I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following:     For any $\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1$, we have a ring isomorphism   $$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.$$   Could someone show me an isomorphism between these rings to prove this?
3951
A very basic ring theory question, which I am not able to solve. How does one show that    $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$. Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.   My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
2317
In one of my assignment, I was told that $(\mathbb Z[x], +)$ and $(\mathbb Q_{&gt;0}, .)$ are isomorphic, with $$\phi (\sum_{k=0}^n a_k x^k) = \prod_{k=0}^n p_k^{a_k}$$ is a one-to-one surjective group homomorphism, where $p_1 = 2$; $p_2 = 3$; $p_3 =5$, and so on.  I know that this proof is based on the decomposition of numbers into a product of primes. But I don't think that it is correct. For instance, for $\frac12 \in \mathbb Q$, I cannot find a corresponding polynomial $f \in \mathbb Z[x]$ such that $\phi(f) = \frac12$.  If this proof is wrong, then are $(\mathbb Z[x], +)$ and $(\mathbb Q_{&gt;0}, \cdot)$ really isomorphic?
122369
Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.  Proof.  Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$. We have $$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$  or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.  Is my proof above correct? Thanks
32893
I've been given a homework problem that requires me to show that the rings $\mathbb{C}[x,y]/(y - x^2)$ and $\mathbb{C}[x,y]/(xy-1)$ are not isomorphic.    This is my attempt at a solution:  For $\mathbb{C}[x,y]/(y - x^2)$, we can parametrize in the following way: $x = t$ and $y = t^2$. Then this ring is isomorphic to $\mathbb{C}[t]$.  For $\mathbb{C}[x,y]/(xy-1)$, we can parametrize $x = t$ and $ y = 1/t$. Then this is isomorphic to $\mathbb{C}[t, 1/t]$.   But $\mathbb{C}[t, 1/t]$ is not isomorphic to $\mathbb{C}[t]$.  Am I on the right track? If not, any helpful hints?
43828
Let $I$ and $J$ be two ideals of a commutative ring $R$ with $1.$ Give a necessary and sufficient condition so that    $$R/IJ\cong R/I\times R/J.$$   Prove your claim. Then decide whether the following ring isomorphisms are true or not:   $$\mathbb{Q}[x]/\langle x^2-1\rangle\cong \mathbb{Q}[x]/\langle x-1\rangle\times \mathbb{Q}[x]/\langle x+1\rangle,$$   $$\mathbb{Z}[x]/\langle x^2-1\rangle\cong \mathbb{Z}[x]/\langle x-1\rangle\times \mathbb{Z}[x]/\langle x+1\rangle.$$   It has been a while since I took Abstract Algebra and I am preparing for the prelims. I am not sure how to tackle this one. Any help/suggestion/hint will be much obliged. This is what I have so far:     This is a simplified version of the chinese remainder theorem. The map   $R\to R/I\times R/J$ defined by $r\mapsto (r+I,r+J)$ is a ring homomorphism with kernel $I\cap J.$ If the ideals $I$ and $J$ are comaximal, then this map is surjective and $I\cap J=I\cdot J$, so    $$R/(I\cdot J)=R/(I\cap J)\cong R/I\times R/J.$$   My question is about the second part of the question, do I have to check whether: $\langle x-1\rangle$ and $\langle x+1\rangle$ are comaximal and $\mathbb{Q}[x]$ and/or $\mathbb{Z}[x]$ are ring homomorphisms? I believe that these ideals are comaximal in both cases but forget how to prove it.
34853
Consider the abelian groups $G_n=(\mathbb R^n,+)$ for $n\geq1$.     Claim: For any $n$ and $m$ the groups $G_n$ and $G_m$ are isomorphic.   This claim is true if one assumes the axiom of choice, and I have sketched a proof below. But this claim seems much weaker than the axiom of choice, which brings me to the following questions:   Are there any milder extensions of ZF that make the claim true? If the claim can be proven with some other axioms, what does the proof look like? Are there any known  axioms that make the claim false? Does the claim have any interesting (set-theoretical) corollaries?   This claim was also considered in  earlier post, but from a different point of view.    A sketchy proof of the claim:  $\newcommand{\Q}{\mathbb Q}$ The groups $G_n$ are naturally vector spaces over $\Q$. It is easy to check that the claim is equivalent with $G_n$ and $G_m$ being isomorphic as vector spaces over $\Q$.  All vector spaces have a Hamel basis; let the basis of $G_n$ be $B_n$. Thus $G_n=\Q^{(B_n)}$ (brackets meaning only finitely many nonzero components). But $|\Q^{(B_n)}|=|B_n|$ since the basis is infinite, so $|B_n|=|G_n|=|\mathbb R|$.  There is a bijection between the bases of the vector spaces $G_n$ and $G_m$ since they have the same cardinality. This gives a linear bijection between them.
21328
The isomorphism $\mathbb{Z}[i]/(a+bi) \cong \Bbb Z/(a^2+b^2)\Bbb Z$ is , when the integers $a$ and $b$ are coprime. But what happens when they are not coprime, say $(a,b)=d&gt;1$?      — For instance if $p$ is prime (which is not coprime with $0$) then $$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &amp;\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &amp;\text{if } p \equiv 1 \pmod 4 \end{cases}$$ (because $-1$ is a square mod $p$ iff $(-1)^{(p-1)/2}=1$).  — More generally, if $n=p_1^{r_1} \cdots p_m^{r_m} \in \Bbb N$, then  each pair of integers $p_j^{r_j}$ are coprime, so that by CRT we get $$\mathbb{Z}[i]/(n) \cong \mathbb{Z}[i]/(p_1^{r_1}) \times \cdots \times \mathbb{Z}[i]/(p_m^{r_m})$$  I was not sure how to find the structure of $\mathbb{Z}[i]/(p^{r}) \cong (\Bbb Z/p^r \Bbb Z)[X] \,/\, (X^2+1)$ when $p$ is prime and $r&gt;1$.  — Even more generally, in order to determine the structure of $\mathbb{Z}[i]/(a+bi)$ with $a+bi=d(x+iy)$ and $(x,y)=1$, we could try to use the CRT, provided that $d$ is coprime with $x+iy$ in $\Bbb Z[i]$. But this is not always true: for $d=13$ and $x+iy=2+3i$, we can't find Gauss integers $u$ and $v$ such that $du + (x+iy)v=1$, because this would mean that $(2+3i)[(2-3i)u+v]=1$, i.e. $2+3i$ is a unit in $\Bbb Z[i]$ which is not because its norm is $13 \neq ±1$.  — I was not able to go further. I recall that my general question is to known what $\mathbb{Z}[i]/(a+bi)$ is isomorphic to, when $a$ and $b$ are integers which are not coprime (for instance $a=p^r,b=0$ or $d=(a,b) = a^2+b^2&gt;1$).  Thank you for your help!
61436
Can someone help me solving the following problems?        $(\mathbb R^2,d_2)$ and $(\mathbb R, d_1)$, $d_2, d_1$ being the respective euclidean norms, are not isometric isomorphic, i.e. there is no distance preserving isomorphism between them.     Let $d$ be the restriction of the euclidean metric on $(-1,1) \times(-1,1).$ Find a metric $d':\mathbb R \times \mathbb R \to \mathbb R$ such that the spaces $((-1,1),d)$ and $(\mathbb R,d')$ are isometric isomorphic.      For 1.)  I thought it suffices to prove that if there was a isomorphism between those two, it wouldn't be distance preserving since it then also must be norm preserving which can't be injective since $||f(x)||_2 = ||x||_2 = ||f(-x)||_2$.    I don't know if that suffices or if it is even true. I would be glad if someone could prove me why they can't be isomorphic in the first place.    For 2.) I don't know what to do.
26386
Let $(\mathbb K,+,\cdot)$ be a field. Is there a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot) $ ?  The answer should clearly be negative. I tried to proceed via contradiction, but this has not led me very far. Since it's some kind of &quot;trick problem&quot;, I'm merely looking for hints.
6343
I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer. I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ &amp; that they're both non-abelian group.  Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ?  Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ?  Thanks in advance
103007
Question. Let $M = \left\{ \begin{bmatrix} a &amp; b \\ -b &amp; a \end{bmatrix} \,\middle\vert\,  a, b \in \mathbb R \right\}$.  Show that $(M,+)$ and $(\mathbb C,+)$ are isomorphic.  Show that $(M^{*},*)$ and $(\mathbb C^{*},*)$ are isomorphic.    (Here, $M^{*}$ is defined as excluding the zero-element) I understand that there exists an isomorphism $\phi:M \rightarrow\mathbb C$ defined by $\begin{bmatrix} \alpha &amp; \beta \\ -\beta &amp; \alpha \end{bmatrix}\mapsto \alpha+i\beta$, ie. I can show that $\phi(M_{1}+M_{2}) = \phi(M_{1})+\phi(M_{2})$ and $\phi(M_{1}*M_{2}) = \phi(M_{1})*\phi(M_{2})$. However, I'm unsure how to go about the two subquestions seperately.
29488
Is easy to show that $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$ as $\mathbb{Q}$-vector spaces and then $(\mathbb{R},+)$ is isomorphic to $(\mathbb{R}^n,+)$ as abelian groups. This result might suggest that we also have $(\mathbb{Q},+) \cong (\mathbb{Q}^n,+)$ as abelian groups, but a similar approach to this dosen't work.  How can I prove or disprove this?
99271
Let $d\in \Bbb{Z}$ be a square free in, I need to find an isomorphism between $\Bbb{Z}[\sqrt{d}]$ and a quotient of $\Bbb{Z}[X].$ I Would guess it's $$\Bbb{Z}[\sqrt{d}]\simeq\Bbb{Z}[X]/\langle X^2-d\rangle.$$ But I need to prove it. So, as $X^2+1$ is monic, I can perform the euclidean algorithm: all polynomial $\Bbb{Z}[X]$  can be written $$P=(X^2-d)Q+(aX+b)$$ for $Q\in\Bbb{Z}[X].$ We have $P(\sqrt{d})=0$ if and only if $a+\sqrt{d}b=0.$  If $d&lt;0$ then $\sqrt{d}=i\sqrt{-d}$ so $a=b=0.$      If $d&gt;0$ I am not sure how can I prove that $a=b=0$, I can do it for some special ("easy") case.    More than a solution, I need to understand what's going on to get the idea of the proof. I can conclude, but
71626
Prove $\mathbb{Z}_3[x]/(x^2+1)$ and $\mathbb{Z}_3[x]/(x^2+x-1)$ are isomorphic by finding an explicit isomorphism. My question is how I can define the map. Here are what I tried:  $\mathbb{Z}_3[x]/(x^2+1)=\{ a+b\alpha \,|\, \alpha^2 + 1 = 0, \quad a,b\in\mathbb{Z} \}$  $\mathbb{Z}_3[x]/(x^2+x-1)=\{ a+b\beta \,|\, \beta^2 + \beta - 1 = 0, \quad a,b\in\mathbb{Z} \}$  Then I am looking for an isomorphism of the form $f(a+b\alpha) = a+bf(\alpha)$ for all $a,b\in\mathbb{Z}$ where $f(\alpha^2)+1=0$. Here I'm stuck.  My professor said $f(a+b\alpha) = a+b(1-\beta)$ so that the restriction of $f$ to $\mathbb{Z}_3$ is the identity morphism. I don't understand why we have $f(\alpha) = 1-\beta$.  He also gave another problem: $\mathbb{Z}_5[x]/(x^2+2)$ and $\mathbb{Z}_5[x]/(x^2-x+1)$, where $f(a+b\alpha) = a+b(2+\beta)$.  Could someone please explain to me why $f(\alpha) = 1-\beta\,$ in the first problem and $f(\alpha)=2+\beta\,$ in the second problem?  Thank you very much.
89299
Let $f \colon A \to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B \otimes_A N$. Show that the homomorphism $g \colon N \to N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g (N)$ is a direct summand of $N_B$.   [Define $p \colon N_B \to N$ by $p (b \otimes y) = by$, and show that $N_B = \operatorname{Im} (g) \oplus \operatorname{Ker} (p)$.]   My idea was to define an isomorphism \begin{align*} \phi \colon N_B &amp; \to \operatorname{Im} (g) \oplus \operatorname{Ker} (p) \\ b \otimes y &amp; \mapsto (p (b \otimes y), b \otimes y - 1 \otimes (g \circ p) (b \otimes y)) \end{align*}  It's easy to check that $\phi$ is injective and surjective. If I take the $B$-module structure of $N_B$ to be $$ b' . (b \otimes y) = (b'b) \otimes y $$ I only manage to show that $\phi$ is a homomorphism of $A$-modules, because $g$ seems to be only a homomorphism of $A$-modules: $$ b.g(y) = b.(1 \otimes y) = b \otimes y \neq 1 \otimes (b.y) = g (b.y) $$ (where $\neq$ means "not equal in general").  However, the way the question is phrased seems to suggest that the isomorphism should be one of $B$-modules.  Should I take the $B$-module structure of $N_B$ to be something else, for example $$ b'.(b \otimes y) = b \otimes (b'.y) $$ or should I show the result only for $A$-modules?  I've been confusing myself about this issue of "as $A$-module" or "as $B$-module" and would appreciate if someone could clarify. (I also wonder if, since homomorphisms are only fancy functions of sets, it somehow doesn't matter.)
101981
Here I am under the impression that $2\mathbb Z$ and $3\mathbb Z$ are the sets of even numbers and multiples of $3$ respectively and the operations are usual addition and multiplication. This is an exercise in Fraleigh Abstract Algebra.  I believe that $2\mathbb Z$ and $3\mathbb Z$ are isomorphic as groups on addition.  I started a proof by supposing that there was a ring isomorphism, $f$,  in hopes of finding a contradiction with $f(x\cdot y) = f(x)\cdot f(y)$.
115398
I need help understanding the following isomorphisms:  $$\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z$$  In general, I am also wondering whether the following are true, and if so briefly why:   $\mathbb Z[\sqrt{D}] \cong \mathbb Z[x]/(x^2 - D)$ for all squarefree integers $D$ $(R/I)/J \cong R/(I +J)$, ie why do we have things like $\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+5) $ where $\mathbb Z[x] / (2) \cong \mathbb Z_2[x]$?   Also for example for something like the following:  $$\,\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+1)$$  Is it true because of the third isomorphism theorem? I'm guessing we set $R = \mathbb Z[x]$, $I = (2)$, and $J = (x^2 + 5)$, but please expand more on each step of the reduction.