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This is a follow-up on the following answer: https://math.stackexchange.com/a/4687845/755010

To the question of how to graph $y^{4}x^{3}+x^{4}-y=0$ without a computer, user Somos suggested to approximate it by easy to graph equations $y^4+x=0$, $(xy)^3-1=0$, and $x^4-y=0$. This is nicely pictured in user Jean Marie's screenshot: enter image description here

It seems to me that this technique could be more generally applied to a variety of difficult to solve quartics. Would there be a general theorem on the conditions under which we could achieve satisfactory results with this technique?

shintuku
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  • User Somos was explicit in his answer: "Another global method is to assume that one of the terms is small compared to the other two and can be removed without great distortion." Keywords "small compared to". – Piita Apr 29 '23 at 09:42
  • Your curve is of degree seven (a septic) and "sparse". It also has the origin on it and has two more rational points at infinity which are singular. Singular points determine the curve more. A "general" septic in the plane would be smooth and only fit tangents, conics, etc along its points with the method of approximating by lowest degree terms and you would have to look for (s)extactic points to get higher contact. – Jan-Magnus Økland Apr 29 '23 at 14:04

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If we look at $y^4x^3+x^4z^3-yz^6=0$ the projective curve, it has singular points at $(1:0:0)$ and $(0:1:0)$ which both are at the line at infinity $z=0$.

So let's investigate the behavior around the singular points:

$(y/x)^4+(z/x)^3-(y/x)(z/x)^6=0$ has $(z/x)^3=0$ as tangent cone and can be approximated by the terms of lowest degree $(y/x)^4+(z/x)^3=0$ or $(y/x)^4=-(1/x)^3=0$ or $y^4=-x.$

$(x/y)^3+(x/y)^4(z/y)^3-(z/y)^6=0$ has $(x/y)^3=0$ as tangent cone and can be approximated by the terms of lowest degree $(x/y)^3=(1/y)^6$ or $x^3y^3=1.$

$(y/z)^4(x/z)^3+(x/z)^4-y/z=0$ is the original curve and has tangent cone $(y/z)=0$ at $(0:0:1)$ which is on the curve, and can be approximated by the terms of lowest degree $x^4=y.$

Added: More generally, if you have a point $(a:b:1)$ on the curve, in your affine, translate the point to the origin by $(y+b)^4(x+a)^3+(x+a)^4-(y+b)=0$ and collect lowest terms to get the tangent line $y-b=\frac{4a^3+3a^2b^4}{1-4a^3b^3} (x-a),$ or $$3ab^4(x-a)^2+12a^2b^3(x-a)(y-b)+(4a^3+3a^2b^4)(x-a)-(1-4a^3b^3)(y-b)=0$$ if you keep the degree less than three terms.

Now, the curve might not have more rational points than the three we investigated. Also note that the conic we found is not the osculating conic, and has lower contact than the five we expect for those. Next, we could search for sextactic points.

Jan-Magnus Økland
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