# Irreducibility of a particular polynomial

I've got this problem for my homework: find out whether the polynomial $$f(x)=x(x-1)(x-2)(x-3)(x-4) - a$$ is irreducible over the rationals, where $a$ is integer which is congruent to $3$ modulo $5$.

It is easy to verify that $f(x)$ has no integer zeros (and no rational zeros too) and, because $f(x)$ is primitive, it is irreducible over $\mathbb{Z}$ iff it is irreducible over $\mathbb{Q}$. It is also clear to me that, since $f(x)$ has no integer zeros, only way to factorize $f(x)$ over $\mathbb{Z}$ is $(x^3 + bx^2 + cx +d)(x^2 + ex +f)$, i.e. as a product of one irreducible cubic polynomial and one irreducible quadratic polynomial, both primitive. I've got pretty ugly system of $5$ equations with $5$ variables. So I decided to put the whole story in $(\mathbb{Z}/5\mathbb{Z})[x]$. Then I've got the polynomial $g(x) = x^5 + 4x +2$. Since I firmly believe that $f(x)$ is irreducible over $\mathbb{Z}$, it means that I must check whether the $g(x)$ is irreducible over $\mathbb{Z}/5\mathbb{Z}$. Is there any other way than brute force?

## 3 Answers

Observation: In ${\mathbb F}_5$ we have $x(x-1)(x-2)(x-3)(x-4) = x^5 - x$.

This follows from the fact that over a finite field $F$ we have $x^{|F|} - x = \prod_{g\in F} (x-g)$ which you can easily check by noticing that all elements are roots and that a polynomial of degree $|F|$ can't have any more roots.

Claim: More generally, we will prove that given a prime $p$ and $t\in{\mathbb F}_p\setminus \{0\}$ we have that $f(x)= \prod_{k\in{\mathbb F}_p} (x-k) - t = x^p - x - t \in{\mathbb F_p}[x]$ is irreducible over ${\mathbb F}_p$.

Proof. Let $\alpha$ be a root of $f$ over some extension field.

Note that more generally over a finite field ${\mathbb F}_q$ we have that, the monic irreducible polynomial $m_\alpha (x)$ for $\alpha$ over ${\mathbb F}_q$, satisfies

$$m_\alpha (x) = \prod_{k=0}^{n-1} (x-\alpha^{q^k})$$

where $n$ is the smallest positive integer such that $\alpha^{q^n} = \alpha$.

This is a well-known result in finite fields. Without getting into too many details, this follows from the fact that $\text{Gal}\left({\mathbb F}_{q^n}/{\mathbb F}_q\right)$ is cyclic, generated by the Frobenius map $x\mapsto x^q$. Thus the roots above are actually the Galois conjugates of $\alpha$, and the result follows from a well-known theorem in Galois Theory.

We need only check now that in our case, with $p=q$, and $f(\alpha)=0$, our $n$ must be $p$. Indeed, since $f(\alpha) = 0$, we have

$$\alpha^{p^n} = (\alpha^p)^{p^{n-1}} = \left( \alpha + t \right)^{p^{n-1}} = \alpha^{p^{n-1}} + t^{p^{n-1}} =\alpha^{p^{n-1}} + t = \ldots = \alpha + n \times t$$

Since $t\neq 0$, the smallest such $n$ is clearly $p$, since otherwise $n\times t \neq 0$. Thus the number of linear factors (Galois conjugates) in $m_\alpha(x)$ is $p$ and we get that our polynomial was irreducible. $\square$

Conclusion: This proves that, given a prime $p$, the polynomial $g_p(x) = x\cdot (x-1)\ldots (x-p+1) - a \in {\mathbb Z}[x]$, with $p\not | a$, must be irreducible over ${\mathbb Q}$.

Indeed in ${\mathbb F}_p$ we have that $g_p(x)$ reduces to $f(x) = x^p - x - \bar{a}$ and $\bar{a}=t\neq 0$.

We may also substitute $k$ in the factors $(x-k)$ in the product by any element $\equiv k(\bmod. p)$.

A link: There has recently been a related post, which is more general, and incredibly interesting.

Final remark: When it comes to Finite Fields, the above result on irreducible polynomials is pretty often a good way to prove that a given polynomial is irreducible.

Note that $f(x)=-f(4-x)$ and since the factorization $f(x)=q(x)r(x)$ is unique, we conclude $q(x)=q(4-x)$ for the quadratic, $r(x)=-r(4-x)$ for the cubic factor. Thus $e=-4$. Then from $e+b=-10$, we have $b=-6$. Then from $f+eb+c=35$, we have $f+c = -5$. And from $fb+ec+d=-50$, we have $-6f-4c+d=-50$. This implies $d\equiv 2f\pmod 5$ and together with $df\equiv -a\equiv 2\pmod 5$: $f\equiv \pm1\pmod 5$, hence $fc\equiv -1\pmod 5$ so that from $fc+ed=24$ we have $ed\equiv 0\pmod 5$, contradiction.

• If $f(x)=-f(4-x)$, then $f(2)=0$. But $f(2)=-a$. – Barry Cipra Nov 26 '13 at 21:56

Note that mod $5$ the polynomial can be written

$$f(x)=(x-m-2)(x-m-1)x(x-m+1)(x-m+2)-3$$

for any integer $m$. Now if there is an irreducible quadratic factor, then $f(x)$ has a root that is a root of that factor. But the roots of the irreducible quadratics mod $5$ are all of the form $m+n\sqrt3$. Plugging this in gives

$$0=(n\sqrt3-2)(n\sqrt3-1)n\sqrt3(n\sqrt3+1)(n\sqrt3+2)-3$$

which implies

$$\sqrt3=(3n^2-4)(3n^2-1)n$$

which is a contradiction.

Added later: The same argument mod $p$ (where $p$ is a prime) shows that any polynomial of the form

$$f(x)=(x-r_1)(x-r_2)\cdots(x-r_p)-a$$

where $a$ is a quadratic non-residue mod $p$ and $r_k\equiv k$ mod $p$ for $1\le k\le p$, has no linear or quadratic factors. For $p\ge7$, this argument leaves open the possibility that the polynomial might still be reducible. Pablo Rotondo's answer, however, shows this is not the case, even when $a$ is a quadratic residue.