Is it necessary to prove this ??? Can it not be shown simply like 2Z is not ring isomorphism to 3Z , when n=2 and m=3.
123548
Show that if $m$ and $n$ are distinct positive integers, then $m\mathbb{Z}$ is not   ring-isomorphic to $n\mathbb{Z}$.   Can I get some help to solve this problem
39656
The full problem is as stated in the title. I am here to check if this is a valid proof. I thought it would be easiest using Linear Algebra.     Recall that an infinite cyclic group is isomorphic to $\mathbb{Z}$. We wish to show that we do not have an isomorphism between $\mathbb{ZxZ\;and\;Z}$. Note that $\mathbb{ZxZ}$ is an infinite group (under addition of course). Now, in order for there to even be potential for an isomorphism, two spaces must have equal dimension. Since the $\mathbb{dim(\mathbb{ZxZ})}=2&gt;\mathbb{dim(\mathbb{Z})}=1$, we know that $\nexists$ an isomorphism between our spaces. Hence, $\mathbb{ZxZ}$ is not a cyclic group.   My question (besides a validity check): was there a better way to prove this? I just found this to be the easiest way.  EDIT Totally wrong with this one. Back to the cutting board.
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?
92708
$R$ is a ring. Why don't we have an isomorphism  $$ R[x] \cong R[[x]]\ ? $$
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.
76898
Let $m$ and $n$ be two nonnegative integers. Assume that there is a group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$. Prove that $m = n$.   I tried using a contrapositive, ($m \neq n$ implies $\mathbb{Z}^m \ncong \mathbb{Z}^n$), and I think the problem is that there won't be a homomorphism, but did not get anywhere. Is there a better approach to this problem?
67383
Is multiplying by a constant m (integer) on group of set of all integers on addition an automorphism?   If so why does the 2nd example in  says that the unique non trivial automorphism is negation?
41643
I have a ring $\mathbf R =(R, +, -, ., 0, 1)$ (note that there is no invers for multiplication, $R$ is not $\mathbb R$, it is any set for the given algebra).  How do you prove that the following does not apply generally (meaning for all $a \in R$)?  $$\mathbf R[x]/(x^2−a)\mathbf R[x] \simeq \mathbf R^2$$  I know that the following does apply:  $\mathbf R[x]/(x−a)\mathbf R[x] \simeq \mathbf R$ , you can prove that using the first isomorphism theorem by finiding a homomorphism by substitution   $\Omega_a : \mathbf R[x] \to \mathbf R$ $\Omega_a(f(x)) = f(a)$  then $Ker(\Omega_a) = (x - a)\mathbf R[x]$ and $Im(\Omega_a) = \mathbf R$ therefore the two are isomorphic.  Please help, Thanks
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.
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.
72407
I need to prove that the center of $A_n$ is trivial for $n \geq 4$.  $Z(A_3) = A_3$, since $A_3 = \mathbb{Z}/3\mathbb{Z}$ is commutative.  One idea is two use "counting" technique. First of, all we count cojugacy classes of even permutations in $S_n$. It's the classes correspodning to the types $[\lambda_1, ..., \lambda_r]$ where $n \equiv r \mod 2$  Now, the conjugacy class in $S_n$ correspoding to the type $[\lambda_1, ..., \lambda_r]$ splits into two equal-sized classes in $A_n$ if $\lambda_1, ..., \lambda_r$ are distinct odd numbers. If they aren't then it is preserved in $A_n$.  Then we can use "the counting formula"(  ) of elements in each "even" class of $S_n$ and divide one by $2$ if needed.  Still, I'm not sure if it can be done. Seems to be a lot of work there. Maybe there are other, he easier ways? Or, maybe, it is the way, in this case, I wouild appreciate any advices.
88985
I think I'm supposed to prove that both $T$ and $I_V$ are isomorphic, but I'm not sure and even if that's the right way to go, how am I supposed to prove that $I_V$ is isomorphic. Or am I going about this all wrong. Please help.
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.
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.
86613
Usually for modules $M_1,M_2,N$ $$M_1 \times N \cong M_2 \times N \Rightarrow M_1 \cong M_2$$ is wrong. I'm just curious, but are there any cases or additional conditions where it gets true?  James B.
111066
The problem is stated as follows:  "Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."  Regardless of the right solution, I don't understand why is the following wrong:     We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.   Harsh criticism will be appreciated. Thanks.
123834
Problem is this:      Suppose for any element $r$ in a ring $R$ with unity $1$, $r$ or $1-r$ is invertible. Then show that non-invertible elements form an ideal of $R$.   It is crystal clear that $-r$ is non-invertible if $r$ is. So I tried to show the sum of two non-invertible elements is non-invertible, and the product of non-invertible elements with any ring element is non-invertible but I'm stuck.  Any help will be appreciated.