# Prove $X^4-5X^3+3X-2$ is irreducible over the rationals.

How can we show that $$P=x^4-5x^3+3x-2$$ is irreducible over the rationals. I tried to use Eisenstein criterion and many shifts $$P(x+a)$$ but did not work. I tried to show it is irreducible modulo $$2$$ or $$3$$ but it did not work.

I tried to use the rational root test to prove that $$P$$ does not have any rational root. Also $$P(-1)=1$$, $$P(0)=-2$$, $$P(1)=-3$$ and $$P(5)=13$$. So we know that there are two irrational roots $$\alpha\in (-1,0)$$ and $$\beta \in (1,5)$$ so $$\beta \not = -\alpha$$, the other two roots are conjugate complex but I don't know how to prove it, is it enough to say that $$P$$ has two irrational non opposite roots and two conjugate complex roots to say it is irreducible over $$\mathbb Q$$?

By Gauss' lemma, irreducibility over $$\mathbb{Q}$$ is equivalent to irreducibility over $$\mathbb{Z}$$ in this case. Since there are no rational roots, the only way $$f$$ can be reducible is being a product of two monic polynomials $$g, h\in\mathbb{Z}[x]$$ of degree $$2$$. But now reduce the equation $$f=gh$$ modulo $$2$$. You get that the polynomial $$\overline{f}=x^4+x^3+x\in\mathbb{F_2}[x]$$ is a product of two monic polynomials of degree $$2$$ in $$\mathbb{F_2}[x]$$. But that is a contradiction, because:

$$\overline{f}=x(x^3+x^2+1)$$

is the factorization of $$\overline{f}$$ into irreducibles in $$\mathbb{F_2}[x]$$, and such factorization is unique. (if it was a product of two polynomials of degree $$2$$, it wouldn't have an irreducible factor of degree $$3$$)

• And the cubic $x^3 + x^2 + 1$ is irreducible over $\mathbb{F}_2$ since it does not have a root. Commented May 4, 2023 at 11:57
• @ronno Yeah, I left some simple details, as based on the question OP probably knows that.
– Mark
Commented May 4, 2023 at 12:06

Using Gauss' lemma, if a factorization exists, it has integer coefficients. The rational root theorem easily shows there is no integer root, so no factorization $$(x-a)(x^3+bx^2+cx+d)$$. There is still the possibility that $$P=(x^2+ax+b)(x^2+cx+d)$$, with integer coefficients. Then we have the equations, after expanding the product:

• $$a+c=-5$$
• $$b+ac+d=0$$
• $$ad+bc=3$$
• $$bd=-2$$

Since coefficients are integers, we can assume $$b=\pm1$$, then $$d=\mp2$$.

If $$b=1$$, the third equation becomes $$-2a+c=3$$, and together with the first we must have $$3c=-7$$, i.e. $$c$$ is not an integer, contradiction.

If $$b=-1$$, the third equation becomes $$2a-c=3$$, and together with the first, we get $$3a=-2$$, and $$a$$ is not an integer, again a contradiction.

Therefore there is no such factorization, and $$P$$ is irreducible.

By Rouché's theorem $$P$$ has exactly one zero outside the complex unit disc (see below). If $$P$$ was reducible into two monic polynomials with integral coefficients (justified by Gauss's lemma), one of the factors, say $$Q(x)$$, would have all roots $$|\alpha_i|<1$$. So we would get $$|Q(0)|=|\prod \alpha_i|<1$$, but that is clearly impossible since $$Q(0)$$ must be an integer that divides $$2$$. So $$P$$ must be irreducible over rationals.

Note to Rouché's theorem: For $$|z|=1$$ we have $$|3z-2|<|z-5|=|z^4-5z^3|$$ (the inequality defines disc with center $$(\frac{1}{8},0)$$ and radius $$\frac{13}{8}$$ which strictly contains a unit disc). Therefore $$x^4-5x^3=x^3(x-5)$$ and $$P(x)$$ have the same number (three) of roots inside the unit disc.