This question was asked in my abstract algebra quiz and professor gave answers  but my answer is wrong.  Question: Is $x^4+x^2 +1 =0$ irreducible in $\mathbb{Z}[x]$?  Root of $x^4+x^2 +1 =0$  doesn't exists in $\mathbb{Z}$ as it is always +ve and as $x^4+x^2 +1 =0$ is irreducible over $Z_2$ so it is irreducible over rationals and hence irreducible over integers. But answer is that it is reducible! Can you please tell what mistake I am making and what is right way to do?
103995
Alright, I met this example solved in a book, but I want to know how did they come up with the answer:  $$ (1+q^2+q^4) = (1+q^2+q)(1+q^2-q) $$  Can you also refer me some internet source where I can learn the rules of factoring. Thank you in advance.
6811
So here is my question: i would like to determine if whether or not the polynomial $x^5+x^2+1$ in $Z[x]$ is irreducible and if not then find the factors.  I tried a lot to find it. I really think it is already irreducible but I don't know how to prove it.  I tried Eisenstein's criteria which doesn't work here, I tried to check this criteria also after changing $x$ as $ax+b$ but everything I check was too complicated.  I tried for the modulus with prime numbers of $Z$, and I'm not sure of my following argument:  The polynomial $\mod 2$ stays $x^5+x^2+1$.  But if $x$ is odd then $x^5$ and $x^2$ are odd too. Then the polynomial is given by $1+1+1=3=1 \mod 2$ which is irreducible (not sure if I have the right to say that).  And if $x$ is even, then $x^5$ and $x^2$ are even too. The polynomial is thus given by $0+0+1=1 \mod 2$ which is irreducible.  Hence the polynomial is irreducible $\mod 2$ and since it's a monic polynomial it implies it is irreducible in $Z[x]$  Is my argument correct? If not if you have any clue please I would like to have some :)  Thank you and good night!
21199
I want to prove that     $p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible.   My attempt.   GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$ iff it's irreducible over $\mathbb Z$ (Gauss's lemma). Let $m\in \mathbb Z$ such that $\varphi(m)=n$ (Euler's totient). Make reduction of $p(x)$ by modulo $m$. Because of $\overline{x^n}=\overline{x^{\varphi(m)}}=\overline{1}$, we get $\overline{p(x)}=\overline{1-x-1}=\overline{-x}$, which is irreducible. Hence, $p(x)$ is irreducible.   Does this proof is correct?  UPDATE. Thanks to Calvin Lin. My mistake is: not for all $n$ we can find such $m$. OK, but as for the rest, does my proof is correct for such $n$, that $n=\varphi(m)$ for an integer $m$? And can it be some changed for all $n$, i.e. can we find such modulo that $\overline{p(x)}$ is irreducible for every $n\ge 2$?
63636
I have a question, I think it concerns with field theory.     Why the polynomial $$x^{p^n}-x+1$$ is irreducible in ${\mathbb{F}_p}$ only when $n=1$ or $n=p=2$?   Thanks in advance. It bothers me for several days.
94703
If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$.  Students and I are unsure how to go about this problem. Also will this be a problem I can solve and prove in front of a class in 20 minutes?
106784
Determine whether $X^4-16X^2+4$ is irreducible in $\mathbb{Q}[X]$.  To solve this problem, I reasoned that since $X^4-16X^2+4$ has no rational roots hence irreducible.  But there is a hint to this question that uses different approach: Try supposing it is reducible, then it must factor into a product of two monic quadratic polynomials with integer coefficients. Then show that it is impossible, then conclude the original polynomial is irreducible.  My questions regarding the hints: 1. Why can't we just show that since $X^4-16X^2+4$ has no rational roots hence irreducible? 2. Why do we have to factorise into a product of two monic quadratic polynomials with integer coefficients? Why monic? And why can't we factorise into polynomial with degree 1 and 3? Also, lastly why the coefficients have to be integers?  Thank you for any explanations!    Edit: Thanks for all the answers.  I saw that Second Gauss Lemma is used. But I only learned the first one in class. Which is: Let $R$ be a Unique Facorisation Domain. If $f,g\in\mathbb{R}[X]$ are primitive, then so too is their product $fg$.  Is it inevitable to use Second Gauss Lemma? Is there any other way around?
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Show that $p(\gamma) = \gamma^{4} + \gamma + 1$ is irreducible over $\textbf Z_2$. Okay, I would just like to see if i'm on the right track for this problem.  $$\gamma^{4}+\gamma+1=(\gamma +b)(c\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$$ So then $a = c = 1$.    So $$\gamma^{4}+\gamma+1=(\gamma +b)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$$  Let $b = 0$ $\gamma^{4} + \gamma+1=(\gamma)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$   Let $\gamma = 0$ Then we see $1 = 0$ which is a contradiction.  Let $b = 1$ $\gamma^{4}+\gamma+1=(\gamma+1)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$  Take $\gamma = 1$ And again $1=0$   So our polynomial is irreducible over $\textbf Z_2$  Is there another way I should be computing this? Maybe with Long division of polynomials?  I'm quite lost..
92545
This is an exercise for the book Abstract Algebra by Dummit and Foote (pg. 519): show $x^{5}-ax-1\in\mathbb{Z}[x]$ is irreducible for $a\neq-1,0,2$  I need help with this exercise, I don't have an idea on how to prove there are no quadratic factors for $a\neq-1,0,2$ (for $a=-1$ there is a quadratic factor).
112767
I have the following question:     Is there an easy way to prove that $x^6-72$ is irreducible over $\mathbb{Q}\ $?   I am trying to avoid reducing mod p and then having to calculate with some things like $(x^3+ax^2+bx+c)\cdot (x^3+dx^2+ex+f)$ and so on...  Thank you very much.
39809
Let $p$ be a prime. How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?   Right now I'm able to prove that it has no roots and that it is separable, but I have not a clue as to how to prove it is irreducible. Any ideas?
40285
I'm trying to show that the polynomial $X^n -2$ ($n \in \mathbb{N}$) is irreducible in $\mathbb{Q}[X]$ but am a bit stuck. Methods I know to show irreducibility:   Gauss' lemma - which says that if I can show it is irreducible in $\mathbb{Z}[X]$ then it will be irreducible in $\mathbb{Q}[X]$. This would allow me to check reduction modulo a prime.    However this doesn't work in this case, because if I reduce mod 2, then the polynomial is just $X^n$ which is reducible.  I'm also aware of Eisensteins criterion where I can check that if a prime divides all the coefficients but its square doesn't divide the constant coefficient, then it is irreducible.  None of these methods are working for this polynomial so help would be much appreciated! Also are there any general methods to look out for when trying to show polynomials are irreducible?  Thanks
33812
Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than $5$.   This is what I tried:  It's evident that $x,x+1$ are irreducible. Then, use these to find all reducible polynomials of degree 2. There ones that can't be made are irreducible. Then use these to make polynomials of degree 3, the ones that can't be made are irreducible. Repeat until degree 5.  Doing this way takes way too long and I just gave up during when I was about to reach degree 4 polynomials.  My question is: is there any easier way to find these polynomials?  P.S.: this is not exactly homework, but a question which I came across while studying for an exam.
47342
Is $x^3-9$ irreducible over the integers mod 31?  I know that one method is to check whether it has any roots or not, but that turns out to be very tedious in this case.  So, is there any other simpler method to find it out?
67525
Show that   $f_p (x^{p^{e-1}})$ is irreducible over $\mathbb{Q}$. I know it is obviously true for $ p=2$. Please suggest me any hints
216
Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ?    Our professor gave us already one, namely to substitute $X$ with $X+1$, but I couldn't make much of that.
71730
Show that $2$ | $x^{2n} - x^n$, $n \in \mathbb{Z^+}$   Let $x^n = a$ then $\frac{a}{2}$ $\implies a = 2q + r$ such that $r = 0 \vee r = 1$.   So $x^{2n} = (x^n)^2 = a^2 = (2q + r)^2$, which means that $2|[x^{2n} - x^n ]= 2| [(2q + r)^2 - (2q + r)]$. If $r = 1$ we get $(4q^2 + 4q + 1) - 2q - 1 = 4q^2 + 2q = 2(2q^2 +1)$. Thus $2$|$x^{2n} - x^n$, $n \in \mathbb{Z^+}$. Is this reasoning correct? Is it adequate?
101390
I want to solve:  Show that, in $\mathbb Q[x]$ (rationals), the polynomial $(x-1)^2$ divides a $zx^{z+1} – (z+1)x^{z}+1$, where $z \geq 1$ is any integer  I am really lost with this one, don't know how to aproach it