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I have a somewhat-long multi-variable degree-three polynomial (call it $f$) and a very long multi-variable degree-five polynomial (call it $g$), both equal to zero. I strongly suspect that a subset of the terms in $g$ form an expression which has the entirety of $f$ as a factor, which would allow me to eliminate all of those terms from $g$ (as they are all equal to zero by virtue of being multiplied by something else equal to zero) and make simplifying the whole polynomial easier. However, I haven't been able to find a way to efficiently search for that subset. Are there methods for approaching a problem like this?

Here are the polynomials in question:

$$ f: 2 X^3 + 3 X^2 Y + X Y^2 + X^2 a + X Y a + X^2 b + X Y b − 4 X a b − 2 Y a b + X^2 c + X Y c − a b c + X^2 d + X Y d − a b d − X^2 e − X Y e + a b e = 0 $$ and $$ g: 54 X^5 + 135 X^4 Y + 122 X^3 Y^2 + 48 X^2 Y^3 + 7 X Y^4 + 55 X^4 a + 110 X^3 Y a + 69 X^2 Y^2 a + 14 X Y^3 a + 18 X^3 a^2 + 27 X^2 Y a^2 + 9 X Y^2 a^2 + 2 X^2 a^3 + 2 X Y a^3 + 55 X^4 b + 110 X^3 Y b + 69 X^2 Y^2 b + 14 X Y^3 b − 92 X^3 a b − 138 X^2 Y a b − 78 X Y^2 a b − 16 Y^3 a b − 42 X^2 a^2 b − 42 X Y a^2 b − 12 Y^2 a^2 b − 8 X a^3 b − 4 Y a^3 b + 18 X^3 b^2 + 27 X^2 Y b^2 + 9 X Y^2 b^2 − 42 X^2 a b^2 − 42 X Y a b^2 − 12 Y^2 a b^2 − 12 X a^2 b^2 − 6 Y a^2 b^2 + 2 X^2 b^3 + 2 X Y b^3 − 8 X a b^3 − 4 Y a b^3 + 103 X^4 c + 206 X^3 Y c + 129 X^2 Y^2 c + 26 X Y^3 c + 72 X^3 a c + 108 X^2 Y a c + 36 X Y^2 a c + 12 X^2 a^2 c + 12 X Y a^2 c + 72 X^3 b c + 108 X^2 Y b c + 36 X Y^2 b c − 168 X^2 a b c − 168 X Y a b c − 48 Y^2 a b c − 48 X a^2 b c − 24 Y a^2 b c − 2 a^3 b c + 12 X^2 b^2 c + 12 X Y b^2 c − 48 X a b^2 c − 24 Y a b^2 c − 3 a^2 b^2 c − 2 a b^3 c + 66 X^3 c^2 + 99 X^2 Y c^2 + 33 X Y^2 c^2 + 24 X^2 a c^2 + 24 X Y a c^2 + 24 X^2 b c^2 + 24 X Y b c^2 − 96 X a b c^2 − 48 Y a b c^2 − 9 a^2 b c^2 − 9 a b^2 c^2 + 14 X^2 c^3 + 14 X Y c^3 − 14 a b c^3 + 103 X^4 d + 206 X^3 Y d + 129 X^2 Y^2 d + 26 X Y^3 d + 72 X^3 a d + 108 X^2 Y a d + 36 X Y^2 a d + 12 X^2 a^2 d + 12 X Y a^2 d + 72 X^3 b d + 108 X^2 Y b d + 36 X Y^2 b d − 168 X^2 a b d − 168 X Y a b d − 48 Y^2 a b d − 48 X a^2 b d − 24 Y a^2 b d − 2 a^3 b d + 12 X^2 b^2 d + 12 X Y b^2 d − 48 X a b^2 d − 24 Y a b^2 d − 3 a^2 b^2 d − 2 a b^3 d + 132 X^3 c d + 198 X^2 Y c d + 66 X Y^2 c d + 48 X^2 a c d + 48 X Y a c d + 48 X^2 b c d + 48 X Y b c d − 144 X a b c d − 72 Y a b c d − 18 a^2 b c d − 18 a b^2 c d + 42 X^2 c^2 d + 42 X Y c^2 d − 18 a b c^2 d + 66 X^3 d^2 + 99 X^2 Y d^2 + 33 X Y^2 d^2 + 24 X^2 a d^2 + 24 X Y a d^2 + 24 X^2 b d^2 + 24 X Y b d^2 − 96 X a b d^2 − 48 Y a b d^2 − 9 a^2 b d^2 − 9 a b^2 d^2 + 42 X^2 c d^2 + 42 X Y c d^2 − 18 a b c d^2 + 14 X^2 d^3 + 14 X Y d^3 − 14 a b d^3 − 55 X^4 e − 110 X^3 Y e − 69 X^2 Y^2 e − 14 X Y^3 e − 36 X^3 a e − 54 X^2 Y a e − 18 X Y^2 a e − 6 X^2 a^2 e − 6 X Y a^2 e − 36 X^3 b e − 54 X^2 Y b e − 18 X Y^2 b e + 84 X^2 a b e + 84 X Y a b e + 24 Y^2 a b e + 24 X a^2 b e + 12 Y a^2 b e + 2 a^3 b e − 6 X^2 b^2 e − 6 X Y b^2 e + 24 X a b^2 e + 12 Y a b^2 e + 3 a^2 b^2 e + 2 a b^3 e − 72 X^3 c e − 108 X^2 Y c e − 36 X Y^2 c e − 24 X^2 a c e − 24 X Y a c e − 24 X^2 b c e − 24 X Y b c e + 96 X a b c e + 48 Y a b c e + 12 a^2 b c e + 12 a b^2 c e − 24 X^2 c^2 e − 24 X Y c^2 e + 24 a b c^2 e − 72 X^3 d e − 108 X^2 Y d e − 36 X Y^2 d e − 24 X^2 a d e − 24 X Y a d e − 24 X^2 b d e − 24 X Y b d e + 96 X a b d e + 48 Y a b d e + 12 a^2 b d e + 12 a b^2 d e − 48 X^2 c d e − 48 X Y c d e + 24 a b c d e − 24 X^2 d^2 e − 24 X Y d^2 e + 24 a b d^2 e + 18 X^3 e^2 + 27 X^2 Y e^2 + 9 X Y^2 e^2 + 6 X^2 a e^2 + 6 X Y a e^2 + 6 X^2 b e^2 + 6 X Y b e^2 − 24 X a b e^2 − 12 Y a b e^2 − 3 a^2 b e^2 − 3 a b^2 e^2 + 12 X^2 c e^2 + 12 X Y c e^2 − 12 a b c e^2 + 12 X^2 d e^2 + 12 X Y d e^2 − 12 a b d e^2 − 2 X^2 e^3 − 2 X Y e^3 + 2 a b e^3 = 0 $$

Lawton
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  • You can fix some term $p$ of $f$, iterate over all terms of $g$, and if term $q$ from $g$ is divisible by $p$, check if all terms of $f \cdot (p / q)$ are in $g$. This can be implemented in $O(nm)$ time, where $n$ and $m$ are number of terms in $f$ and $g$. Is this efficient enough? (to fully simplify this way, you probably want instead $p$ such that $g - fp$ has minimal number of non-zero terms, but this looks much harder) – mihaild Jul 03 '23 at 14:54
  • @mihaild is there any chance the procedure you describe can be implemented in a computer algebra system like WolframScript or SageMath? I've been looking for a way to do that, but nothing jumps out in my searches. – Lawton Jul 03 '23 at 15:13
  • I am pretty sure it can be implemented in any computer algebra system. We need 4 base operations: split polynomial in list of terms, check if one term is divisible by another and finding quotient, multiple one term by another, and checking if polynomial includes term (if we already can split polynomial into list of terms, this is the same as checking if two terms are equal). – mihaild Jul 03 '23 at 16:01
  • Unlike the single variable case, there is no Euclidean division (or unique remainder) in multivariate polynomial rings, see for example 1, 2. You can still try different orderings of the monomials and hope one of them gives a simple form. – dxiv Jul 03 '23 at 16:42

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