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How to show that a rational polynomial is irreducible in $\mathbb{Q}[a,b,c]$? For example, I try to show this polynomial $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c)(*)$$ is irreducible, where $a,b,c\in \mathbb{Q}$.

The related problem is Ask for the rational roots of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$. Could I consider the points $(*)$ intersect with $L_{\infty}$ are three? $L_{\infty}$ is the infinity line in a projective space $\mathbb{C}P^2$.

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    Irreducible in $\mathbb{Q}[a,b,c]$? – mathworker21 Jun 16 '18 at 06:39
  • @mathworker21 Yes! –  Jun 16 '18 at 07:02
  • It doesn't make sense to ask about irreducibility of $p(a,b,c)$ as a polynomial if $a,b,c\in \mathbb Q$, because assigning rational values to $p(a,b,c)$ makes it a rational number in $\mathbb Q$, and hence a polynomial only in the sense of being a constant term. – hardmath Jun 21 '18 at 02:10

2 Answers2

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Suppose contrary that $p(a,b,c)$ is reducible over $\mathbb{Q}$. You can write $p(a,b,c)$ as $$a^3+b^3+c^3-3(b+c)a^2-3(c+a)b^2-3(a+b)c^2-5abc\,.$$ It suffices to regard $p(a,b,c)$ as a polynomial over $\mathbb{F}_3$ (why?). Over $\mathbb{F}_3$, $$p(a,b,c)=a^3+b^3+c^3+abc=a^3+(bc)a+(b+c)^3\,.$$ Since $p(a,b,c)$ is homogeneous of degree $3$ and reducible, it has a linear factor $a+ub+vc$ for some $u,v\in\mathbb{F}_3$. Clearly, we must have $ub+vc \mid (b+c)^3$, whence $u=v=1$ or $u=v=-1$. However, both choices are impossible via direct computation.

Batominovski
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  • Very nice:+1 ${}{}{}$ – Georges Elencwajg Jun 16 '18 at 08:37
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    To show that the polynomial $p(a,b,c)$ in the indeterminates $a,b,c$ is irreducible over $\mathbb F_3$ it suffices to prove that the curve $V(p)\subset \mathbb P^2_{\mathbb F_3 }$ given by $p(a,b,c)=0$ is smooth. But this is clear because the system of equations $p(a,b,c)=bc=ca=ab$ has no non-trivial solution over any extension of $\mathbb F_3$. – Georges Elencwajg Jun 16 '18 at 08:50
  • What is the $\mathbb{F}_{3}$? –  Jun 16 '18 at 13:53
  • @GeorgesElencwajg What is the $\mathbb{F}{3}$ and $\mathbb{P}{\mathbb{F}_{3}}$? –  Jun 16 '18 at 13:54
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    Dear Tinzoe-Yui, the notation $\mathbb F_3=\mathbb Z/(3)$ stands for the field with 3 elements. And $\mathbb P^2_{\mathbb F_3}$ is the projective plane over that field. My comment is only a little variant for users interested in algebraic geometry. It is certainly not indispensable for understanding Batominovski's excellent and purely algebraic proof. – Georges Elencwajg Jun 16 '18 at 14:38
  • @GeorgesElencwajg Schinzel's book has a concrete test. I put an answer. Note that Schinzel is talking about reducing over the complexes, while the question is about the rationals. – Will Jagy Jun 16 '18 at 16:13
  • @GeorgesElencwajg Thanks for your explanation. Is $\mathbb{Z}/(3)$ =$3\mathbb{Z}$ right? –  Jun 17 '18 at 08:17
  • The notation $\mathbb Z/(3)$ means the quotient field $\mathbb Z/3\mathbb Z$ with three elements, not the infinite ideal $3\mathbb Z\subset \mathbb Z$. – Georges Elencwajg Jun 17 '18 at 21:17
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To ask about complete (three linear factor) reducibility over the complexes, we take the Hessian matrix of second partials. The entries are linear in the named variables. Next, let $\Delta$ be the determinant of the Hessian. This $\Delta$ is once again a cubic form. The original cubic (homogeneous) ternary form factors completely if and only if $\Delta $ is a constant multiple of it. I have finished this first test, your cubic does not factor completely.

More difficult if the cubic might be a linear times an irreducible quadratic. In that case, there is still a conclusive test:

enter image description here

For your problem, the coefficients in order from $0$ to $9$ are $$ 1,-3,-3,-3,-5,-3,1,-3,-3,1 $$ where the best looking diagram is a triangle as in bowling. The $-5$ refers to $-5abc \; ,$ the $1$s refer to $a^3,b^3,c^3 \; .$

Alright, I wote a little program to correctly type the 8 by ten matrix, then put that into gp-Pari. The original polynomial is irreducible over the complex numbers.

? 
?  sch = [ 0, 3, 0, -6, -6, 0, -3, -5, -3, 0; 0, 0, 3, 0, -6, -6, 0, -3, -5, -3; -3, -6, -5, 3, -6, -3, 0, 0, 0, 0; 0, 0, -3, 0, -6, -5, 0, 3, -6, -3; -3, -5, -6, -3, -6, 3, 0, 0, 0, 0; 0, -3, 0, -5, -6, 0, -3, -6, 3, 0; 1, 0, -3, 3, 0, -3, -2, 3, 0, 1; 1, -3, 0, -3, 0, 3, 1, 0, 3, -2]
%25 = 
[ 0  3  0 -6 -6  0 -3 -5 -3  0]

[ 0  0  3  0 -6 -6  0 -3 -5 -3]

[-3 -6 -5  3 -6 -3  0  0  0  0]

[ 0  0 -3  0 -6 -5  0  3 -6 -3]

[-3 -5 -6 -3 -6  3  0  0  0  0]

[ 0 -3  0 -5 -6  0 -3 -6  3  0]

[ 1  0 -3  3  0 -3 -2  3  0  1]

[ 1 -3  0 -3  0  3  1  0  3 -2]

? matrank(sch)
%26 = 8
? 
? 
?
Will Jagy
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