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Let $\phi$ be the $\mathbb{C}$ algebra homomorphism from $\mathbb{C}[U,V,W] \to \mathbb{C}[X,XY,XY^2]$ that sends $U$ to $X,$ $V$ to $XY$ and $W$ to $XY^2.$ Let $I:= (V^2-UW).$ I am trying to show that $I$ is the kernel of $\phi.$ The hard direction is $\ker \phi \subseteq I.$

Suppose $p(U,V,W)\in \ker \phi,$ so $p(X,XY,XY^2)$ is the zero polynomial. I need to show $p(U,V,W)\in I.$ Note that by always writing $V^2=UW + (V^2-UW)$ we can write $p(U,V,W) = g(U,W) + h(U,W)V +q(U,V,W) (V^2-UW).$ Therefore the problem boils down to showing that if $g(X,XY^2) + h(X,XY^2) XY=0$ then $g(U,W)+h(U,W)V\in I.$ Can someone help me with that?

Katie Dobbs
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2 Answers2

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Basically you want to show that $k[U,V,W]/(V^2-UW) \to k[X,Y]$, $U \mapsto X$, $V \mapsto XY$, $W \mapsto XY^2$ is injective. But the elements of $k[U,V,W]/(V^2-UW) \cong k[U,W]\bigl[V\bigr]/(V^2-UW)$ are easy to describe: They can be written as $f + V g$ for some (unique) $f,g \in k[U,W]$. If such an element lies in the kernel, we have $f(X,XY^2) + XY g(X,XY^2)=0$ in $k[X,Y]$. In the first summand the exponents of $Y$ are even, in the second one they are odd. It follows that $f(X,XY^2)=0$ and that $g(X,XY^2)=0$. If $f=\sum_{ij} a_{ij} U^i V^j$, then $0=f(X,XY^2)=\sum_{ij} a_{ij} X^{i+j} Y^{2j}$, and these monomials are pairwise distinct by looking at $Y$ again, so that $a_{ij}=0$ for all $i,j$, i.e. $f=0$. Likewise we get $g=0$.

Martin Brandenburg
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Hint: It is not possible for non-zero $g(U,W)+Vh(U,W)\in (V^2-UW)$, since all multiples of $V^2-UW$ have exponent at least $2$ for $V$. So you really need to show that if $g(X,XY^2) + XY h(X,XY^2) = 0$ then $g(U,W)=0$ and $h(U,W)=0$.

Basically the proof goes in two steps. First prove:

$$g(X,XY^2)+XYh(X,XY^2) = 0\implies g(X,XY^2),h(X,XY^2)=0$$ Then show: $$g(X,XY^2)=0\implies g(U,W)=0$$

Thomas Andrews
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