# Show that a polynomial is irreducible in $\mathbb{Q}$

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 $\left( {\mathbb{Z}/5\mathbb{Z}} \right)\left[ X \right]$?

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## 2 Answers

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 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450995#p2450995.

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 $n | p$.

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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.

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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