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As the title says, I want to prove the following fact: If $x^3-x+1$ is irreducible in $\mathbb{Q}[x]$ then $u^3-uv^2+v^3$ is irreducible in $\mathbb{Q}[u,v]$.

$x^3-x+1$ is irreducible since we find by the rational root theorem that the only possible rational roots of $x^3-x+1$ are $\pm 1$, and neither is a root, hence $x^3-x+1$ has no rational roots. But if $x^3-x+1$ was reducible, then it would have to decompose into a product of a second-degree polynomial multiplied by a linear factor, and this linear factor have to be on the form $x-\alpha$ for $\alpha \not \in \mathbb{Q}$ hence $x-\alpha \not \in \mathbb{Q}[x]$. It follows (I believe) that $x^3-x+1$ is irreducible.

Furthermore, we note that if $p(u,v) = u^3-uv^2+v^3$ then $p(u,1) = u^3-u+1$. Assume that $p(u,v)$ is reducible, so that $p(u,v) = q(u,v)s(u,v)$ were neither $s(u,v)$ nor $q(u,v)$ are units.

We have $p(u,1) = q(u,1)s(u,1)$. We note that we can rewrite $q(u,v)$ and $s(u,v)$ as $s(u,v) = s_0(u)+s_1(u)v+s_2(u)v^2 + \ldots s_n(u)v^n$ and $q(u,v) = q_0(u)+q_1(u)v+ \ldots + q_m(u)v^m$ for $q_i,s_i \in \mathbb{Q}[u]$. We then have $p(u,1) = (q_0(u)+q_1(u) + \ldots + q_n(u))(s_0(u) + s_1(u) + \ldots + s_m(u)) = u^3-u+1$. We know that since we showed that $x^3-x+1$ is irreducible, we need $s(u,1)$ or $q(u,1)$ to be a unit. Hence up to unit $s(u,1) = u^3-u+1$ or $q(u,1) = u^3-u+1$. Here is where I am currently stuck.

Ben123
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2 Answers2

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By the rational root theorem for $\mathbb Q[u,v]=\mathbb Q[v][u]$, a root of $p(u)=u^3-uv^2+v^3$ is of the form $av^i$ for $i\in\{0,1,2,3\}$. When $i\ne1$ then clearly $p(av^i)\ne0$. Thus we need $p(av)=(a^3-a+1)v^3=0$. But $a^3-a+1$ is never zero.

Kenta S
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    I think I understand your argument, thanks. Although I believe you want $p(av) = (a^3-a+1)v^3$. – Ben123 Jan 09 '24 at 23:07
  • @Servaes I mean the rational root theorem for the UFD $A=\mathbb Q[v]$. – Kenta S Jan 10 '24 at 01:29
  • @KentaS Then where does this argument fail for $q(u)=u^4+u^2v^2+v^4$? It should then also have a root of the form $av^i$, but clearly $q(av^i)\neq0$ when $i\neq1$, and $q(av)=(a^4+a^2+1)4v^4=0$ is not possible either as $a^4+a^2+1$ is never zero. I guess I just don't understand what you mean by the rational root theorem for a UFD. – Servaes Jan 10 '24 at 01:36
  • @Servaes see: https://math.stackexchange.com/questions/2816207/concerns-with-the-proof-of-the-rational-root-theorem. Indeed your example $q(u)$ also does not have a root in $A=\mathbb Q[v]$, I don't see what the problem is. Of course, $q(u)$ is reducible, but since $q$ is of degree $4$ this does not necessarily imply $q(u)$ has a root. – Kenta S Jan 10 '24 at 01:41
  • @KentaS Ah I see, you're using the fact that if $p$ is reducible, then it has a factor of the form $f(v)u+g(v)$ for some $f,g\in\Bbb{Q}[v]$. Thanks for the clarification! – Servaes Jan 10 '24 at 01:43
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    @Servaes yes, there is a small caveat though, having such a factor does not immediatly imply that it has a root in $\mathbb{Q}[v]$, for example $uv+1$. In this case it works though, since if you look at the $u^3$-term you get from $(f(v)u+g(v))(a(v)+b(v)u+c(v)u^2)=p(u)$ that $f(v)c(v)=1$ so $f(v)$ is a unit and then $u+\frac{g(v)}{f(v)}$ is a factor and thus $-\frac{g(v)}{f(v)}$ is a root. – Kilian Jan 10 '24 at 01:51
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Suppose $f(u,v),g(u,v)\in\Bbb{Q}[u,v]$ are polynomials satisfying. $$f(u,v)g(u,v)=u^3-uv^2+v^3.\tag{1}$$ Then evaluating at $(u,v)=(x,1)$ shows that $$f(x,1)g(x,1)=x^3-x+1.$$ Because $x^3-x+1$ is irreducible, this implies that either $f(x,1)$ or $g(x,1)$ is constant. Without loss of generality $f(x,1)$ is constant, and this implies that $f(u,v)\in\Bbb{Q}[v]$. Plugging in $u=0$ shows that $$f(0,v)g(0,v)=v^3,$$ and so $f(u,v)=f(0,v)$ divides $v^3$, meaning that $f(u,v)=cv^k$ for some $c\in\Bbb{Q}$ and some $k\in\{0,1,2,3\}$. Of course $(1)$ then shows that $k=0$ and hence that $f(u,v)=c$. This shows that $u^3-uv^2+v^3$ is irreducible.

Servaes
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