# Elementary proof of the irreducibility of $T^4 – a T – 1$ in $\mathbf{Q}$ when $a\in\mathbf{Z}-\{0\}$

This is from the exercises of Bourbaki, Algèbre, Chapitre V, first exercise of the exercises concerning the second paragraph of the fifth chapter. (p. 140.)

As Gauss Lemma (“if your gcd is invertible then you are irreducible in $\mathbf{Q}[T]$ if and only you are irreducible in $\mathbf{Q}[T]$”) is not stated in any chapter of Bourbaki’s Algèbre, I would like to prove the polynomial’s irreducibility without it, and without using any gcd of coefficients of a polynomial, in an elementary way.

Below is what I did using Gauss Lemma, plus what I tried to do without it, without success (case where the polynomial is product of two degree $2$ rational polynomials)

One has to show that $P = T^4 – a T – 1$ is irreducible in $\mathbf{Q}[T]$ when $a\in\mathbf{Z}-\{0\}$. Obviously, as the gcd of $P$’s coefficients is $1$, it suffice to show that $P$ is irreducible in $\mathbf{Z}[T]$.

Now as $P$ as no integer root. Let $r = \frac{p}{q}$ be a rational root, with $q\not=0$ and $p\wedge q = 1$. Then, multiplying both members of the equation $P(r)=0$ by $q^4$ gives $p^4 – a p q^3 – q^4 = 0$, that is, $p^4 = q^3 (ap+q)$. Let $p’$ be a prime dividing $p$. Then $p’$ divides either $q^3$ and then $q$, either $ap+q$ and then $q$. So $p’$ divides $q$. If there is a prime number dividing $p$, it divides $q$ and then $p\wedge q \not= 1$ which is absurd, showing that no prime number divides $p$, that is that $p\in\{0,\pm 1\}$. If $p=0$, then $r=0$ and $0 = P(r) = P(0) = 1$, absurd. If $p = \pm 1$ then $1 = \pm q^3(a+q)$ showing that $q\in\mathbf{Z}^{\times}$, that is, that $q = \pm 1$. This shows that $r = \pm 1$. But then $0 = P(r) = 1 \pm a – 1 = \pm a$, absurd.

If $P$ is reducible, then $P = P_1 P_2$ with the $P_i$’s in $\mathbf{Z}[T]$, of degree $2$ and with leading coefficient $1$. Say $P_1 = T^2 + \alpha T + \beta$ and $P_2 = T^2 + \gamma T + \delta$. This leads to the system $$(S) \left\{\begin{array}{rcl} \alpha + \gamma & = & 0 \\ \beta + \delta + \alpha \gamma & = & 0 \\ \alpha \delta + \beta \gamma & = & -a \\ \beta\delta & = & -1 \\ \end{array}\right.$$ Fourth equation implies that $\beta, \delta = \pm 1$ and that they have opposite sign, implying also that $\beta = – \delta$. Then second equation implies that $\alpha \gamma = 0$. Then the first equation implies that both $\alpha,\gamma$ are equal to $0$. Up to a permutation of $1$ and $2$, we have therefore $P_1 = T^2 +1$ and mainly $P_2 = T^2 – 1$, leading to a contradiction, as we saw that $P$ has no root in $\mathbf{Z}$. I am a bit puzzled by the fact that $a$ didn’t play any role in the second case, but fine, $P = T^4 – a T – 1$ is irreducible in $\mathbf{Q}[T]$. But…

I have tried to write $P = P_1 P_2$ as before, but with the $P_i$’s with rational coefficients, getting a system (S) with this time $\alpha,\beta,\gamma,\delta\in\mathbf{Q}$, and then trying to get a contradiction, but without success… Any advice ? Thx !

#### Solutions Collecting From Web of "Elementary proof of the irreducibility of $T^4 – a T – 1$ in $\mathbf{Q}$ when $a\in\mathbf{Z}-\{0\}$"

Lemma 2. — Let $f,g,$ and $h$ be unitary polynomials in $\mathbf{Q}[X]$ such that $f = gh$. Si $f$ has integer coefficients, so do $g$ and $h$.

This is in Algèbre, Bourbaki. Chapitre 5, p A.5.78, before your exercise. So that your proof becomes “Bourbaki-legal”, so to speak, modulo the use of this lemma at the beginning of your proof, no worries !

As you may see reduction modulo a prime $p$ is used, as well as integrity of $\mathbf{F}_p [X]$. The translation goes as follows :

Let $a$ (resp. $b$) the smallest integer $\alpha \geq 1$ (resp $\beta\geq 1$) such that $\alpha g$ (resp. $\beta j$) has integer coefficients ; let us set $g’ = ag$ and $h’ = bh$ and let us show ab absurdo that $a = b = 1$. If not, a prime divisor $p$ of $ab$ would exist. If $u\in\mathbf{Z}[X]$ note $\overline{u}$ le polynomial in $\mathbf{F}_p [X]$ obtained by reducing modulo $p$ the coefficients of $u$. We have $g’h’ = abf$, so that $\overline{g’} \overline{h’} = 0$. As the ring $\mathbf{F}_p [X]$ is a domain (IV, p.9, prop. 8), we have therefore $\overline{g’} = 0$ or $\overline{h’} = 0$. Put differently, $p$ divides all coefficients of $g’$ or all coefficients of $h’$, a contradiction with the hypothesis we made.

Finally, IV, p.9, prop. 8 is just the proposition saying that if $A$ is a domain, so is $A[(X_i)_{i\in I}]$