# Irreducible Polynomial over $\mathbb{Q}$

I'm trying to see if the following polynomials are irreducible over $\mathbb Q$:

$f(x) = x^4 - x^2 + 2x -1$

$g(x) = x^3 + 7x^2 -8x +1$

$h(x) = x^4 + x^3 + x^2 + x + 1.$

Now, for $h(x)$, I can write $(x+1)$ for $x$ and get : $x^4 + 5x^3 + 10x^2 + 10x +5$, for which I can set $P=5$ and by Eisenstein, this is irreducible over $\mathbb Z$, therefore irreducible over $\mathbb Q$.

For $g(x)$, since it is cubic and primitive, there is either a root or it is irreducible. I get to the stage $g(x)=(x-a)(x^2 + bx +c)$ but then cannot equate the coefficients.

Similar situation for $f(x)$, I get to the stage $f(x)=(x^2 + ax +b)(x^2 +cx + d)$ but get stuck when trying to figure out the coefficients. Am I doing something wrong here? I would really appreciate it if someone could explain this to me please, since I am wondering if I am overlooking a silly mistake. Thanks.

-
For $g$, any integer root must divide the constant term $1$. –  Chris Eagle Apr 20 '12 at 16:30
If you have a look at $g(x)$ in $\mathbb Z_2$, then it is $x^3+x^2+1$. It has no root and degree is 3, hence it is irreducible in $\mathbb Z_2$. Whenever you can write polynomial as a product of two polynomials of lower degree, the same must be true if you look at these polynomials modulo a prime $p$ such that $p\nmid a_n$ (i.e. $p$ does not divide the highest coefficient). This criterion is described e.g. in Prasolov's book Polynomials, p.51. –  Martin Sleziak Apr 20 '12 at 16:35
Thank you, Martin, this has helped a lot! Is there a way to achieve this result without looking at the poly in Z2? –  user29553 Apr 20 '12 at 16:45
The technique of looking for irreducibility over finite fields is fundamental, and is often the very quickest route to proving irreducibility over $\mathbb{Q}$. –  Lubin Apr 20 '12 at 18:09

## 2 Answers

If $f(x)=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$, then we have $a+c=0$, $ac+b+d=-1$, $ad+bc=2$ and $bd=-1$. By Gauss's theorem of primitive polynomial, we have $a,b,c,d$ must all be integers, hence $c=-a$, and $d=1$, $b=-1$ or $d=-1$, $b=1$. We may assume $b=1$, $d=-1$ by symmetry of these two factors, then $-a^2=-1$, $-2a=2$ hence $a=-1$, $c=1$, so we have $f(x)=(x^2-x+1)(x^2+x-1)=x^4-(x-1)^2=x^4-x^2+2x-1$. Hence $f$ is reducible in $\mathbb{Q}[t]$.

-
Thank you! I would not have come across using Gauss's theorem of primitive polynomials unless after many hours! So, can we use this for any primitive polynomial, i.e. can we use this principle to find the factorisation of g(x), a primitive cubic poly? Many thanks. –  user29553 Apr 20 '12 at 16:43
@user29553: Of course yes, since if $g$ is reducible, then $g(x)=(x-a)(x^2+bx+c)$ where $a,b,c$ are all integers, by looking at constant term we have $1=-ac$, so $a=\pm 1$. But $\pm 1$ are not a root of $g$, since the sum of all coefficients of $g$ is a odd number, so $g$ is irreducible in $\mathbb{Q}[t]$. –  Yuchen Liu Apr 21 '12 at 0:48

$f(x) = x^4 - (x-1)^2$ which is the difference of two squares, hence reducible.

-