 15b$$ $$\phi_{5, 8}(a, b) = 16a + 25b$$ $$\phi_{5, 9}(a, b) = 36a + 10b$$ $$\phi_{6, 7}(a, b) = 7a + 36b$$ $$\phi_{7, 8}(a, b) = 8a + 49b$$ $$\phi_{7, 9}(a, b) = 36a + 28b$$ $$\phi_{7, 10}(a, b) = 50a + 21b$$ $$\phi_{8, 9}(a, b) = 9a + 64b$$ $$\phi_{9, 10}(a, b) = 10a + 81b$$ If you want to try and generate a few yourself, here is the Mathematica function I implemented. I'm sure it can be optimised, but for the first 10 coprime integers it sufficed.  isomorphism[{m_, n_}] :=       Module[{x = Null},       For[i = 1, i &lt; m*n,           i++, (If[#[[2]] == Table[i, {i, 0, m*n - 1}],           x = ToString[#[[1]][[1]]] &lt;&gt; "a+" &lt;&gt; ToString[#[[1]][[2]]] &lt;&gt;          "b"; Break[]] &amp;) /@       Table[{{i, j},       Mod[i #[[1]] + j #[[2]], m*n] &amp; /@       Table[{Mod[i, m], Mod[i, n]}, {i, 0, m*n - 1}]}, {i, 1,      m*n}, {j, 1, m*n}][[i]]]; x];   I'd appreciate if anyone can help me find the pattern to determine $\phi_{m, n}$ in general.
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?
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
72862
Refer to localization of rings in Lang's Algebra p. 108. In particular, let $A$ be a commutative ring, $S$ a subset of $A$ that is a submonoid of the multiplicative monoid structure of $A$, containing the identity. Then  $S^{-1}A$ is the set of elements $\frac{a}{s}$ with $a \in A$ and $s \in S$.  We have $\frac{a_1}{s_1}=\frac{a_2}{s_2}$ if there exists $\bar{s} \in S$ such that $\bar{s}(s_2a_1 - s_1 a_2)=0$. Now consider the injection of $A$ into $S^{-1}A$ given by $x \mapsto \frac{x}{1}$. This is a ring homomorphism. Let $J$ be an ideal of $A$. Ring homomorphisms preserve ideals and so the injection of $J$ into $S^{-1}A$, call it $J^{*}$, will be an ideal. My problem is that i don't see why for any $\frac{a}{s} \in S^{-1}A$ and any $\frac{j}{1} \in J^{*}$ we will have that $\frac{aj}{s}$ will be in $J^{*}$, i.e. why there must exist an element $\kappa \in J$ such that $\frac{aj}{s}=\frac{\kappa}{1}$.
14188
Let $M$ and $N$ be graded $R$-modules (with $R$ a graded ring). $\varphi:M\rightarrow N$ is a homogeneous homomorphism of degree $i$ if $\varphi(M_n)\subset N_{n+i}$. Denote by $\mathrm{Hom}_i(M,N)$ the group of homogeneous homomorphisms of degree $i$. We define $^*\mathrm{Hom}_R(M,N)=\bigoplus_{i\in\mathbb{Z}}\mathrm{Hom}_i(M,N)$. This is a (graded) $R$-submodule of $\mathrm{Hom}_R(M,N)$.      How can I prove that these two modules are equal if $M$ is finite? And do you know a counterexample if $M$ is not finite?
112259
There's a small detail in a proof of the Chinese Remainder Theorem for modules I don't understand when it comes to showing the normally constructed homomorphism is a surjection.     Suppose $R$ is a commutative unital ring, with $I_1,\dots,I_k$ pairwise comaximal ideals. How then for any $j$ can I find $r\in R$ such that $r\equiv 1\pmod{I_j}$ but $r\equiv 0\pmod{I_i}$ for $i\neq j$?    Thank you!
98026
Let $A$ be a commutative ring and $S$ a multiplicative subset of $A$ generated by $s\in A$ which is not a zero-divisor. Consider the polynomial ring $A[x]$.      Given a polynomial $f\in A[x]$, suppose that $f(1/s)=0$ for $s\in S$. Does it follow by some application of the first isomorphism theorem that $f(x)=(sx-1)g(x)$, with $g(x)\in A[x]$?   On the one hand I think this can be proven by constructing an isomorphism from $A[x]$ modulo the relation $x=1/s$ (really the replacement of $x$ with $1/s$). This is evidently an isomorphism from $A[x]/(x=1/s)$ into $S^{-1}A$. It is tempting to conclude (though I'm not certain that this should work), that there must be some surjective homomorphism $\phi: A[X]\to S^{-1}A$ such that $\ker(\phi)=(sx-1)$. [Since generally $(A/B\cong A/C) \nRightarrow (B\cong C)$, it is not sufficient to conclude that, say, the kernel of $\psi: f\to f(1/s)$ is $(sx-1)$.]  I am especially interested in answers as to whether, say, $\psi$ has kernel $(sx-1)$. But for anyone who can follow my own attempt at a proof, I'm also interested in whether my argument breaks down or can be made to work rigorously.
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 :)
95567
I am wondering the following question.     Let $A$ be a ring. Is it always true that End$_A(A)\cong A$. Is true when $A$ is semisimple?   I tried to give an isomorphism $\phi$ between End$_A(A)$ and  $A$. $\phi(f)=f(1)$, but I don't know if I am right.  Thanks for your help.
91972
If $\gcd(m,n)=1$, then $\mathbb{Z}_n \times \mathbb{Z}_m$ is a cyclic group.    Let's denote $\mathbb{Z}_n=\langle1_n \rangle$ and $\mathbb{Z}_m=\langle1_m \rangle.$ My proof goes as follows: since $|\mathbb{Z}_n \times \mathbb{Z}_m|=mn$ and $\gcd(m,n)=1$, $|\langle 1_n,1_m\rangle|=\text{lcm}(|\langle 1_n\rangle|,|\langle 1_m\rangle|)=mn$. Hence the direct product is cyclic. Is my proof correct?
28394
Consider the quotient $\mathbb{Z}-\text{module}$ $$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$ Prove that $M$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$.  I think the first step is to find a basis of $M$. Clearly $(1,0,0)$ and $(0,0,1)$ are in a basis of $M$. I don't know how to find the third element in that basis. Also the basis in $\mathbb{Z} \oplus \mathbb{Z}_4$ are $(1,0)$ and $(0,1)$.
1531
If $R$ is a local ring and $M$ and $N$ are finitely generated $R$-modules such that $M\otimes N=0$ then how does it follow from Nakayama's lemma that either $M=0$ or $N=0$?    This is an exercise in Atiyah and Macdonald. The part I could not show in the hints is $({M{\otimes}_R N)}_{k}=0$ implies $M_{k}{\otimes }_{k} N_{k}=0$, where $k=R/\mathfrak m$ and $\mathfrak m$ is the maximal ideal of $R$.
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!