Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that polynomial $f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) - a$ is irreducible in $\mathbb{Q}$, where $a \in - 2 + 5\mathbb{Z}$.

I have shown that $f$ doesn't have zeroes in $\mathbb{Q}$. We also know that since $f$ is primitive, by Gauss lemma, it is irreducible in $\mathbb{Q}$ iff it is irreducible in $\mathbb{Z}$.

I figured we could use $\pi :\mathbb{Z}\left[ X \right] \to \left( {\mathbb{Z}/5\mathbb{Z}} \right)\left[ X \right]$, show that it is a homomorphism and that irreducibility of $\pi \left( f \right)$ implies irreducibility of $f$.

But, assuming all these statements are true, which I haven't bothered yet to check, how can we see that $\left( {\pi \left( f \right)} \right)\left( x \right) = {x^5} + 4x + 2$ is irreducible in $\left( {\mathbb{Z}/5\mathbb{Z}} \right)\left[ X \right]$?

share|cite|improve this question
up vote 7 down vote accepted

In $\mathbb{F}_p[x]$, we have $\prod_{i=0}^{p-1} \left ( x-i \right ) = x^p - x$.

Now it is known that if $p \nmid a$ ,then $x^p - x + a$ is irreducible over $\mathbb{F}_p[x]$.

Proof: Let $\alpha \not\in \mathbb{F}_p$ be a root of $x^p - x + a$ over $\mathbb{F}_p$. Then all the elements $\alpha, \alpha + 1 ,\ldots, \alpha + p-1$ are roots of $x^p - x + a$ over $\mathbb{F}_p$.So $\mathbb{F}_p (\alpha)$ is the splitting field of $x^p -x + a$ over $\mathbb{F}_p$.Now its easy to see that the degree of the minimal polynomial of $\alpha$ over $\mathbb{F}_p$ divides $p$.(*) Since $p$ is a prime, degree of the minimal polynomial of $\alpha$ over $\mathbb{F}_p$ is $p$. (It can't be $1$ since $\alpha \not\in \mathbb{F}_p$.) This proves that the minimal polynomial is actually $x^p - x + a$. So it must be irreducible and we are done.

An elementary proof of it for the case $p=5$, can be found in

Edit: (*)

The degree of the minimal polynomial $g_0(x)$ of $\alpha$ over $\mathbb{F}_p[x] $ is equal to $[\mathbb{F_p}(\alpha) : \mathbb{F_p}] = n$ (let). Since $\mathbb{F_p}(\alpha) = \mathbb{F_p}(\alpha + t)$, minimal polynomial $g_k(x)$ of $\alpha + t$ also has degree $n$. Note that $g_k(x) | x^p - x + a$.So roots of $g_k(x) \in \left \{ \alpha, \ldots, \alpha + p-1 \right \} $. Also note that from the uniqueness of minimal polynomial $g_r(x)$ and $g_s(x)$ has no common root for $r \ne s$. So roots of $g_k(x)$ partitions $\left \{ \alpha, \ldots, \alpha + p-1 \right \} $ with $n$ elements in each class. So $n | p$.

share|cite|improve this answer
Thank you for the general proof, I'm reading it now and will accept your answer as soon as the option becomes available – Alen Nov 20 '12 at 13:26
How can we see that degree of the minimal polynomial divides $p$? – Alen Nov 20 '12 at 17:04
Hello, I have edited my answer with explanation. – Shubhodip Mondal Nov 20 '12 at 17:48
Suppose ${\mu _\alpha }\left( {\alpha + j} \right) = {\mu _{\alpha + k}}\left( {\alpha + j} \right) = 0$, then ${\mu _\alpha },{\mu _{\alpha + k}}$ are both irreducible monics over $\mathbb{Z}/p\mathbb{Z}$, have a common zero $\alpha + j$ and $\deg {\mu _\alpha } = \deg {\mu _{\alpha + j}} = \deg {\mu _{\alpha + k}} = n$ which implies ${\mu _\alpha } = {\mu _{\alpha + j}} = {\mu _{\alpha + k}}$. Thank you, everything is clear now – Alen Nov 20 '12 at 19:33
Yes. Actually in my answer it should be " ...For $r \ne s$, either $g_r(x) = g_s(x)$, or they have no common root." – Shubhodip Mondal Nov 20 '12 at 20:01

An alternative, possible, but really terrible, approach, is to prove that there no polynomials of the form $q(x)=x^2-Ax-B$ that divide $p(x)=x(x^2-1)(x^2-4)-a$. If we explicitly compute $p(x)\pmod{q(x)}$, we get: $$ (B^2+(3A^2-5)B+(A^2-1)(A^2-4))\,x+AB(2B+A^2-5)-a, $$ and four times the coefficient of $x$ is: $$ (2B+3A^2-5)^2-(5A^4-10A^2+9).$$ Since the last quantity must be zero, it is necessary that $(5A^4-10A^2+9)$ is a square, say: $$(\clubsuit)\qquad 5(A^2-1)^2 + 4 = Q^2.$$ Using the theory of Pell's equations we can write down the entire family of integer solutions to $$ 5X^2-Y^2 = -4 $$ and look for the solutions $(X,Y)$ in which $X+1$ is a square, in order to prove that the only integer solutions to $(\clubsuit)$ occur for $A=0,\pm 1,\pm 2,\pm 3$. Do not try this at home.

share|cite|improve this answer
Isn't the set of solutions to your Pell equation (a) infinite and (b) given by a fairly complicated recurrence relation/explicit formula involving quadratic surds raised to arbitrary powers? How do you show the non-squarehood of $X+1$ from there? (I appreciate that this is a How Not To Solve; I'm just wondering how you would even go from there...) – Steven Stadnicki Nov 20 '12 at 16:15
The key is that $5X^2-Y^2=-4$ is not a "generic" Pell equation, here the $X$ are the Fibonacci numbers with even index, so the problem is to find the natural numbers $m$ such that $F_{2m}+1$ is a square, that is not so terrible. – Jack D'Aurizio Nov 20 '12 at 16:22
Yes, I was hoping to avoid this approach in order to better understand the technique with $\mathbb{Z}/p\mathbb{Z}$ – Alen Nov 20 '12 at 17:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.