Consider the polynomial $P=X^5-X-1\in\Bbb{F}_3[X]$. I want to show that $P$ is irreducible. We can easily check it has no roots, so the only way it could not be irreducible is by being a product of two polynomials of degree $2$ and $3$ respectively. So I determine all the irreducible polynomials of degree $2$ in $\Bbb{F}_3[X]$. These are $$X^2+1;{~~} X^2+X-1 {~~}\text{and}{~~}X^2-X-1.$$ Finally I check using the Euclidean division algorithm that none of those polynomials divides $P$, which concludes the proof.  My question is this:  is there a more efficient way to do this? I had to first determine all the polynomials of a given degree ($2$) and then apply Euclidean algorithm to each of them. That is quite some work, and we're still considering polynomials of relatively small degree, and fields of small cardinal. I can't imagine applying the same reasoning to determine whether $Q=X^9-X^2-1\in\Bbb{F}_{17}[X]$ is irreducible.  So, is there a more efficient method to determine irreducibility, at least in the case of polynomials of small degree over small fields?
123169
Using the affine linearized polynomial $f(x)=x^p-x-b$ is irreducible over $\mathbb{F}_q$ if and only if $\operatorname{Tr}_{\mathbb{F}_q/\mathbb{F}_p} \neq 0$. But here $\mathbb{F}_3$ is not a field extension of $\mathbb{F}_5$.  Also we have that: Let $a \in \mathbb{F}_q$ and let $p$ be the characteristic of $\mathbb{F}_q$. Then the trinomial $x^P - x - a$ is irreducible in $\mathbb{F}_q[x]$ if and only if it has no root in $\mathbb{F}_q$.
120366
this is a really easy question but I cant find an answer;  In need to see if $x^4+x^2+x+1$ is an irreducible polynomial over $\mathbb{Q}$
47441
Is there a general or elegant way to approach this problem? One can show that $\sqrt 5+\sqrt[3] 2$ is a root of the hexic $x^6-6x^4-10x^3+12x^2-60x+17$, which should then be its minimal polynomial to get the result. I see no straightforward way to show this is irreducible, though. It has a root in $\mathbf{F}_4$, and to show it's irreducible over $\mathbf{F}_3$ already seems to require a lot of grunt work.  One thing I can do is rule out the possibility that $[\mathbf{Q}[\sqrt 5+\sqrt[3] 2]:\mathbf{Q}]=2$, because $[\mathbf{Q}(\sqrt 5,\sqrt[3] 2):\mathbf{Q}]=6$ and $\mathbf{Q}(\sqrt 5,\sqrt[3] 2)$ is the compositum of $\mathbf{Q}[\sqrt 5+\sqrt[3] 2]$ and $\mathbf{Q}[\sqrt 5]$, where the latter has degree two. But I don't seem to be able to play an analogous trick to rule out $3$, because I know of no general lower bound on the degree of a compositum.   Could there be some way to bring Galois theory into the picture? These extension aren't Galois, and so it's by no means clear to me how to reason from their normal closures, but perhaps that's the direction to go in. Suggestions are very much appreciated.
87223
Assume that $F$ is a field and $\operatorname{char}(F)=p$. Let $a$ be an element in $F$ without $p$th root, then the polynomial $$x^{p^n}-a$$ is irreducible and inseparable over $F$ for all $n$.    I have proved the inseparable part by considering the derivative of the polynomial, but I'm having trouble with the irreducible part. Any help?
41057
Find all primes $p&gt;2$ for which $x^2+x+1$ is irreducible in $\mathbb{F}_p[x]$   Attempt. Since $x^2+x+1$ is of degree 2, it is reducible iff it has a root in $\mathbb{F}_p$. It has a root in $\mathbb{F}_p$ iff $x^3-1=(x-1)(x^2+x+1)$ has a root other than 1. The latter has a root other than $1$ iff $\mathbb{F}_p$ has elements of order $3$, which happens iff $3\mid |\mathbb{F}_p^{\times}|=p-1$, which happens iff $p\equiv 1\bmod 3$. So $x^2+x+1$ is irreducible iff $p\not \equiv 1\bmod3 $.   I am getting a different answer than my professor, who has $p\equiv 1\bmod 3$. If I am incorrect, were am I going wrong?  Update. Now I am unsure about my own reasoning. What if $1$ is a root of $x^2+x+1$? I don't know why I am considering roots of $x^2+x+1$ other than $1$ when $1$ can be a root.
54804
I want to prove that for any set of distinct integers $a_1,\ldots,a_n$, the polynomial $$h = (x-a_1)\cdots(x-a_n) + 1$$ is irreducible over the field $\mathbb{Q}$, except for the following special cases which are reducible:  $$\left.\begin{cases} a_1 = a\\ a_2 = a+2 \end{cases}\right\} \implies h = (x-a-1)^2$$  and $$\left.\begin{cases} a_1 = a\\ a_2 = a+1\\ a_3 = a+2\\ a_4 = a+3 \end{cases}\right\} \implies h = ((x-a-1)(x-a-2)-1)^2$$
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?
73197
Question is to prove :   Irreducibility of $(x-1)(x-2)\cdots (x-n)- 1$ over  $\mathbb{Z}$ for all $n\geq 1$ Irreducibility of $(x-1)(x-2)\cdots (x-n)+ 1$ over  $\mathbb{Z}$ for all $n\geq 1$ and $n\neq 4$   Hint for first bullet is      If the polynomial factors consider the value of the factors at $x=1,2,\dots,n$   For second bullet :  Suppose $p(x)=(x-1)(x-2)\cdots (x-n)-1$ is reducible we have:  $p(x)=q(x)r(x)$ with $\text {Max {degree of p(x), degree of r(x)}}&lt;n$  Hint is suggesting me to use that $p(i)=-1$ for all $1\leq i\leq n$   i.e., $q(i)r(i)=-1$ for all $1\leq i\leq n$  i.e., $q(i)=-1; r(i)=1$ or $q(i)=1;r(i)=-1$  for all $1\leq i\leq n$  For second bullet :  Suppose $p(x)=(x-1)(x-2)\cdots (x-n)+1$ is reducible we have:  $p(x)=q(x)r(x)$ with $\text {Max {degree of p(x), degree of r(x)}}&lt;n$  Hint is suggesting me to use that $p(i)=1$ for all $1\leq i\leq n$   i.e., $q(i)r(i)=1$ for all $1\leq i\leq n$  i.e., $q(i)=r(i)=1$ or $q(i)=r(i)=-1$  for all $1\leq i\leq n$  I am getting some vague ideas but could not bind them to prove this.  I would be thankful if some one can help me to clear this.  Thank you.  P.S : Please give "just hints". Do not write whole answer at once. This is a "request". Thank you :)  Edit : I have changed the title from   Irreducibility of $(x-1)(x-2)\cdots (x-n)\pm 1$ over $\mathbb{Z}$  to   Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree  for two reasons :   Irreducibility of $(x-1)(x-2)\cdots (x-n)\pm 1$ over $\mathbb{Z}$ for any $n\geq 1$ implies Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree The title looks atractive
79858
Prove that $P_n(X) = X^n - X^{n-1} - X^{n-2} - ... - X - 1$ is irreducible over $\mathbb{Z}$ for all $n$.  I was able to prove the result for $n=2^k-1$ by applying  to $P_n(X+1)$. But for other values of $n$, I'm stuck. Has anyone an idea on this ?
53386
I have to find $P(0)$ from the polynomial with minimum degree given that  $$(x-1)^3|(P(x)+1)$$  $$(x+1)^3|(P(x)-1)$$  Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively:  $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$ $$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$  Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$  and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.
115644
I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field.  seems to show that irreducible polynomials over $\Bbb F_p$ have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me. Does the result hold for $\Bbb F_q$ where $q$ is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots? Please help.
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!
20598
Determine the number of irreducible monic polynomials of degree 3 in $\mathbb F_p[x]$ where $p$ is a prime.  I'd like to start off by acknowledging that I know there are many posts relating to similar problems both on here and mathoverflow, however none address my problem.  In class, my professor went over the monic polynomials of degree 2 by reasoning that there are $p^2$ distinct polynomials, the reducible ones look like $(x - a)^2$ and $(x - a)(x - b)$ for the first one there are $p$ of them and for the second one there are $p(p-1)/2$ of them, subtracting the reducible ones from the total ones gives $$p^2 - p - \frac{p(p - 1)}{2} = \frac{2p^2 - 2p - p(p-1)}{2} = \frac{2p^2 - 2p - p^2 + p}{2} = \frac{p^2 - p}{2}.$$  I'm trying to use similar reasoning for polynomials of degree three using the fact that there are $p$ irreducible monic linear polynomials and $p(p-1)/2$ irreducible monic quadratic polynomials. But I can never get the result shown  and .  My reasoning is as follows: Observe that reducible polynomials would look like $(x - a)(x - b)(x - c)$, $(x - a)^2(x - b)$, $(x - c)^3$ and $(x^2 + ax + b)(x - c)$ formed from irreducible linear and quadratic monic polynomials.    For the first type where all three are distinct we can surmise that there are $p$ choices for the first, $p-1$ choices for the second and $p - 2$ choices for the third. So we have $p(p-1)(p-2)$. (is it $p$ choose $3$? Do I have to account for order not mattering?) For the second type there are $p$ choices for the square, and $p-1$ choices for the linear polynomial, so we have $p(p-1)/2$. For the third type, we have shown that there are $p(p-1)/2$ choices for the quadratic polynomials and there are $p$ choices for the linear polynomial. So we have $p^2(p-1)/2$.   Subtracting these values from $p^3$ should give me what I'm looking for, which is $p(p^2 - 1)/3$. But I'm not getting that.  Can someone tell me where my reasoning is wrong? Or tell me how to go about counting the number of irreducibles? (I know there are other methods of getting the answer, even more general ones, but I'm looking for a simple counting method that can help me better understand where my reasoning is wrong and help me correct it.)
105931
Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result:    For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.   My thoughts so far: The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:   For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof. For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case?
36969
So I am trying to show that $f(x) = x^4-3$ will factor over $\Bbb{F}_p$ when $p\equiv 3\mod 4$, with $p\neq 3$. Specifically, I am trying to show that $f$ will factor into irreducibles such that there is at least one factor of degree 2.  My Work So Far  So if I assume that it can be factored, we would have $$ x^4-3 = (x^2+ax+b)(x^2-ax+d) = x^4 + (b+d-a^2)x^2 + a(d-b)x + bd.$$ Because we want the linear term to be 0, we need either $a=0$ or $d-b = 0$. Now if $a=0$ we get that $b=-d$, so $-b^2 = -3$. That is, $3$ is a square. If $d-b = 0$, then we would have $b^2 = -3$, so $-3$ is a square. Because $-1$ is not a square when $p\equiv 3\mod{4}$, we know that one of $-3$ or $3$ must be a square, but not both, so $a=0$ or $d=b$, but not both.  If $a=0$, that is $3$ is a square, then we have $f(x) = (x^2 +b)(x^2-b)$. Then one of $b$ or $-b$ is a square, but not the other, so $f$ will factor into two linear terms and a quadratic. So this case works.  The trouble I am having is with the second case. If $d=b$, then we have  $$ f(x) = (x^2 + ax + b)(x^2 -ax +b)$$ with $2b = a^2$. This is where I get stuck. How do I know that such an $a$ can exist?  Edit: If there is a better approach to solving this, that would be great to know as well. I haven't been taught any of the relevant number theory to solve this, so I'm kind of out of my element here.
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 $n\ge 2$?
73277
What algorithm can be used find an irreducible polynomial of degree $n$ over the field $GF(p)$ for prime $p$. The reason I ask is I want to make a program for finite field arithmetic, but creating a field $GF(p^k)$ requires finding a irreducible polynomial. (I'm mostly interested in $GF(2^k)$.)