Let $\mathbb{Z}[\frac{{1+\sqrt{-3}}}{2}]$ be the ring of integers where $\alpha=\frac{{1+\sqrt{-3}}}{2}$. I wish to show that $1-\alpha$ is a prime and check whether 3 is a prime or not.  To show that $1-\alpha$ is prime, suppose $1-\alpha$ divides $(a+b\alpha)(c+d\alpha)$. Then I wish to show that it divides one of them. I wrote down the division algorithm for $1-\alpha$ divides $(a+b\alpha)(c+d\alpha)$ but it's a big mess that I can't seem to make sense of. Any help is appreciated.
124222
$\mathbb{Z}[\omega] = \{a + b\omega : a, b\text{ are integers}\}$  $\omega = e^{2πi/3} = -\frac12 + \frac{\sqrt3}2i$  My task is to factor the numbers $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ into irreducibles in $\mathbb{Z}[\omega]$.  I don't know if I'm overthinking this, but I do not even know where to start.  I'm used to doing this in $\mathbb{Z}[i]$ and $\mathbb{Z}[\sqrt2]$, but I have no idea how to approach $\mathbb{Z}[\omega]$.  EDIT: I am not supposed to use the norm in this exercise.  My biggest struggle right now is just figuring out how to factor these numbers.  Using guess and check I was able to find:  $$3 = (2 + \omega)(1 - \omega)$$  $$7 = (3 + \omega)(2 - \omega)$$  I'm plugging and chugging into Wolfram Alpha right now. Is there any better way to do this?
53923
Let $n$ and $a$ be positive integers with $a &gt; 1$. I need to show that $n$ divides $\phi(a^n -1)$.   Here, $\phi$ denotes the .  Could any one give me a hint?
102629
Let $\alpha = 2^{1/3}$.          Prove that the set $\mathbb{Q}[\alpha] = \{a+b\alpha+c\alpha^2 \mid a, b, c \in\mathbb{Q}\}$ is closed under addition and multiplication.    Prove that if $z \in \mathbb{Q}[\alpha]$ is nonzero then there exists $z^{-1} $such that $z*z^{-1} = 1$      I have proved that $\mathbb{Q}[\alpha]$ is closed under addition, but I am having difficulty proving that it is closed under multiplication.   With all coefficients in $\mathbb{Q}$, let $a_1 + b_1\alpha + c_1\alpha^2 $ and $a_2 + b_2\alpha + c_2\alpha^2 \in \mathbb{Q}[\alpha]$. When multiplied, these two elements yield a fourth degree polynomial, and I do not know how to prove that it is an element of $\mathbb{Q}[\alpha]$
105287
I am trying to prove $(a)$ is a maximal ideal in $\mathbb{Z}$,  if and only if $a$ is prime number.  Now I wrote: assume $a \in\mathbb{Z}$, while it's not prime number we can write as $a=xy$, for some integers $x$ and $y$. then $(a)\subset (x)$ and $(a)\subset (y)$  while if $(a)$ is maximal ideal, it's not exist $(k)$ such that $(a)\subset (k)\subset\mathbb{Z}$. so $(a)$ cannot be maximal ideal. we can say by contradictory if $a$ is prime, $(a)$ is maximal ideal.  is this correct?  and how to prove the other way.
34203
Let $n \in ℕ$. Show that the ring $ℤ/nℤ$ is a field if and only if $n$   is prime.   Let $n$ prime. I need to show that if $\bar{a} \neq 0$ then $∃\bar b: \bar{a} \cdot \bar{b} = \bar{1}$. Any hints for this ?  Suppose $ℤ/nℤ$ is a field. Therefore: for every $\bar{a} \neq 0$ $∃\bar b: \bar{a}\cdot \bar{b}=1$. How can I show that $n$ must be prime ?
69736
Show that if $a+bi$ is irreducible over $\mathbb{Z} [i]$ then $a^2 + b^2$ is prime over $\mathbb{Z} $ . (a,b are non-zero integers)  I know that $\mathbb{Z} [i]$ is a Euclidean domain and so  $a+bi$ is prime in it. Can I say that as the norm of  $a+bi$ is $a^2 + b^2$ that $a^2 + b^2$ must be prime over $\mathbb{Z} $? This doesn't seem correct to me but it's the best try I have been able to come up with.
83575
Does anyone know how to show this preferable without using modular  For any prime $p&gt;3$ show that 3 divides $2p^2+1$
49005
This is a problem out of the text Introduction to the Theory of Numbers by Niven, Zuckerman, and Montogmery and I am having quite a bit of trouble with it.  I tried to prove it directly, but that didn't work.  The question is:     Let $p$ be a prime and $\alpha$ a positive integer.   Show that $\binom{p^{\alpha}-1}k \equiv (-1)^k \pmod p$ for $ 0 \leq k \leq p^\alpha -1$.   If relevant, I did prove that ${{p^{\alpha}}\choose k} \equiv 0 \pmod p$ for $ 0 &lt; k &lt; p^\alpha $ but I can't seem to incorporate it.
17873
So far, I know if $p$ is a rational prime, then $(1)$ if $p\equiv 3\mod4$, then $p$ is prime in $\mathbb{Z}[i]$. $(2)$ If $p\equiv1\mod4$  then $p=π_1 π_2$ where $π_1 $ and $π_2$ are conjugate, Then $π_1 $ and $π_2$ are primes in $\mathbb{Z}[i]$. $(3)$ $2=(1+i)(1-i)$, then $(1+i)$and$(1-i)$ are primes in $\mathbb{Z}[i]$.  What's are all the prime elements in Gaussian integers $\mathbb{Z}[i]$? For example, $-3$ are prime in $\mathbb{Z}[i]$, but not in the above $3$ cases.
88197
Let $p\gt 3$, be a prime number and  $1+\frac{1}{2} +\frac{1}{3} +...+\frac{1}{p-1} =\frac{a}{b}$ when $a,b\in \mathbb N$ and $gcd(a,b)=1$.      prove that $p^2|a$.   I proved that $p|a$, but I cant prove $p^2|a$, and my idea is as follow:  By multiplying $(p-1)!$, we get:  $(p-1)!+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-2}+\frac{(p-1)!}{p-1}=\frac{a(p-1)!}{b}$ $\,\,\,\,$ $(1)$  So, the number $\frac{a(p-1)!}{b}$ should be integer. consider the left hand of the sum of $(1)$ in the ring $Z_p$. Since in $Z_p$ the inverses are unique and what we have is indeed a rearrangement of all the congruence classes $mod\,p$. So:  $(p-1)!+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-2}+\frac{(p-1)!}{p-1}\equiv 1+2+...+p-1\equiv \frac{p(p-1)}{2}\equiv 0$$\,\,\pmod p$. And from here, $p|\frac{a(p-1)!}{b}$, and that's mean: $p|a$.   I tried to prove the statement by considering the ring $Z_{p^2}$ but I get nothing.  Thanks in advance…
43469
I have a doubt on something. I have a theorem that says that if $R$ is a ring, then $R/I$ is a domain iff $I$ is prime. Since $X^2-1$ is note prime, $\mathbb Z[X]/(X^2-1)$ is not a domain. But, $X^2-1=(X-1)(X+1)$, and since $X-1$ and $X+1$ are prime in $\mathbb Z[X]$, there are coprime, and thus, by the chinese remainder $$\mathbb Z[X]/(X^2-1)\cong \mathbb Z[X]/(X-1)\times \mathbb Z[X]/(X+1),$$ since $\mathbb Z[X]/(X-1)\times \mathbb Z[X]/(X+1)$ is a domain, then so is $\mathbb Z[X/(X^2-1)$. What is the mistakes here ?
66110
Let $R = \mathbb{Z} + x\mathbb{Q}[x] \subset \mathbb{Q}[x]$. Find the irreducibles of $R$. Show that the irreducible elements in $R$ are $\pm p$ for prime integers $p$ and the irreducible polynomials $p(x) \in \mathbb{Q}[x]$ whose constant coefficient is $\pm 1$. Prove these irreducibles are prime in $R$.  The first part is pretty clear. ($\pm p$ for prime integers $p$), if $\alpha \in R$ s.t. $\deg(\alpha)=0$ then $\alpha$ is a constant polynomial thus in $\mathbb{Z}$. If $\alpha$ is a prime then it is irreducible. Having trouble showing that for $p(x) \in R$ s.t. $\deg(p(x)) &gt; 0$ with constant term $\pm 1$ are irreducible.
36435
Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements  The field portion I understand. $\mathbb{Z}[i]$ is a PID and because $q$ is congruent to 3 mod 4 it is irreducible thus prime. and hence the ideal $(q)$ is prime and therefore is maximal because $\mathbb{Z}[i]$ is a pid. Thus $\mathbb{Z}[i]/(q)$ is a field  I am not totally clear on the number of elements being $q^2$. I think perhaps the division algorithm would help but I am still not totally clear on how to use it in a case like this.
72037
I am doing some exercises on commutative algebra and came along the following expressions, which were not elaborated on. Is someone familiar with them?  The first is for $p$ a prime number $\mathbb{Z}_{(p)}$, where the exercise is to determine all ideals. Do you think this is just $\mathbb{Z}/p\mathbb{Z}$?  The second one is for $P$ a prime ideal the expression $P_P$, where $P_P$ lies in $R_P$ for $R$ a ring.  I never came across this notation before so I have to ask here.
74924
I know that $1$ is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$.  However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=0$ is a prime ideal in $\mathbb Z$. Isn't $(p)$ being a prime ideal the very definition of $p$ being a prime element?  (I know that this would violate the .)    Edit: Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see ).  This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see ).  So, my question is now: Why do we make this awkward convention?
41301
Suppose $p$ is a prime number such that $(p-1)/4$ and $(p+1)/2$ are also primes. Show that $p=13$.   My approach:   Since $p$ is prime, it is of the form $6k+1$ or $6k-1$. Now if $p$ is of the form $6k-1$ then $(p+1)/2 = 3k$ which is not possible as $3|3k$.   So $p$ is of the form $6k+1$. Now $(p-1)/4 = 3k/2$ so $2\mid k$. So $p$ is of the form $12k+1$. But now, how to prove $k=1$?  Please help!
96366
I am self studying algebra, and have come across this exercise:     Show that $\mathbb{Z}[i]/(2+3i)$ has 13 elements and is a field.   My proof: Since Z[i] is a euclidian domain with euclidian norm function $f(c+id)= c^2+d^2$, for any element $a+bi \in Z[i]$, we can "divide" it by $2+3i$ to get $a+bi= q (2+3i) + r$ where either $r=0$ or $r=c+di$ for some integers $c,d$ not both $0$ with $c^2 + d^2=f(c+di) &lt; f(2+3i)= 13$. But then $a+bi$ belongs to the coset $ r + (2+3i) $. So every element in the quotient ring belongs to some $r + (2+3i)$ for $r=0$ or $r=c+di$ with $c^2 + d^2&lt;13$. Now we can count all the possible pairs of integers satisfying this condition, ie $(0,0),(0,1),(0,2),(0,3), ..., (3,1)$ to get 13 and so there are 13 elements in the quotient ring.   Is this a correct argument? Also is there an elegant way of showing it is a field rather than showing its an integral domain by showing each element has an inverse by brute force? Thanks in advance!