Let $\phi: \mathbb{R^1}\rightarrow \mathbb{R^2}$ be the map given by $t \mapsto (t^2,t^3)$; prove directly that any polynomial $f\in \mathbb{R}[X,Y]$ vanishing on the image $C=\phi(\mathbb{R^1})$ is divisible by $Y^2-X^3$.
Determine that property of a field $k$ will ensure that the result holds for $\phi: k\rightarrow k^2$ given by the same formula.
Here is how I did the first part:
By division algorithm, I can write
$$f(X,Y)=(Y^2-X^3)f_1(X,Y)+Yf_2(X)+f_3(X)$$
Since $f(t^2,t^3)=0$ for all $t\in \mathbb{R}$, we have, for all $t$,
$$t^3f_2(t^2)+f_3(t^2)=0$$
Comparing the odd and even power terms of $t$ gives
$$f_2(X)=f_3(X)=0$$ This shows $f$ is divisible by $Y^2-X^3$.
I am not sure how to do the second part though. It seems to me the argument could work for all fields, unless the division algorithm fails. But when does it fail? Or will the odd and even power argument fail at some point?
Thank you for your help!
As it has been pointed out, a sufficient condition is taking the ground field to be infinite. This can be traced back to the following very general considerations: if $f\in A[(X_i)_{i\in I}]$ is a polynomial ($A$ a commutative ring), then we get an induced map $\tilde{f}:A^I\rightarrow A$ obtained by substitution. It turns out that the map $f\mapsto\tilde{f}$ is injective whenever $A$ is an infinite integral domain (e.g. a field).