# Playing with plane curves

Let $\phi: \mathbb{R}^1 \longrightarrow \mathbb{R}^2$ be the map given by $t \mapsto (t^2,t^3)$. I'm trying to show that any polynomial $f \in \mathbb{R}[X,Y]$ vanishing on the image $C = \phi(\mathbb{R}^1)$ is divisible by $Y^2-X^3$. And what property of a field $k$ will ensure that the result holds for $\phi: k \longrightarrow k^2$ given by the same formula?

Also, I am trying to do it for $t \mapsto (t^2-1,t^3-t)$.

Thanks.

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Dear Mary, The necessary and sufficient criterion is that $k$ should be infinite. Regards, – Matt E Sep 10 '12 at 1:59
Can you please explain that and try to give another proof of this question. What Rod has looks good, but isn't what I'm looking for – mary Sep 10 '12 at 4:54
Dear Mary, Georges has elaborated on my comment in his answer below. Best wishes, – Matt E Sep 11 '12 at 0:28
P.S. If you want to see whether or not you understand his technique, you could try applying it to a similar but different map, such as $t \mapsto (t^2, t^5)$. – Matt E Sep 11 '12 at 0:28

By Euclidean division we may write any polynomial $f(X,Y)\in k[X,Y]$ as $f(X,Y)=(Y^2-X^3)g(X,Y)+u(X)Y+v(X)$ with $f(X,Y)\in k[X,Y]$ and $u(X), v(X)\in k[X]$. Hence on the curve we have ( by substituting $X=t^2, Y=t^3$) $0=f(t^2,t^3)=((t^3)^2-(t^2)^3)g(t^2,t^3)+u(t^2)t^3+v(t^2)=u(t^2)t^3+v(t^2)$.
The equality $u(t^2)t^3+v(t^2)=0$ for all $t\in k$ forces the polynomial $u(T^2)T^3+v(T^2)$ to be zero if $k$ is infinite (as Matt told you in his comment).
And this in turns forces the polynomials $u(X),v(X)\in k[X]$ to be zero, because in $u(T^2)T^3+v(T^2)=0$ the first summand has only odd monomials and the second only even monomials.

Conclusion
If $k$ is an infinite field, any polynomial vanishing on the curve $y^2-x^3=0$ is a multiple of $Y^2-X^3$

A counterexample
The result is false for finite fields: for example if $k=\mathbb Z/2\mathbb Z$, check that the polynomial $X+Y$, which is definitely not a multiple of $Y^2-X^3$, vanishes on all points of the curve $y^2-x^3=0$ (that is on the points ...?)

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Dear @Matt, I have taken the liberty to spell out a proof of your comment because I had a few minutes of free time . I hope you don't mind? – Georges Elencwajg Sep 10 '12 at 6:53
Dear Georges, Not at all; thank you! Best wishes, Matt – Matt E Sep 11 '12 at 0:27

Let $x (t) = t^2$ and $y (t) = t^3$, so that we can write $\phi (t) = (x (t), y (t))$. Note that $y (t) = \left( x(t) \right)^{3/2}$, which yields $\left (y (t)\right)^2 = \left( x (t)\right)^3$ or, equivalently $\left (y (t)\right)^2 - \left( x (t)\right)^3 = 0$. Hence, we have that $\phi (t)$ travels along the curve $C = \{ (x,y) \in \mathbb{R}^2 \mid y^2 - x^3 = 0\}$, which is depicted below

Note that for $t < 0$, $\phi (t)$ will lie on the 4th quadrant and approach the origin from infinity, whereas for $t > 0$ it will lie on the 1st quadrant and travel from the origin to infinity.

Every polynomial $f \in \mathbb{R}[X,Y]$ that vanishes on the curve $C$ can be factored as follows

$f (X, Y) = (Y^2 - X^3)^n \, g (X, Y)$

where $n \geq 1$ and $g \in \mathbb{R}[X,Y]$ does not vanish on $C$, and is thus divisible by $(Y^2 - X^3)$. I have no idea what property of a field $\mathbb{K}$ is required for the result to hold for the case $\phi : \mathbb{K} \to \mathbb{K}^2$.

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Thanks, but this is not the approach I had it mind though – mary Sep 10 '12 at 0:37
Just not by proceeding graphically. I think that there might be a theorem – mary Sep 10 '12 at 0:42