How to find an ideal of variety? I've been going through Ideals, Varieties, and Algorithms book by D. Cox et al., and have been stuck on exercise 4.12. I've solved it in a half, but have stuck on the next question: how to find an ideal $I(V)$ of an variety $V$? Mainly, in the exercise there is an affine variety $$ V = \mathbf{V}(y^2-xz, z-x^2), $$
which I'm particularly interested in.
I've already seen the pages on M.SE with similar question, e.g. this one, but I don't really understand why we just assume that the field is algebraically closed and then use Nullstellensatz. In the exercise there's nothing on this, and it seems it's assumed to be solved without Nullstellensatz. So how can one find an ideal of variety?
Earlier in the book, D. Cox solves a similar exercise (p. 33). There, D. Cox wants to show that $$ I(V) = I(\mathbf{V}(y-x^2, z-x^3)) = \left<y-x^2, z-x^3\right>. $$
In order to do this, he factors $f \in I(V)$ by the generators, so $$ f = h_1(y-x^2) + h_2(z-x^3) + r, $$ such that $r$ is actually from $k[x]$. Then, since $f \in I(V)$, on $V$ we have following (using the parametrization of $V$, for which $x=t, y=t^2, z=t^3$): $$ 0 = 0 + 0 + r(t), $$ and then $r=0$, and we are done (here $k$ is assumed to be infinite).
In the exercise it's said that the polynomial division algorithm will help in tackling this task, meaning that one can perform the same manipulations as above. But I can't still see how this can be done. The only intuition I had were the following:

Since the generators are $y^2-xz, z-x^2$, it seems that in ideal $\left<y^2-xz, z-x^2\right>$ any power of $x$ and $z$ can be obtained, whileas not every power of $y$ can be obtained (e.g. I can't get odd power). So the reminder in my case will be in $k[y]$, and then the same steps as in the book applies.

But I don't feel like it's the right direction.  Moreover, I know that it's not right: consider $f = x^2 + yx + 5z$, then the reminder is going to be $xy + 6z$, which breaks everything stated above.  (given $f \notin I(V)).$
So, can anyone help me to figure things out? Thanks!
 A: Many thanks to @red_trumpet for the helpful suggestion in the comments, which led to the solution I find appropriate.
Let's try to prove, that $$ I(V) = \left<y^2 - xz, z - x^2\right>. $$
It's obvious in one direction: $\left<y^2 - xz, z - x^2\right> \subseteq I(V)$, but what about the converse?
Let's suppose $f \in I(V)$, then if we prove that $f$ factors by both $y^2 - xz$, $z - x^2$, then, surely, $f \in \left<y^2 - xz, z - x^2\right>$. Let's then divide $f$ by generators. Clearly, one can obtain the next:
$$ f = q_1(z - x^2) + r, $$
where $r \in k[x,y]$. In other words, we can eliminate all terms that contain $z$ via dividing by $z - x^2$, and then a polynomial in $x,y$, the reminder $r$, is only left.
Now, we need to divide $r$ by $y^2 - xz = y^2 - x^3$ (since $z$ is eliminated). Any power of $y$, except $1$, can be eliminated by that division, and therefore what is left is just
$$ f = q_1(z - x^2) + q_2(y^2 - x^3) + r_1y + r_2, $$
where $r_1, r_2 \in k[x]$. As it's seen, that $r_1y$ term contains exactly the last $y$ term, which was not eliminated by the division by $y^2 - x^3$, while $r_1$ and $r_2$ are now the polynomials in $x$ only.
So, what's next? We need to show, that the reminder $r_1y + r_2$ is zero. Let's recall that $f \in I(V)$, so $f(v) = 0$ for $v \in V$. We have a parametrization for $V$:
$$ \begin{aligned}
x = t^2, \\
y = t^3, \\
z = t^4,
\end{aligned} $$
so, putting this into $f$ gives us the next:
$$ f(x,y,z) = f(t^2, t^3, t^4) = 0 = 0 + 0 + r_1(t^2)t^3 + r_2(t^2). $$
Or, shortly,
$$ 0 = r_1(t^2)t^3 + r_2(t^2). $$
Next, consider that $r_2(t^2)$ contains only even powers of $t$, but $r_1(t^2)t^3$ contains only odd, so, if we put $R(t) = r_1(t^2)t^3 + r_2(t^2)$, we will get that $0 = R(t)$. This holds for any $t$, and since $k$ is infinite, $R = 0$, as in the book example by D.Cox. So, $f \in \left<y^2 - xz, z - x^2\right>$, and therefore
$$ I(V) = \left<y^2 - xz, z - x^2\right>.  $$
