This is exercise 12 from chapter 1.4 of Ideals, Varieties and Algorithms.
Let $V$ the affine variety parametrized by $(t^2,t^3,t^4)$ for $t\in\mathbb{R}$. I'm trying to find the ideal $\mathcal{I}(V)$. By the way it is asked, I'm assuming that I should do something similar to what is done on page 34.
I have found that $V=\mathcal{V}(y^2-xz, z-x^2)$, so I'm trying to prove that $\mathcal{I}(V)=\langle y^2-xz, z-x^2\rangle$. One inclusion is clear. Therefore, given $f\in\mathcal{I}(V)$, I must show that $f\in \langle y^2-xz, z-x^2\rangle$. So, following page 34, I should write $f$ as $h_1(y^2-xz)+h_2(z-x^2)+r$, for some suitable $h_1,h_2,r\in\mathbb{R}[x,y,z]$. By the division algorithm I could get $r\in\mathbb{R}[x,y]$ but the exponent of $y$ is at most $1$ (Am I right?).
If I evaluate $f(t^2,t^3,t^4)=0$, I get $r(t^2,t^3)=0$. But, unlike on page 34, now $r$ is a bivariable polynomial, so I cannot deduce that $r=0\in\mathbb{R}[x,y]$ just because it vanishes along the curve $(t^2,t^3)$. What can I do then?
Edit:
I've got $r(x,y)=\sum_{i=0}^na_ix^i+y\sum_{j=0}^mb_jx^j+c$. Evaluating,
$r(t^2,t^3)=\sum_{i=0}^na_it^{2i}+\sum_{j=0}^mb_jt^{2j+3}+c=0$.
The terms in the first sum have all even exponent and the terms on the right have odd exponent. Hence, I can conclude $a_0+c=0$, $a_i=0$ ($1\leq i\leq n$) and $b_j=0$ ($0\leq j\leq m$). Thus, $r$ is in fact the zero polynomial.
Is it right?
So you know the degree of $y$ in $r$ is at most one. Let $r(x, y) = p(x) + y q(x)$. One knows that $f(t) = r(t^2, t^3) = p(t^2) + t^3 q(t^2) \in \mathbb{R}[t]$ is the zero polynomial. This implies that each of $p(t^2)$ and $q(t^2)$ are zero, since they each have mutually exclusive degrees. But these are polynomials in $x$, and of course the polynomial being zero when substituting $x \mapsto t^2$ implies that the original polynomial was zero. So $r(x, y) = 0$.