Question. Prove that $f(x)=x^{5}-x+a$ is irreducible in $\Bbb Z[x]$ if $5\nmid a$.  My approach. If we let $f(x)=(x+a_0)(x^4+b_3x^3+b_2x^2+b_1x+b_0)$, we know that $a=a^5-a$. So $5 \mid a$ and it's contradictory to condition.  But can we let $f(x)=(x^2+a_1x+a_0)(x^3+b_2x^2+b_1x+b_0)$? I can't prove we can't. I've used Eisenstein's criterion, but it was helpless. I couldn't use it anywhere. Can somebody help me? (I don't want a generalized proof.)
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?
68768
Show that $2,3, 1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]$, but not prime and that 1 and -1 are the only units.   Let $N$ be the norm map into $\mathbb{Z}$ and let u denote the unit, then because it is a homomorphism it follows that :   $N(xy)=N(x)N(y)$ and thus with $N(u)N(u^{-1})=N(uu^{-1})=N(1)$ it follows that the norm of u is a divider of 1 and thus also a unit of $\mathbb{Z}$. Therefore we have in $\mathbb{Z}[\sqrt{-5}]$: $\forall a,b \in \mathbb{Z}: N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^{2}+5b^{2}=1 \ \ \Rightarrow u= \pm 1$  Now it is to show that $2,3,1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]: $  Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$ But since $1-\sqrt{-5}$ is not a quadratic remainder of $5$, there doesn't exist a solution for the equations $a^{2}+5b^{2}= \pm (1-\sqrt{-5})$ And the exactly same argument also works for $2,3$ and $1+\sqrt{-5}$.  Thus we have shown that $2,3,1\pm \sqrt{-5}$ are not reducible over $\mathbb{Z}[\sqrt{-5}]$.   Now we show that they are not prime:   Assume that 2 is a prime in $\mathbb{Z}[\sqrt{-5}]$, then because of $2\cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5})=6$ it must hold that $2|(1-\sqrt{-5})$ or $2|(1+\sqrt{-5})$. But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :   $(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.   Assume that 3 is a prime in $\mathbb{Z}[\sqrt{-5}$, then: $(1\pm \sqrt{-5}) = 3(a+b\sqrt{-5}) \Rightarrow 3b= \pm 1$ it follows that 3 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.  Assume that $1\pm \sqrt{-5}$ is a prime in $\mathbb{Z}[\sqrt{-5}]$, then: $3 = (1\pm \sqrt{-5})(a+b\sqrt{-5}) \Rightarrow (1\pm \sqrt{-5} )a = 3$ which is not solvable in $\mathbb{Z}$. And also $2=(1\pm \sqrt{-5})(a+b\sqrt{-5})=(1\pm \sqrt{-5})a$ which is also not solvable in $\mathbb{Z}$ and thus $(1\pm \sqrt{-5})$ can not be a prime in $\mathbb{Z}[\sqrt{-5}]$.   Tell me if this proof is correct. Please.
25044
I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I'm trying to construct a polynomial with root $\sqrt{2} + \sqrt[3]{5}$ and coefficients in $\mathbb Q$ of degree $6$, and then show that it is irreducible over $\mathbb Q$.   I managed to find that it is a root of the polynomial $x^6 - 6x^4 - 10x^3 +12x^2 - 60x +17$. This is where I run in to some problems. I don't know how to show that this is irreducible over $\mathbb Q$ (or $\mathbb Z$). The only criteria we have learned is Eisenstein's criteria, which clearly does not apply here. How else can I show that the degree of this number over $\mathbb Q$ is $6$?
18769
Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?
25461
Suppose that $a_1,a_2, \cdots, a_n$ are $n$ different integers. Then $[(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1$ is irreducible over $\mathbb Q$.   I've no idea why it is true. Thanks very much.
113998
I'm having a hard time showing this:     If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.   I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.  Any help would be appreciated.
76053
Given that  $f(x)=\sum\limits_{k=0}^n a_kx^k$ is an irreducible polynomial, show that $g(x)=\sum\limits_{k=0}^n a_kx^{n-k}$ is irreducible too. I don't see how this specific coefficient exchange would preserve the irreducibility. Is there something else I should be looking at?
22569
Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:     Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field  and let $c \in  F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$   has no root  in $F$. [Hint: consider two cases: char$(F) = p$ and   char$(F)$ different of $p$.]   I have attempted this a lot. Anyone has an answer?
84054
Two part question:  1) There are three irreducible cubic polynomials $\mathbb F_2$  Aproach: Bruteforce, there are not as much cubic polynomials, just 8 of them:  $x^3$  $x^3+1$  $x^3+x$  $x^3+x+1$  $x^3+x^2$  $x^3+x^2+1$  $x^3+x^2+x$  $x^3+x^2+x+1$  Only three of them are irreducible(??): $x^3+1$,$x^3+x+1$ and $x^3+x^2+1$  Reality check - do i miss point of irreducible polynomial here, and should consider adding $x^3+x^2+x+1$ to irreducible list? It is reducible, bit different way then other polynomials: $x^3+x^2+x+1=(x+1)^3$  2) Only irreducible polynomial of degree 2 over $\mathbb F_2$ is $x^2+x+1$. Does it help, with conjuction of part 1, to find all irreducible polynomials of degree 5?
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!
108313
I'm trying to find a necessary and sufficient condition for $x^n - y^m$ to be irreducible in $\Bbb C[x,y]$.  What I have now is rather trivial sufficient conditions: if $m\mid n$ then $m=1$ and if $n\mid m$ then $n = 1$.  There are similar questions previously posted to this site. To me the answers to those questions seem overkill to this particular problem, and I am looking for an elementary solution.  I would be grateful for your help.
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.
81764
Let $f(x) \in \mathbb{Z}[x]$. If we reduce the coefficents of $f(x)$ modulo $p$, where $p$ is prime, we get a polynomial $f^*(x) \in \mathbb{F}_p[x]$. Then if $f^*(x)$ is irreducible and has the same degree as $f(x)$, the polynomial $f(x)$ is irreducible.   This is one way to show that a polynomial in $\mathbb{Z}[x]$ is irreducible, but it does not always work. There are polynomials which are irreducible in $\mathbb{Z}[x]$ yet factor in $\mathbb{F}_p[x]$ for every prime $p$. The only examples I know are $x^4 + 1$ and $x^4 - 10x^2 + 1$.  I'd like to see more examples, in particular an infinite family of polynomials like this would be interesting. How does one go about finding them? Has anyone ever attempted classifying all polynomials in $\mathbb{Z}[x]$ with this property?
15942
Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$.   How to prove? I really have no idea... Thank you a lot.
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.
75159
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.
96390
I know that $\mathbb Z[\sqrt{7}]$ is a UFD, and I can write the equation $(2 + \sqrt{7})(3 - 2\sqrt{7}) = (5 - 2\sqrt{7})(18 + 7\sqrt{7}) $. So clearly these are not all irreducibles. How do I determine which of these aren't irreducible? How do I factor those into irreducibles?  It seems difficult when dealing with $\mathbb Z [\sqrt{d}] $ with $d$ positive, since the norm map isn't as nice (it's not always positive).  Thanks