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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 anelementary 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)

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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 !

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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 absurdothat $a = b = 1$. 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 $A[(X_i)_{i\in I}]$

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