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The following problem appeared on an exam I had yesterday. I was unable to solve it, but I would like to know the solution.

Let $k$ be algebraically closed, and let $f\in k[X,Y]$ be a polynomial with $\deg f = 3$. Assume that $V(f)\subseteq \mathbb{A}^2$ has a triple point $p$. Show that $V(f)$ is reducible.

Eventually after an affine transformation, we can assume that $p=(0,0)$ Then each monomial in $f$ has dimension 3, so $f=aX^3+bX^2Y+cXY^2+dY^3$ is homogeneous, which means that $V(f)$ is a collection of lines through the origin. If we look at the projection onto $P^1$, we get $V(f)\subseteq \mathbb{A}^1=V(aX^3+bX^2+cX+d)$ for the points at finite distance, which is a collection of at most 3 distict points. However, how do we exclude the case where $f(X,1)$ has a triple root, say at $X=x_0$, in which case $V(f)$ is a single ray through the origin and $(x_0,1)$ ? Then, $f(X,1)=a(X-x_0)^3$ so $b=3x_0,c=3x_0^2$ and $d=x_0^3$.

And here I'm stuck. I would appreciate a hint to bring this home.

Espen Nielsen
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1 Answers1

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(Turning my comment into an answer...)

OK, I get the point now — sorry for my previous bumbling.

The point is that the situation you describe can't happen. If $f$ is the cube of a linear form $l$, which is what happens in the case you want to exclude, then $V(f)=V(l)$ doesn't have a triple point at the origin. Indeed, to say that $V(f)$ has a triple point (for $f$ of degree 3) actually implies that $f$ must be reduced, because otherwise $V(f)=V(g)$ for some $g$ of degree lower than 3.