I have no idea how to prove that $\mathbb{Z}[i]^{\times}$ is finite (unit group of $\mathbb{Z}[i]$). I don't know any theorem and never studied properties of unit groups, so I don't even know where to start here. This fact fact appeared to me as an example while I was studying modules. Any help or hint would be appreciated. Thanks in advance.
43565
How can we show that $\pm 1, \pm i$ are the only units in the ring of Gaussian integers, $\mathbb Z[i]$?  Thank you.
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?
1549
I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$  But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.
22479
Let $m,n$ be positive integers bigger than $1$. Show that $\mathbb{Z}_m \times \mathbb Z_n$ is cyclic if and only if $\gcd(m,n)=1$.    I have no idea on how to start. Anyone hints are much helpful.
37217
How would one go about proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where  $l=\mathrm{lcm}(m,n)$ and $d=\gcd(m,n)$?    I am attempting to use the fundamental theorem of finitely generated abelian groups but am struggling. In the interest of honesty, this is a past exam question that I am attempting for which the solutions are not available. EDIT: As the question was asked in the paper, the Chinese Remainder Theorem would not be permitted since this is only proved in the follow-up course.
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How do you describe the quotient group $\mathbb{Z} \times \mathbb{Z} / \langle (1, 2) \rangle$? It is easy to get it when it is $\mathbb{Z}_n$, but it is difficult when it is $\mathbb{Z}$.  How should I start and solve the problem?
67836
I am attempting to understand and compute: $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$   I know $(0,2)$ generates $H = \{(0,0),(0,2),(0,4)\}$, which has  order 3 because there are 3 elements. Now, I must find all the cosets  which there should be 8. The reason there should be 8 is because $4\cdot 6/3 = 8$.  This is where I run into the issue. I have a hard time generating the cosets, but furthermore, once I actually do get the cosets, I have a hard time analyzing them and knowing what it means. If someone could walk me through a step-by-step solution that would be great.  The answer is that the factor group is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$.   Reference: Fraleigh, A First  Course in Abstract Algebra, p. 147, Example 15.10.
110292
I am trying to prove that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic. I was thinking of arguing the following:  Suppose there exists an isomorphism $\varphi: \mathbb{Z}[x]\rightarrow\mathbb{Q}[x]$. Because isomorphisms are by definition surjective, there exist $x, y \in\mathbb{Z}[x]$ such that $\varphi(x) = c \in \mathbb{Q}[x]$ and $\varphi(y) = d \in \mathbb{Q}[x]$ for any $c, d\in\mathbb{Q}[x]$. Because $\varphi$ is an isomorphism we must have $\varphi(x+y) = \varphi(x) + \varphi(y)$ for all $x, y \in \mathbb{Z}[x]$. Namely, because polynomial addition is defined componentwise, we must have that the constant term of $\varphi(a + b) = c_{0} + d_{0}$ (where $c_{0}, d_{0}$ are the constant terms of $c$ and $d$ respectively. I would then argue that because $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as additive groups, no such isomorphism $\varphi$ exists. Is this a valid proof?  I've seen proofs that argue that because $\varphi(1) = 1$ for any homomorphism we have $1 = \varphi(2(1/2)) = 2(\varphi(1/2))$ so $\varphi(1/2)$ must be contained in $\mathbb{Z}[x]^{\times}$. Then because $\mathbb{Z}[x]^{\times} = \mathbb{Z}^{\times} = \{\pm1\}$ we have $2 \times \pm1 \neq1$, a contradiction. Is this any different than arguing that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ each have a different number of units?
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How to show  that $\mathbb{Z}[i]/I$ is a finite field whenever $I$ is a prime ideal? Is it possible to find the cardinality of $\mathbb{Z}[i]/I$ as well?  I know how to show that it is an integral domain, because that follows very quickly.
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I'm doing some sheets for my Abstract Algebra class and I can't seem to remember the group defined as $\mathbb{R}^{\times}$. It's obviously some variation of $\mathbb{R}$ but I'm away from college on reading week so can't ask my tutor. If someone could clear up the confusion I'd be grateful.
103876
The units of $\mathbb Z_4[x]$ should be $$\{f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n\mid a_0\in\{1,3\},a_i\in\{0,2\}, 1\le i\le n\}.$$ But I don't know how to prove. Any suggestion?
100805
The motivation for this question is in my answer to this post :       (1) If I take $A$, $B$ and $C$ three groups such that $A\times B$ is isomorphic to $A\times C$ then do we have $B$ is isomorphic to $C$ ?   In the post a counter example is given when $A$ is an infinitely generated group. However I cannot get a counter-example when $A$, $B$ and $C$ are assumed to be finite groups. Note that there is no reason a priori if $\phi$ is an isomorphism from $A\times B$ to $A\times C$ that $\phi(A\times\{1_B\})=A\times\{1_C\}$.   In the case of finite groups it is true because of the Krull-Schmidt theorem : .  A slightly more general question is the following :     (2) If $B$ and $C$ are two finitely generated groups, can I always find a finitely generated group $A$ such that $A\times B$ is isomorphic to $A\times C$?   If $A$ is not assumed to verify any property but to be a group we can take :  $$A=B^{\mathbb{N}}\times C^{\mathbb{N}} $$  Furthermore replacing :$ \prod_{\mathbb{N}}$ by $\bigoplus_{\mathbb{N}}$ (or more likely the restricted product for groups but I am not sure of a well known notation, it is the subgroup of the product with all but a finite number components being trivial) will give you the same result for countably infinite groups.   $\underline{\text{Edit}}$ : To be clear about what I am asking. (1) is kown to be true for finite groups whereas (2) is known to be true when we are considering groups without restriction (even just for countably infinite groups). Let us pretend we are working on the smaller class of groups : the finitely generated groups. Is there any counter example to $(1)$? Is there any counter example to $(2)$?   $\underline{\text{Edit 2}}$: Here is a partial answer, from the classification of finitely generated abelian groups $(1)$ is always true if all groups are assumed to be abelian finitely generated groups. Now assume that $A\times B=G=A\times C$. Then you have (derived groups) :  $$D(G)=D(A)\times D(B)=D(A)\times D(C) $$  So the abelianizations verify :  $$A^{ab}\times B^{ab}=A^{ab}\times C^{ab}$$  Because we are now in the realm of finitely generated abelian groups we get :  $$B^{ab}=C^{ab} $$  So $(2)$ cannot be true in general for instance you can take $B:=\mathbb{Z}$ and $C=\mathbb{F}_5$, those two groups have different abelianizations.   The new question can be found here :
100847
When $d$ is in $\mathbb{Z}$, $\mathbb{Z}[\sqrt{d}] = \{a_0 + a_1 \sqrt{d} \mid a_0, a_1 \in \mathbb{Z}\}$  However, when $d$ is in $\mathbb{Q} \setminus \mathbb{Z}$, I cannot prove it.  Of course, when I see an example, I can understand that this is not a finitely generated abelian group, but I want to know the proof in general.
53094
How to show that $\mathbb Z_{mn}$ is isomorphic to $\mathbb Z_m \times\mathbb Z_n$ when $m$ and $n$ are coprime? It is easy to show that the natural map from $\mathbb Z_{mn}$ to $\mathbb Z_m \times\mathbb Z_n$ is a ring homomorphism. How to show that it is bijective?
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I know this has been asked before, in a bunch of places, so I expect it to be closed as a duplicate at some point, but I'd like to know if my own proof works. If the proof is incorrect, please limit answers to pointing out the errors. If the proof is correct, then thoughts on improving it would be appreciated.  Prove that $\Bbb Q$ is not the direct product of two non-trivial groups.  —Aluffi exercise II.3.5. My proof Suppose that $\Bbb Q = G \times H$, neither $G$ nor $H$ trivial. First note that $0_G$ is the only element of $G$ with finite order: If $m\ne 0$ and $m\cdot g=0_G$, then $m\cdot(g,0_H)=0_{G\times H}$, contradicting the fact that $\Bbb Q$ has no non-zero elements of finite order. Suppose that for some $m\ne 0$ and some $n\ne 0$, $\pi_G(m/n)=0_G$. Then $0_G=n\cdot\pi_G\left(\frac m n\right)=\pi_G\left(n\cdot\frac m n\right)=\pi_G(m)=m\cdot\pi_G(1)$. Since $\pi_G(1)$ has finite order, we conclude that $\pi_G(1)=0$. Since $\Bbb Z=\langle 1\rangle$, $\pi_G$ maps every integer to $0_G$. Now let $a$ and $b$ be non-zero integers. Then $b\cdot\pi_G\left(\frac a b\right)=\pi_G\left(b\cdot \frac a b\right)=\pi_G(a)=0_G$. As above, this shows that $\pi_G\left(\frac a b\right)=0_G$. Thus if $\pi_G$ is not an injective function, then it must map everything to $0_G$, so $G$ will be a trivial group. The same result holds for $\pi_H$. Since by assumption, $G$ and $H$ are non-trivial, $\pi_G$ and $\pi_H$ are both injective functions, which is of course impossible.
85655
Let $\mathbb{Z}[i]=\left\{a+bi:a,b \in \mathbb{Z}\right\}$  be the ring of Gaussian integers $(i^{2}=-1)$ and let $\mathbb{Q}[i]=\left\{a+bi:a,b \in \mathbb{Q}\right\}$ be its fraction field.      1) Show that the tensor product $\mathbb{Q}[i] \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ is isomorphic to the product $\mathbb{Q}[i] \times \mathbb{Q}[i]$ of two copies of the field.       2) Let $S=\left\{1,2,2^2,2^3,...\right\} \subseteq \mathbb{Z}$ be the multiplicatively closed subset generated by 2. Find two distinct idempotents in the algebra $S^{-1}(\mathbb{Z}[i] \otimes_{\mathbb{Z}} \mathbb{Z}[i])$ other than $0$ and $1$.    For the first part, I thought I would try to use the presentation $\mathbb{Q}[i] \simeq \mathbb{Q}[T]/(T^2+1)$ and see what comes from that. Any help would be much appreciated!
82895
If $m|n$. Why the map $f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times$ given by $a \mod{n}\mapsto a \mod m$ is a surjective homomorphism of groups?  Attempt: I proved it is well a well defined homomorphism because the canonical projection $\overline{f}\colon \mathbb{Z}_n \to \mathbb{Z}_m$ is a surjective ring homomorphism, it extends $f$ and maps units to units.  But I can't prove surjectiveness. Given $a\in\mathbb{Z}$ such that $(a,m)=1$ I'm looking for a $k\in\mathbb{Z}$ such that $(a+k\cdot m,n)=1$. I don't know why one of the $n/m$ candidates for such $k$ makes the number $a+k\cdot m$ an unit modulo $n$.