How to I use the Eisenstein Criterion to show that for any prime $p$, $1+x^{p}+x^{2p}+...+x^{p(p-1)}$ is irreducible in $\mathbb{Z}[x]$? I tried using the substitution $x = y+1$ and I guess this works but this method seems too complicated. Is there a more efficient/elegant way?
121230
It is well known that the $p$th cyclotomic polynomial $$\Phi_p(x)\ =\ x^{p-1}+\cdots+x+1\ =\ \frac{x^p-1}{x-1}$$  is irreducible over $\mathbb Q$ for prime $p$. The standard trick is to make the transformation $x\mapsto x+1$ and turn the polynomial to one of the form $x^{p-1}+p\cdot q(x)$, where $q(x)$ is polynomial of degree $p-2$ and constant term 1, and apply Eisenstein's criterion.  On p.35 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition) the authors state that the following polynomial is also irreducible over $\mathbb Q$: $$\Phi_p(x^p)\ =\ x^{p(p-1)}+\cdots+x^p+1\ =\ \frac{x^{p^2}-1}{x^p-1}$$  (where $p$, as before, is prime). According to the book the result can also be proved by applying the transformation $x\mapsto x+1$. I tried it but I'm afraid it got me nowhere; neither have I been able to prove it by any other method.  I would be grateful if someone could show me how it is done. Thanks.
47687
How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input.
102222
If $R$ is a UFD and $f \in R[x]$, then we can't always use Eisenstein's Rule directly to show that $f$ is irreducible in Frac$(R)[x]$. However if we have $\phi$, an automorphism of Frac$(R)[x]$, then $f$ is irreducible if and only if $\phi(f)$ is irreducible. In order to apply Eisenstein to the resulting polynomial, you need $\phi(f) \in R[x]$, then you can deduce that your original $f$ was irreducible (we are using a 'shifted' version of the rule).  For example, for the cyclotomic polynomials $x^{p-1} + \dots + x + 1$, you cannot use Eisenstein directly, however if you consider $\phi : \mathbb{Q}[x] \rightarrow \mathbb{Q}[x]$ which sends $x$ to $x+1$ and does nothing on $\mathbb{Q}$, then using the binomial the resulting polynomial is Eisenstein at $p$.      What I am wondering is given an irreducible polynomial, can you always find an automorphism such that you can apply Eisenstein to show it's irreducible?   If not in general then what about the case where $R = \mathbb{Z}$ and $\phi$ is an automorphism of $\mathbb{Q}[x]$?
80506
Every irreducible polynomial $f$ over a perfect field $F$ is separable.   Can you check my proof? Assume that $f$ is irreducible but inseparable. So we have $f=\sum_i h_ix^i$ and $f^p=\sum_i h_i^px^{ip}$. Now I use Frobenius mapping $\phi (f)=f^p=(\sum_i h_ix^i)^p=\sum_i h_i^px^{ip}$ so it is reducible. Contradiction. Should I also show for char(F)=0?
106715
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.    The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.  $0^2 + 0 + 4 = 4 \\ 1^2 + 1 + 4 = 6 \\ 2^2 + 2 + 4 = 10 \\ 3^2 + 3 + 4 = 5 \\ 4^2 + 4 + 4 = 2 \\ 5^2 + 5 + 4 = 1 \\ 6^2 + 6 + 4 = 2 \\ 7^2 + 7 + 4 = 5 \\ 8^2 + 8 + 4 = 10 \\ 9^2 + 9 + 4 = 6 \\ 10^2 + 10 + 4 = 4 $   But this process is very long one. I am seeking a short method by which I can solve such problem.   Now suppose that it is reducible, then the factors should be linear. So how can we solve the problem based on this kind of observation? Any help will be great. Thanks.
26125
I want to show that $p(x,y)=x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. There is already a similar problem posted:    but this question uses the irreducibility of $p(x)$ to show that it is prime and therefore an integral domain. The part of the problem I am stuck on is showing that $x^2+y^2-1$ is irreducible over $\mathbb{Q}$ in the first place.  I have considered taking the approach used in the above link except for backwards, but I am unsure of whether or not the converses of the theorems used still apply.  Edit: I see that this question has been marked as a duplicate, but I think I have explained how I am asking a different question than the one that was posted previously.
44204
Let $\pi \in \mathbb{Z}[i]$ irreducible and $\pi$ divides $p$ ($p\in\Bbb Z$ prime). Then show that $N(\pi)=p$.   ($N(\alpha)= a^2 +b^2$ where $\alpha = a+ib$.)
74901
I'm trying to show that $f(x)=1+x^p+x^{2p}+\dots +x^{p(p-1)}$ is irreducible over $\mathbb{Q}[X]$. I'm well aware that cyclotomic polynomials are irreducible, however the (many) proofs of this statement are quite involved. Since this exercise appears in a textbook that only covers basic algebra, there should be a more direct approach to this exercise.  I tried all criterions that I know to solve this (Eisenstein, linear substitutions, reducing to $\mathbb{Z}_p[X],\dots$). Only one remains. We know that $f(x)=\frac{1-x^{p^2}}{1-x^p}$, or equivalently $1-x^{p^2}=(1-x^p)f(x)$. We know that the roots of $1-x^{p^2}$ are given by $$e^{\frac{2\pi i k}{p^2}}, \mbox{ for $0\leq k\leq p^2-1$ }. $$ Similarly, the roots of $1-x^{p}$ are given by $$e^{\frac{2\pi i kp}{p^2}}, \mbox{ for $0\leq k \leq p-1$}.$$  Hence the roots of $f(x)$ are given by $e^{\frac{2\pi i k}{p^2}}$ where $p\not|  \;k$ and $0\leq k\leq p^2-1$. One way to show that $f(x)$ is irreducible over $\mathbb{Q}$ is to show that any product of linear factors of $f(x)$ is not a polynomial over $\mathbb{Q}$. I tried to show this, but it is too difficult.  Does anyone know a elementary way of solving this exercise?
16306
Let $\pi$ denote a prime element in $\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4, p$ is a prime.   I know that $\pi$ is prime in $\mathbb Z[i]$ implies it is irreducible. Also if $\pi$ has a prime norm, that is $N(\pi) = p$ then $\pi$ is irreducible.  For any $z \in \mathbb Z[i]$ we have $N(z) = a^2 + b^2$. Since $\pi \notin \mathbb Z, i \mathbb Z$ both $a$ and $b$ must be non-zero in the case of $\pi$. By Fermat's Two Square theorem every prime number $\equiv 1 \pmod 4$ can be written as a unique sum of two squares. I have some idea I should utilize this theorem here, but no success so far.  However I've come to a dead end. I don't know how to go on from here. Could someone help me out ?  Thanks.
29681
$$x^n-a$$  So $n$ is any integer greater than 1, and $a$ is any integer.  $a$ being any integer is where I am running into trouble.  I have already shown and worked out a proof for this being irreducible when a is prime.  Now I am just working through examples, trying to figure out a pattern.  The exercise is asking for application of Eisenstein's Criterion only, so I am assuming that the value of n should have no effect on the irreducibility.
49495
Are there infinitely many irreducibles in the ring of integers of any algebraic number field ?  I tried to follow the same argument as we usually do for integers. Suppose there are finitely many irreducibles, say $p_1,\ldots ,p_n$ and let $\alpha :=1+ p_1\cdots p_n$. Now if $\alpha $ is not a unit then it must have an irreducible $p$ such that $p|\alpha$ but then $p$ can not be any of the $p_i$'s and we have a contradiction. $\textit{What if $\alpha$ is an unit ? Is it possible for $\alpha$ to be an unit ?}$   Of course, one can replace $p_1\cdots p_n$ by $p_1^{k_1}\cdots p_n^{k_n}$ for any $k_1,\ldots ,k_n\in\mathbb{N}$ and the same argument would go through.
107600
Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ which is called the coordinate ring of the unit circle. I am trying to prove that $y$ is irreducible in $B.$ I have the following information about $B$: 1- $B$ is an integral domain. 2- $B = \mathbb R[x] \oplus \mathbb R[x]y$ and multiplication by $y$ is a map of $B$ as an $\mathbb R[x]$-module. And $y$ is integral over $\mathbb R[x]$ and so there is a ring inclusion $B \rightarrow M_2(\mathbb R[x]),$ called a representation of $B$ described below: $$\phi:B \rightarrow M_{2}(\mathbb R[x]), \phi (a_{1} + a_{2}y) =  \begin{pmatrix} a_{1} &amp; a_{2}(1-x^2) \\ a_{2} &amp; a_{1}  \end{pmatrix} $$ 3- Also, I know that $B^* = \mathbb R^*$ I am trying to use  Eisenstein criterion for showing that $y$ is irreducible. Here is my try: We need to find a prime $p$ such that $p$ divides $0$ because we have only 1 coefficient which is the leading coefficient and every other coefficients are $0$ and any prime can divide zero trivially. Also any prime does not divide 1. But now I do not know why $p^2$ does not divide $a_0$ which is zero in our case ...I need $p^2$ to divide $a_0$ to fullfill the conditions for Eisenstein Criterion. Could anyone explain this to me please? Also, is there any use for the fact that $y$ is integral in this proof?
69147
So we all know about Eisenstein's Criterion and how useful it turns out to be. In particular, if we have a monic cubic $x^3+ax^2+bx+c$, it is irreducible if there exists a prime number $p$ such that $p\mid a,b,c$ and $p^2\nmid c$.   But suppose that a monic cubic is irreducible, and all its coefficients are positive, but there does not exist such a prime. Sometimes we can find an integer $k$ such that we can substitute $x=y+k$ into the cubic, and the resulting polynomial in $y$ will satisfy the criterion. For example, if we have the cubic $x^3+3x^2+5x+1$, then no $p$ exists, but if we substitute $x=y+3$ into it, we end up with $y^3+12y^2+50y+70$, which, taking $p=2$, satisfies the criterion, hence it is irreducible. My question is whether or not this stronger version of the criterion applies to every irreducible polynomial. That is to say, if $x^3+ax^2+bx+c$ is irreducible, then we can always find some integer $k$ such that $(y+k)^3+a(y+k)^2+b(y+k)+c$, when simplified, will satisfy the criterion.   Thank you all in advance!
62474
I want to prove that if $p$ is a prime and $x,y \in \mathbb{Z}$, then $$(x+y)^p \equiv x^p+y^p \pmod{p}$$  So far I know that $$(x+y)^p = \sum_{k=0}^{p} \dbinom{p}k x^{p-k} y^k$$  A part of the above equation is supposed to cancel, I think, but I can't figure out a way how to make it cancel.
42333
How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?   For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$  gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$?
60574
If $p$ is an irreducible element of an integral domain $D$, and if $e$ belongs to $D^\ast$, prove that $ep$ is also ireducible .  To me it seems so profound, but I cannot get the proof . I'm trying to say that it is not irreducible, so there exists $a,b$ so that $ep=ab$ where $a,b$ don't belong to $D^\ast$....
16690
Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?  One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely  $ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$  where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.  This trick doesn't work for $n &lt; |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.  I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence.