# 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.

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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

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 $\mathbb{Q}[t]$.
@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.