Show that $h(x) = x^5 + x^2 +1$ is irreducible in $\mathbb{Z}[x]$.   I have tried applying Eisenstein's criterion to $h(x+1)$ but this didn't work How would I go about showing it is irreducible?
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!
71194
So, this is Hungerford problem 9 on page 166. Here's the problem in full:     Let $f(x) = \sum_{i=0}^n a_i x^i \in \mathbb{Z}[x]$. Suppose that for some $k$, $0 &lt; k &lt; n$, and some prime $p$ such that $p \nmid a_n$, $p \nmid a_k$, and $p \mid a_i$ for all $i = 0, \dots, k-1$, but $p^2 \nmid a_0$. Show that $f$ has a factor of degree at least $k$ that is irreducible in $\mathbb{Z}[x]$.   Here's my progress:  If we construct a new polynomial $g$ where $g$ is $\sum_{i=0}^k a_i x^i$ , we know by Eisenstein's criterion that $g$ is irreducible in $\mathbb{Q}[x]$. If we somehow knew that $g$ was irreducible in $\mathbb{Z}[x]$ too (that would follow if $g$ were primitive), then we could say that the smallest thing we could possibly factor out of $f$ that isn't a unit has to be of degree $k$ and that $g$ must be a factor of it. I tried writing $f$ as $C(f) \cdot f_1$ where $C(f)$ is the content of $f$, so that we'd be dealing with a primitive $f_1$ (and of course, the nice things about $g$ don't go away since $p$ doesn't divide every term), but this doesn't necessarily result in $g$ being primitive (right?)!  So, here's where I'm stuck: how can I show I can even factor $f$? How can I show that $g$ is irreducible in $\mathbb{Z}[x]$, whether by showing it's primitive or by some other means?  Thanks so much, everyone!
109631
So first of all, I used the following from my lecture notes:   If $f \in \mathbb{Z}[x]$ is primitive (gcd of all the coefficients is 1) - $f$ irreducible in $\mathbb{Z}[x] \Leftrightarrow$ f irreducible in $\mathbb{Q}[x]$.   So now I only have to show that $f_n = x^n + x + 3$ is irreducible in $\mathbb{Z}[x]$ for all $n \geq 2. $  Is there a pattern I should be appealing to in some way?
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$$x^n-a$$  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.
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Question:  If $p(x) \in F[x]$ is of degree $3$, and $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, show that $p(x)$ is irreducible over $F$ if there is no element $r\in F$ such that $a_0+a_1r+a_2r^2+a_3r^3 =0$.  My understanding: If $p(x)$ is reducible, then there exists $ax + b$ such that $a, b \in F$ and $a\ne 0$. And $p(x) = (ax + b)(cx^2 + dx + e)$. Then an $r$ exists such that $p(r) = 0$. Is that how you're supposed to tackle it? I understand that I am supposed to first explain the argument for a polynomial of degree 1, but I do not understand how to go about that, and why that helps.
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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?
91733
Let $p,q$ be primes, estimate the degree $[\Bbb Q(\sqrt[p]{2}\cdot\sqrt[q]{2}):\Bbb Q]$ and prove that the polynomial $X^q-2$ is irreducible in the ring $\Bbb Q(\sqrt[p]{2})[X]$  I found this problem on mathematical competition without any solution and it would be very interesting if we can find a proof for this.
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I proved that if $n$ is a prime, then $p(x)=x^{n-1}+\cdots+x+1$ is irreducible over $\mathbb Z$. But, I don't know how to prove that if $p(x)$ is irreducible over $\mathbb Z$, then $n$ is prime. Can you give me a hints?
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Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$.   This  to a related question mentions the result is due to Capelli.  I can prove the result if $F$ has characteristic $p$ as follows.  Suppose $f$ is reducible: $f(X)=g(X)h(X)$ with $g(X)$ an irreducible factor of degree $m$, $1\le m&lt;p$. Then if $\alpha$ is a root of $g$ in some extension field $K$ of $F$, we have $$f(X)=X^p-\alpha^p=(X-\alpha)^p$$ so its divisor $g(X)$ must be of the form $(X-\alpha)^m$. Since the coefficient of $X^{m-1}$ in $g$ is in $F$, we have $m\alpha\in F$.  So $\alpha\in F$ because $m$ is invertible modulo $p$.  How would you show the result in other characteristics?
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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
14967
I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks          Show that the polynomial $h(x) = (x-1)(x-2)\cdots(x-n) + 1$ is irreducible over $\mathbb Z$ for all $n\ge1$ and $ n\ne4$.      Suppose $h(x) = f(x) g(x)$, then we must have $f(i)g(i) = 1$ for all $i = 1,2,...n$. So both $f(i)$ and $g(i)$ are $1$ or $-1$. In either case, $m(x) = f(x) - g(x)$ has degree smaller than $n$ and have $n$ different roots ($1,2,...,n$). So we must have $m(x) = 0$. Then $h(x) = f(x)^{2}$. So $n$ must be even. Let $n = 2k$. Because $f(x)$ has degree $k$, there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.  This is where I got stuck. Hope some one can help me solve this. Thanks.
4072
How can I prove that $2+\sqrt{-5}$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$? I tried to show by $2+\sqrt{-5}=(a+b\sqrt{-5})(c+d\sqrt{-5})$ but I could not get a contradiction.
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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.
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A polynomial $p$ over a field $k$ is called irreducible if $p=fg$ for polynomials $f,g$ implies $f$ or $g$ are constant. One can consider the determinant of an $n\times n$ matrix to be a polynomial in $n^2$ variables. Does anyone know of a slick way to prove this polynomial is irreducible?  It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that $GL_n$ is the complement of a hypersurface in $M_{n}$. Thanks.
48542
Is there any criterion (theorem or algorithm) to find out a given polynomial $f(x)$ is irreducible over the field $\mathbb{F}_2$?
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Problem statement:     Show that $x^3+x+1$ is irreducible over $\mathbb{F}_2$ and let $\theta$ be a root. Compute the powers of $\theta$ in $\mathbb{F}_2$.   I am having trouble computing the powers of $\theta$. $x^3+x+1$ is irreducible, since any cubic polynomial over a field is reducible if and only if it has a root in the field. Since $\mathbb{F}_2=\{0,1\}$, we see that none of these is a root of $x^3+x+1$.  To compute the powers of $\theta$, note that $\theta^3+\theta+1=0$. Hence $\theta^3=-(\theta+1)$. Is this a good enough answer for $\theta^3$?  To compute $\theta^2$, note that $\theta = -(\theta^3+1)$. Therefore, $$\theta^2=(\theta^3+1)^2 = \theta^6 + 2\theta^3 + 1 = (-\theta-1)^2-2\theta-2+1,$$ which just reduces to $\theta^2$. Any ideas for computing these powers?  Thanks!
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?