Let $\mathbb{Z}[\zeta]=\{a+b\zeta:a,b\in\mathbb Z\} $, and $\zeta=e^{\frac{2\pi i}{3}}=\frac{-1}2+\frac{\sqrt{3}}{2}i$, ($\zeta^3=1$)  Define the norm $N:\mathbb Z[\zeta]\rightarrow \mathbb Z$ by $N(a+b\zeta)=a^2+b^2-ab$  I need to show that $\alpha$ is a unit iff $N(\alpha)=1$.($\alpha=a+b\zeta$) and finding all the units  I showed $\Rightarrow$, now if $N(\alpha)=1$, then $a^2+b^2-ab=1$, so clearly if $a=b=\pm 1, a=\pm 1\ \&amp; \ b=0, and \ a=0\  \&amp; \ b=\pm 1$  I need another method that is not involving  the calculations of the units. In fact, I need to write the general form of $\alpha^{-1}$, so I can confirm it is a unit.
123962
On page 230 of Dummit and Foote's Abstract Algebra, they say: the units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ are determined by the integers $a,b$ with $a^2+ab+b^2=\pm1$ i.e. with $(2a+b)^2+3b^2=4$, from which is is easy to see the group of units is a group of order $6$ given by $\{\pm1,\pm\rho,\pm\rho^2\}$ where $\rho=\frac{-1+\sqrt{-3}}{2}$.  First, why change the characterization of unit from integers solutions of $a^2+ab+b^2=\pm1$ to integers solutions of $(2a+b)^2+3b^2=4$? How did they arrive at their answer?
99008
Is there a method to find the units in $\mathbb{Z}_{n}[x]$?    For instance, take $\mathbb{Z}_{4}$. How do we find all invertible polynomials in $\mathbb{Z}_{4}[x]$? Clearly $2x+1$ is one. What about the others? Is there any method?
50951
Suppose it asks to show $\mathbb{Z}[x]/(2x-1) \cong \mathbb{Z}[\frac{1}{2}]$  Cand I do like this ? First of all elements of $\mathbb{Z}[x]/(2x-1)$ is of form $\frac{m}{2^n}+(2x-1)$ and also all elements of $\mathbb{Z}[\frac{1}{2}]$ are of form $\frac{m}{2^n}$. Now define a normal projection $\mathbb{Z}[\frac{1}{2}] \to \mathbb{Z}[x]/(2x-1)$ where $a \to a+(2x-1)$ for all $a\in \mathbb{Z}[\frac{1}{2}]$. So kernal is srt of all $a$ such that $2x-1|a \implies a=0$ so both are isomorphic.
10216
Sorry. This might probably be a really easy question, but I am only a beginner in algebraic number theory. So, please bear with me.  Let $K$ be an algebraic number field and $\mathcal{O}_K$ its ring of integers.  Preludium   In my lecture we have defined the norm $N_{K/\mathbb{Q}}(x)$ for an element $x \in K$ in the usual way. In a following lecture we "extended" this to the norm of an ideal $\mathfrak{a}$ of $\mathcal{O}_K$ by setting $$ \mathfrak{N}(\mathfrak{a}) := [\mathcal{O}_K : \mathfrak{a}] := \left| \mathcal{O}_K / \mathfrak{a} \right|$$ and showed that for the principal ideal $(a)$ with $a \in \mathcal{O}_K$ we always have $\mathfrak{N}((a)) = \left|N_{K/\mathbb{Q}}(a)\right|$. Later, we introduced the notion of a fractional ideal in $K$ being a non-zero, finitely generated $\mathcal{O}_K$-submodule of $K$. And we showed that a non-zero $\mathcal{O}_K$-submodule $\mathfrak{a}$ of $K$ is a fractional ideal iff there is a $c \in \mathcal{O}_K \setminus \{0\} $ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (thus $c\mathfrak{a}$ is an ideal of $\mathcal{O}_K$). We now extended the norm $ \mathfrak{N}$ by defining for each fractional ideal $\mathfrak{a}$ of $K$ the norm $$ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|}, $$ $c$ being an element of $\mathcal{O}_K \setminus \{0\}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (exists according to 2.).   Question  How can I show that this latter definition is independent of the chosen $c$?  Edit  I changed the definition in (3) from $ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{N_{K/\mathbb{Q}}(c)} $ to $\mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|} $ and added the absolute value everywhere else, too. I think that was a typo in my lecture notes.  Thoughts (Is it better to add them as comments? I was not sure.)   Let $c, c' \in \mathcal{O}_K \setminus \{0\}$ such that $c \mathfrak{a} \subseteq \mathcal{O}_K $ and $c' \mathfrak{a} \subseteq \mathcal{O}_K$. At the moment I am contemplating if and in which way $c$ and $c'$ are related. I was wondering whether one always has $c'=bc$ (or vice versa) for an $b \in \mathcal{O}_K$. Then we had $$ \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c')\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(bc)\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(b)\right| · \left|N_{K/\mathbb{Q}}(c)\right|}$$ and it remained to show that $[\mathcal{O}_K : bc\mathfrak{a}] = \left|N_{K/\mathbb{Q}}(b)\right| · [\mathcal{O}_K : c\mathfrak{a}]$ (maybe because $[\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}]$ ???).
107495
Question: Is $\mathbb{Z}= \{\dots, -3, -2, -1, 0 ,1 ,2 , 3, \dots \}$ countable?  My attemp so far: Let us create the following one-to-one correspondence between $\mathbb{Z}$ and $\mathbb{N}$. $$\begin{matrix}1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; \dots \\ \updownarrow &amp; \updownarrow &amp; \updownarrow &amp; \updownarrow &amp; \updownarrow &amp; \updownarrow &amp; \updownarrow &amp; \dots\\0 &amp; 1 &amp; -1 &amp; 2 &amp; -2 &amp; 3 &amp; -3 &amp; \dots\end{matrix}$$ In order for $\mathbb{Z}$ to be countable we must define a function $f:\mathbb{N} \to \mathbb{Z}$ so that $\mathbb{Z} \sim \mathbb{N}$. $$\displaystyle f(n) = \begin{cases} \frac{n}{2} &amp; n \text{ is even} \\ -\frac{n-1}{2} &amp; n\text{ is odd} \end{cases}$$ To show that $\mathbb{Z} \sim \mathbb{N}$ we require $f$ to be bijective. From the picture above, we can clearly see that $f$ is surjective since $\forall z \in \mathbb{Z}\  \exists n \in \mathbb{N}$ such that $f(n) = z$. Hence we must now show that $f$ is injective. We need to consider three cases:  $n_1, n_2$ are odd $n_1, n_2$ are even $n_1$ is odd and $n_2$ is even  Case 1: \begin{align}f(n_1) &amp;= f(n_2) \\ \implies -\frac{n_1 -1}{2} &amp;= -\frac{n_2 -1}{2} \\ \implies n_1 -1 &amp; = n_2 -1 \\ \implies n_1 &amp;= n_2\end{align} Case 2: \begin{align}f(n_1) &amp;= f(n_2) \\ \implies \frac{n_1}{2} &amp;= \frac{n_2}{2} \\ \implies n_1 &amp;= n_2\end{align} I am, however, experiencing some difficulty showing the injective property for the $3^{rd}$ case. Can anyone please give me some assistance with that specific case?
47857
I'm trying to do the following exercise from Neukirch's Algebraic Number Theory (exercise 2, $\S 1$, chapter 1):  Show that, in the ring $\mathbb{Z}[i]$, the relation $\alpha\beta=\varepsilon\gamma^n$, for $\alpha,\beta$ relatively prime numbers and $\varepsilon$ a unit, implies $\alpha=\varepsilon^{\prime}\xi^n$ and $\beta=\varepsilon^{\prime\prime}\eta^n$, with $\varepsilon^{\prime},\varepsilon^{\prime\prime}$ units.  My attempt: once $\alpha$ and $\beta$ are relatively prime, $\alpha\nmid\beta$ and $\beta\nmid\alpha$. Therefore $N(\alpha)$ and $N(\beta)$ are relatively prime too (where $N(a+ib)=a^2+b^2$). Therefore,  $$N(\alpha\beta)=N(\alpha)N(\beta)=N(\varepsilon)N(\gamma^n)=N(\gamma)^n=(p_1\cdot\ldots\cdot p_r)^n=p_1^n\cdot\ldots\cdot p_r^n$$  And once $N(\alpha)$ and $N(\beta)$ are relatively prime, we have (if necessary, we reorder the $p_i$'s):  $$N(\alpha)=p_1^n\cdot\ldots\cdot p_k^n=(p_1\cdot\ldots\cdot p_k)^n=N(\xi)^n=N(\xi^n)$$ and $$N(\beta)=p_{k+1}^n\cdot\ldots\cdot p_r^n=(p_{k+1}\cdot\ldots\cdot p_r)^n=N(\eta)^n=N(\eta^n)$$  And the result follows.  But I'm not sure about one step. It is $\alpha$ and $\beta$ relatively prime $\Rightarrow$ $N(\alpha)$ and $N(\beta)$ relatively prime. I know that if $\alpha|\beta$ then $N(\alpha)|N(\beta)$, but the previous assertion is not necesseraly true, and if it is, and don't know how to prove it. If it is true, the result follows, unless I made something wrong in the rest of the proof.  That's it. If you know another way of doing the exercise, please show me.
51147
Q: Find a polynomial of degree > 0 in $\mathbb Z_4[X]$ that is a unit.  I know (2x+1) is a unit. Is there any other units in $Z_4[X]$ if there are infinite units, could you generalize it in a certain from such as (2nx+1) s.t n is an integer?
4380
I have to compute $Z(t)=\sum_{n=1}^\infty e^{-\lambda_nt}$, $t \in \mathbb{R}_{&gt;0}$, with $\lambda_n=\pi^2(\frac{m^2}{a^2}+\frac{n^2}{b^2})$. So $\sum_{m,n=1}^\infty e^{-(\pi^2(\frac{m^2}{a^2}+\frac{n^2}{b^2}))t}=\sum_{m=1}^\infty e^{-\pi^2(\frac{m^2}{a^2})t}\sum_{n=1}^\infty e^{-\pi^2(\frac{n^2}{a^2})t}$.  Is anyone could give me a good hint how to compute $\sum_{n=1}^\infty e^{-\pi^2(\frac{n^2}{a^2})t}$?
44290
After numerical analysis it seems that   $$ \frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} $$  Could someone prove the validity of such identity?
109318
We know that for $m=p^r, 1-\zeta_m$ is a prime.Now suppose that m has at least 2 distinct primes appearing in its prime factorization,we need to show that $1-\zeta_m$ is a unit in its ring of integers $\mathcal{O}_{\mathbb{Q}(\zeta_m)}=\mathbb{Z}[\zeta_m]$.  I tried proving that $N_{\mathbb{Q}}^{\mathbb{Q}(\zeta_m)} (1-\zeta_m)=\pm 1$ but got stuck in finding norm of $\zeta_m$. Some hint would be nice.
47622
Suppose $z\neq -1$ is a complex number of norm 1m $(|z| =1)$. I know that $z= n + mi$, what is the most straightforward way of solving this problem? I was also given the following sketch for a possible approach.
41740
That's it. I want to do this quotient,  $\mathbb Z [\sqrt{-11}] / (1+\sqrt{-11})$. My first idea was to see which elements are in the ideal:  $(a+b\sqrt{-11})(1+\sqrt{-11}) = a+ a\sqrt{-11} + b\sqrt{-11} -11b$  So I want to make and application who send this elements to $0$, and then see the Kernel of the application and apply the Isomorphism Theorem.  Any ideas?
87766
Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost.
70109
$$ z = 1+2i \ (complex \ number) \\ z^n = a_n + b_ni \ \ \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) $$ Prove that $ b_{n+2} - 2b_{n+1} + 5b_n = 0$  How can I solve this? Thank you!  EDIT: Or please tell me your ideas.
56179
Problem :  Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.  My approach : Since : $(a^3+b^3)=(a+b)^3-3ab(a+b)$ $\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})$ Now I don't know further about this problem whether or not this will help here. Please guide to solve this problem will be of great help, thanks.
88451
A complex number $z=\cos(\alpha)+i\sin(\alpha)=e^{i\alpha}$ is called an $n$th root of unity if $z^n-1=0$, and is a primitive $n$th root of unity if in addition $z^k-1\neq0$ for $1\leq k&lt;n$. The $n$th cyclotomic polynomial, $\Phi_n(x)$, is the monic polynomial of which the zeros are the distinct primitive $n$th roots of unity. Take $n&gt;1$.      (a) Show that $\zeta=e^{2\pi i/n}$ is a primitive $n$th root of unity.      (b) Show that   $$\Phi_n(x)=\prod_{\substack{1\leq j\leq n\\ (j,n)=1}}(x-\zeta^j),$$   and that $x^n-1=\prod_{d\mid n}\Phi_d(x)$.      (c) By applying the inversion formula to $\log \Phi_n(x)$, or otherwise, show that   $$\Phi_n(x)=\prod_{d\mid n}(x^{n/d}-1)^{\mu(d)}.$$   In particular, conclude that $\Phi_n\in\mathbb{Z}[x]$.      (d) Suppose that $p$ is prime, $p\nmid n$, and $\Phi_n(a)\equiv 0\pmod p$. Show that   $\qquad$ (i) $\hskip0.11in p\nmid a$;   $\qquad$ (ii) $\hskip0.07in t\mid n$, where $t=\operatorname{ord}_p a$;   $\qquad$ (iii) $\hskip0.03in$if $a^{n/d}\equiv 1\pmod p$, then $d\mid\frac{n}{t}$;   $\qquad$ (iv) $\hskip0.03in$if $p^s\| (a^t-1)$, then $p^s\|(a^{mt}-1)$ when $p\nmid m$;   $\qquad$ (v) $\hskip0.07in$if $n/t&gt;1$, then $p\nmid \Phi_n(a)$ [use (c)];   $\qquad$ (vi) $\hskip0.03in$the congruence $\Phi_n(x)\equiv0\pmod p$ is solvable if and only if $p\equiv 1\pmod n$.      (e) Show that there are infinitely many primes $p\equiv 1\pmod n$. [Hint: Consider $\Phi_n(np_1\cdots p_ry)$, where the $p_i$ are $\equiv1\bmod n$.]   These are a bunch of exercises on cyclotomic polynomial from Fundamentals of Number theory, p. 131, William J. Leveque. I have thought really hard, but still cannot prove most of them. Could anyone talk something about them or give me some reference books on that? I am really curious about their proofs. Thanks in advance.
91590
Find $|\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle|$ and determine if its cyclic.   $\mathbb{Z} \oplus \mathbb{Z} = \{(1,1), (1, 2),... (2, 1),...(3,1),...\}$  $\langle(3,6)\rangle = \{(0,0),(3,6),(6,12),...\}$  $\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle = \{(a,b) + \langle(3,6)\rangle : a,b \in \mathbb{Z}\}$  How would I go by finding the order of the quotient group? Would it be infinite? Since both sets that make up the quotient group are infinite, does that mean that the order of the quotient group is also infinite? What are some things I'm not noticing that could be helpful?