Let $f(x) = x^4 -10x^2 + 1$ a polynomial. Prove that $f$ has a factorization over $(\Bbb Z/p\Bbb Z)[X]$, $\forall$p prime number.  I have proved that this polynomial is irreducible over $\Bbb Z[X]$ using rational root theorem, but factor over $(\Bbb Z/2\Bbb Z)[X]$ because of $\overline 1$ is a root.  Now I'm not able to prove the generalization.
120574
So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with.  I managed to prove it for the primes such that $p \equiv 1, 7 \pmod 8$, by noting that $2$ is a square modulo those primes and thus $x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)$ for those $\mathbb{Z}_p$... however, trying to get a similar result for $3 \pmod 8$ and $5 \pmod 8$ has been stumping me for a long time, I am having a hard time making $q^2 = -1$ and $q^2 = -2$ give me something factorable...  I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it?
36429
Let $F$ be a field and $F[x]$ a polynomial ring. Let $p(x)$ be an irreducible polynomial.  Show that $\gcd(p(x),q(x)) = 1\Longrightarrow \exists r(x),s(x)$ such that $r(x)p(x)+s(x)q(x) = 1$.  I know the proof for Bezout's identity for , but this proof uses the notion of absolute value, which cannot be applied to a polynomial ring.   I am thinking about a way to prove the statement: Use the Euclidean algorithm to compute $\gcd(p(x),q(x))$ and show that the result is $1$, and somehow go backwards to obtain $r(x)p(x)+s(x)q(x) = 1$, but it is a bit messy, and I'm not yet able to see it completely.  Please offer some help. If you have some other methods, that would be great, too.
113379
I must find the irreducible factors of $f(x) = x^8 - x$ in $Z/2Z[x]$ and that's what I did:  $f(x) = x(x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$  but of course the correct solution is:  $f(x) = x(x-1)(x^3 + x^2 + 1)(x^3 + x + 1)$  I do not understand how I can reach the correct solution by logical reasoning (without going on attempts). The solution of the exercise says:  1) "The irreducible factors are all the irreducible polynomials of degree 1 and 3 alone" Why?  2) "You find exactly $6/3 = 2$ irreducible polynomials of degree 3". Does this happens because the 6th degree polynomial I found is not reducible (since the only ones have degree 1 or 3) and it has no zeros?
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 ?
40010
Let $\mathbb{F}_2 \cong \mathbb{Z}/2\mathbb{Z}$ denote the field of 2 elements   (a) Is $x^4+x^2+1$ irreducible in $\mathbb{F}_2[x]$? Find a complete factorization.  (b) How many irreducible polynomials of degree 4 are there in $\mathbb{F}_2[x]$?  For the first part of the problem, I just plug in $0$ and $1$ into $x^4+x^2+1$, and know that there is no root for that polynomial, so this polynomial is irreducible.   For the second part, I just list polynomials of degree 4 and having no zero in $\mathbb{F}_2[x]$, which are $x^4 + x+1, x^4 +x^2 +1, x^4 + x^3 +1$ , and $x^4 + x^3 +x^2 + x +1$.   But I am not sure about whether these approaches are right, so could you help me to figure it out? Thank you!
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
100951
I'm trying to show that the minimal polynomial of a linear transformation $T:V \to V$ over some field $k$ has the same irreducible factors as the characteristic polynomial of $T$. So if $m = {f_1}^{m_1} ... {f_n}^{m_n}$ then $\chi = {f_1}^{d_1} ... {f_n}^{d_n}$ with $f_i$ irreducible and $m_i \le d_i$.  Now I've managed to prove this using the primary decomposition theorem and then restricting $T$ to $ker({f_i}^{m_i})$ and then using the fact that the minimal polynomial must divide the characteristic polynomial (Cayley-Hamilton) and then the irreducibility of $f_i$ gives us the result.  However I would like to be able to prove this directly using facts about polynomials/fields without relying on the primary decomposition theorem for vector spaces. Is this fact about polynomials true in general?  We know that $m$ divides $\chi$  and so certainly $\chi = {f_1}^{m_1} ... {f_n}^{m_n} \times g$ but then how do we show that $g$ must have only $f_i$ as it's factors? I'm guessing I need to use the fact that they share the same roots. And I'm also guessing that it depends on $k$, i.e. if $k$ is algebraically closed then it is easy because the polynomials split completely into linear factors.  Help is much appreciated, Thanks
91399
let $f$ be a polynomial function such that $f(\mathbb{Q})\subset \mathbb{Q}$ and $f(\mathbb{R}-\mathbb{Q}) \subset \mathbb{R}- \mathbb{Q}$ . Prove that $f(x)=ax+b$ for some $a,b\in\mathbb{Q}$   My approach :  I started by defining $g(x)=f(x)-f(0)$ .   Now $g(x)$ satisfies the properties $g(\mathbb{Q})\subset \mathbb{Q}$ ,  $g(\mathbb{R}-\mathbb{Q})\subset\mathbb{R}-\mathbb{Q}$ and $g(0)=0$ . $g$ is a polynomial and $g(0)=0$ . $0$ is root of $g$ and hence $g(x)=x h(x)$ for some polynomial $h(x)$ . It will be enough if one can prove that $h(x)$ is constant . I'm stuck here .   Any form help will be highly appreciated.
26962
Let $A$ be a field and let $p(X)$ be a polynomial over $A$. Let $a\in A$.   Want to show:     $p(X)$ is irreducible if and only if $p(X+a)$ is irreducible.    I suspect that I should use the substitution principle somehow, but that's as far as I've come. Completely stumped.
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.
21335
Let $f(x)$ and $g(x)$ be irreducible polynomials over a field $F$ and let $a,b \in E$ where $E$ is some extension of $F$. If $a$ is a zero of $f(x)$ and $b$ is a zero of $g(x)$, show that $f(x)$ is irreducible over $F(b)$ if and only if $g(x)$ is irreducible over $F(a)$.   Attempt:  Since $f(x),g(x)$ are irreducible over $F \implies a,b \notin F$.  $f(x)$ is irreducible over $F(b) $ and $f(x)$ is irreducible over $F \implies a \neq b$ (As, $a$ is the zero of $f(x)$)  Which means $b \notin F(a)$ either $\implies g(x)$ is irreducible over $F(a)$.  Similarly, the other half can be proved in a similar way.  Is my solution attempt correct?  Thank you for your help..
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.
89703
Question is :     Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]   What I have done so far is :  Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :  $p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$  i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.  so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$  I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(  Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$   i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$   $p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$ $p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$  $p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$    Once I prove that those are the only divisors then I would consider :    $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$ $p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$ $p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$ $p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$ $p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$ $p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$   So, no divisor of $2$ is a root..  Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$  suppose I have something like :  $$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$  But then, $bd=2\Rightarrow \text { b,d  are  a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$  But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  So, only possibilities are   $x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$ $x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$   But these are not possible...  This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...  So, please help me to clear this..  Thank you...   P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain.
44175
Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$.   Is it possible to prove this without using induction on degree? If so, how can I do this?
68892
This is what I'm trying to prove:     Let $f(x)\in\mathbb{Q}[x]$ and $\deg(f(x))&gt;1$. Prove that $f(x)$ is irreducible in $\mathbb{Q}[x]$ iff its reciprocal polynomial $f^*(x)$ is irreducible in $\mathbb{Q}[x]$.    Note: The reciprocal polynomial $f^*(x)=x^nf(1/x) \in \mathbb{Q}[x]$, where $n=\deg f$.   So my thoughts are to prove the contrapositive, i.e. $f^*(x)$ is reducible iff $f(x)$ is reducible.   So to prove the ($\Rightarrow$) direction I assume that $f^*(x)=g(x)h(x)$, so I get that $g(x)h(x)=x^nf(1/x)$ which implies $f(1/x)=(1/x^n)g(x)h(x)$. I want to say this somehow makes $f(x)$ reducible but I am unable to proceed as $f(1/x)\notin \mathbb{Q}[x] $. I thought about substituting $y=1/x$, which gives $f(y)=y^ng(1/y)h(1/y)$, but I am unable to show that this is irreducible in $\mathbb{Q}[y]$. If I could, then it would be irreducible in $\mathbb{Q}[x]$ since $\mathbb{Q}[y]\cong \mathbb{Q}[x]$   Any suggestions or hints will be appreciated.
98034
I'm trying to understand the following explicit construction of a finite field with $p^n$ elements where $p$ is prime, $n \geq 2$:  Take any irreducible polynomial $f(X) \in \mathbb{F}_p$ of degree $n$. Then $\mathbb{F}_p[X] / (f(X))$ is a field with $p^n$ elements.  Now, I understand that if $f$ is irreducible, the ideal generated by $f$ is maximal, so $\mathbb{F}_p[X] / (f(X))$ is indeed a field.  I don't understand the following points:   Why does for every $n \geq 2$ an irreducible $f(X) \in \mathbb{F}_p[X]$ of degree $n$ exist? Why does the field $\mathbb{F}_p[X] / (f(X))$ have $p^n$ elements?   Thanks in advance for any help!
61503
Problem: Is there an irreducible $f\in \mathbb{Z}[x]$, whose image in every $(\mathbb{Z}/p\mathbb{Z})[x]$ has a root for $p$ prime? If there is, what is the minimal degree possible?   I can only prove that $x^2-c$ is impossible, by quadratic reciprocity and Chinese remainder theorem. Even the case of $a x^2 - c$ is unknown to me.  Meanwhile, if $f$ is not required  to be irreducible, but only have no root in $\mathbb{Z}$, then $(x^2-a)(x^2-b)(x^2-ab)$ is a solution for non-sqaure $a,b,ab$, since if both $a,b$ are not squares in $\mathbb{Z}/p\mathbb{Z}$, then $ab$ is a square. Yet I still don't know if this is of minimal degree.  Also, it is natural to pose generalizations   When $p$ is not necessarily prime (equivalently for all prime powers). When $p$ is odd prime. When $p$ represents sufficiently large primes (equivalently all but a finite number left).