Factorizations and roots of a bivariate polynomial

Question

Problem: Show that for $g(x,y)\in \mathbb{C}[x,y]$ and $t$ an indeterminate, $g(t^2,t^3)=0$ if and only if $g$ is divisible by $y^2-x^3$.

Attempt: If $g$ is divisible by $y^2-x^3$, then obviously $g(t^2,t^3)=0$.

To prove the converse we use the following theorem, which is a special case of Hilbert's Nullstellensatz.

Theorem $2.2$ (in $\S2.3$ of An Introduction to Algebraic Geometry by Kenji Ueno): A homogeneous polynomial $G(x_0,x_1,x_2)\in \mathbb{C}[x_0,x_1,x_2]$ vanishes identically on the irreducible curve $$ C: F(x_0,x_1,x_2)=0 $$ (i.e., $F\in\mathbb{C}[x_0,x_1,x_2]$ is homogeneous and irreducible) if and only if $G(x_0,x_1,x_2)$ is divisible by $F(x_0,x_1,x_2)$.

Proof: Nullstellensatz states that $G\in \sqrt{(F)}$. Since the ideal $(F)$ is prime in $\mathbb{C}[x_0,x_1,x_2]$, it follows that $G\in \sqrt{(F)}=(F)$.

We homogenize $g(x,y)$ to get a homogeneous $G(x,y,z):=g(x/z,y/z)z^{\deg(g)}$, and similarly homogenize $f(x,y)=y^2-x^3$ to $F(x,y,z)=zy^2-x^3$, which is irreducible. Let $C$ be the set of zeros of $F$.

• For $(x,y,z)\in C$, if $z=0$, then $x=0$. So $G(0,y,0)=0$, for otherwise there is a nonzero monomial of the form $a y^{\deg(g)}, a\neq 0,$ in $g(x,y)$, which contradicts the assumption $g(t^2,t^3)=0$ (just consider its degree in $t$).

• If $(x,y,z)\in C$ and $z\neq 0$, then we have $(y/z)^2-(x/z)^3=0$ and hence $G(x,y,z)=z^{\deg(g)}g(t^2,t^3)=0$, where $t=\sqrt{x/z}$.

By Theorem $2.2$, $F$ divides $G$. Setting $z=1$, we get that $f$ divides $g$.

Question: Is there a simpler solution that does not use the theorem above? I would like to see one.

Any hints and answers are welcome.

Answer 1

Though in general $R[x]$ for a general ring has no euclidean algorithm, one may always divide by a monic polynomial (at least if $R$ is commutative, there is an indication in the comments of the cited post that the result may hold with greater generality).

In particular we can apply this to $\mathbb C[x,y] = (\mathbb C[x])[y]$ and the monic polynomial $y^2 - x^3$. Suppose that $f \in R[x,y]$ satisfies $f(t^2,t^3) = 0$. Write

$$ f = (y^2-x^3)q(x,y) + r(x,y). $$

Evaluating at $(t^2, t^3)$ we have $r(t^2, t^3) = 0$. Observe that $r$ has degree less than $y^2-x^3$ as a polynomial in $y$; that, is, we have

$$ r(x,y) = yp(x) +s(x). $$

Thus $t^3p(t^2) + s(t^2) = 0$, i.e. $t^3p(t^2) = -s(t^2)$. Unless both $s$ and $p$ are zero, we have reached a contradiction: the right hand side has even degree, whereas the left hand side must have odd degree.

Hence $r = 0$ and $f \in \langle y^2-x^3\rangle$ as desired.