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I met a curious problem on factorizing a homogenerous polynomial of degree 9.

Problem: Show that the following polynomial can be divided by $(a_1+a_2+a_3)$:

\begin{align} &\quad\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{array} \right|^3 \\ & + \left( a_1^3 \left(b_2 c_3-b_3 c_2\right) +a_2^3 \left(b_3 c_1-b_1 c_3\right) +a_3^3 \left(b_1 c_2-b_2 c_1\right) \right)\left(b_1+b_2+b_3\right)^2\left(c_1+c_2+c_3\right)^2 \\ &- \left( b_1 \left(c_2 a_3-c_3 a_2\right)^3 +b_2 \left(c_3 a_1-c_1 a_3\right)^3 +b_3 \left(c_1 a_2-c_2 a_1\right)^3 \right) \left(b_1+b_2+b_3\right)^2 \\ &- \left( c_1 \left(a_2 b_3-a_3 b_2\right)^3 +c_2 \left(a_3 b_1-a_1 b_3\right)^3 +c_3 \left(a_1 b_2-a_2 b_1\right)^3 \right) \left(c_1+c_2+c_3\right)^2 \end{align}

Mathematica tells me it is true(one factor is $(a_1+a_2+a_3)$ while the other factor consists of more than 200 terms), but I can not figure out a human proof.

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    $\begingroup$ can you disclose how this lengthy expression originates? that will likely be helpful to answering your question. $\endgroup$
    – Carlo Beenakker
    Commented Mar 14 at 8:56
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    $\begingroup$ It's unclear what you are asking. What "clever way" you look for, and why using Mathematica is not one? $\endgroup$
    – Max Alekseyev
    Commented Mar 14 at 11:16
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    $\begingroup$ I guess the whole expression can be written as the determinant of a $3\times 3$ block matrix with $3\times 3$ elements similar to the given $3\times 3$ matrix. It looks like a (block) Laplace expansion. Btw, I LaTeX'd the expression. $\endgroup$
    – Fred Hucht
    Commented Mar 14 at 12:31
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    $\begingroup$ @LichenSDU A human proof for a non-human polynomial ...? $\endgroup$
    – Peter Mueller
    Commented Mar 14 at 14:11
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    $\begingroup$ Did you try using the following? $$\left|\begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{array}\right|=\left|\begin{array}{ccc} a_1+a_2+a_3 & a_2 & a_3 \\ b_1+b_2+b_3 & b_2 & b_3 \\ c_1+c_2+c_3 & c_2 & c_3 \\ \end{array}\right|\equiv -(b_1+b_2+b_3)(a_2c_3-a_3c_2)+(c_1+c_2+c_3)(a_2b_3-a_3b_2) \mod a_1+a_2+a_3$$ $\endgroup$
    – Yaakov Baruch
    Commented Mar 18 at 11:04

1 Answer 1

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Here is an idea that should lead to a human way to verify this.

Take a triple $(a_1,a_2,a_3)$ for which $a_1+a_2+a_3=0$. For example, $a_1=a=-a_2$, and $a_3=0$ or $a_i=a\cdot\omega^i$ where $\omega^3=1$, $\omega\neq 1$, etc. Substitute these to your equation. If you can verify that that's zero, then you are in business. This should be a calculation that can be reasonably carried out by a human if needed. It involves expanding a few binomials to the third power and sorting out the terms. These should work for an indeterminate $a$, which shows that the zero set of the polynomial $a_1+a_2+a_3$ is contained in the zero set of your big polynomial over any field. Then you get that the polynomial $a_1+a_2+a_3$ divides your big polynomial over any algebraically closed field by the Hilbert Nullstellensatz (because $a_1+a_2+a_3$, having degree $1$ is irreducible over any field). Finally that (should?) imply that it divides it over $\mathbb Z$ as well (this should be checkable directly given that the dividing poly is linear with all (non-zero) coefficients being $1$). One could also try to use the Euclidean algorithm for checking this.

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    $\begingroup$ But $a_1+a_2+a_3=0$ is a plane, whereas you suggest to verify the vanishing only on a line contained in that plane. Surely that's a bit too much simplification? $\endgroup$
    – Joachim König
    Commented Mar 18 at 10:58
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    $\begingroup$ Well, that was preceded by the words "for example"... Also, you can surely rotate the conditions around, so that means at least 3 lines... :) You can use some other lines until you get more than 9. $\endgroup$
    – Sándor Kovács
    Commented Mar 18 at 21:10

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