I'm taking an algebraic geometry course and I'm trying to solve the next problem: Take the curve $V$ parametrized by $(t,t^3,t^4) \in \mathbb{R}^3$. Show that $V$ is an algebraic set and find $\mathbb{I}(V)$, where $\mathbb{I}(V)$ denotes the ideal of all polynomials that are zero over $V$.
So far, I've shown that $V$ can be seen as $Z(\langle y-x^3, z-x^4 \rangle)$, where $Z(I)$ is the set of zeros of the ideal $I$. Now I'm trying to calculute the ideal $\mathbb{I}(V)$ but I don't know how to do so, the only thing I've shown is that this ideal must contain the ideal $\langle y-x^3, z-x^4 \rangle$. I suspect those ideals coincide but I can't seem to prove it. Any suggestions on how to do so?
The most straightforward way to deal with these statements is to use the division algorithm. Indeed, let $f \in \mathbb{I}(V)$. We may then write $f = q*(y - x^3) + g$ where $q$ is a polynomial and $g \in \mathbb{R}[x,z]$. Similarly, $g = h*(z - x^4) + r$ where $r(x) \in \mathbb{R}[x]$. To conclude, we show $r = 0$.
Indeed, since $f(t, t^3, t^4) = 0$ it follows that $r(t) = 0$ for all $t \in \mathbb{R}$ so $r$ is the zero polynomial.
EDIT: I wanted to quickly describe what I meant by the "division algorithm". Let $A$ be a ring and $f \in A[x]$ a polynomial, written as $f = \sum f_i x^i$ for $f_i \in A$. Then, let $g = x^n + g_{n - 1} x^{n - 1} + \cdots + g_0$ be a monic polynomial. We claim that we can write $f = q*g + r$ where the $x$-degree of $r$ is less than $n$.
Indeed, if $\deg f < n$ then we're already done. Otherwise, let $f_mx^m$ be the highest term. Then, $f - (f_m x^{m - n})* g$ has strictly lower degree than $f$. Repeating we eventually arrive at a polynomial of lower degree than $g$. Note that each step we are subtracting a polynomial multiple of $g$, so all together we may write $f = q*g + r$ where $r$ is of lower degree.
In my above calculation, when we divide $f$ by $(y - x^3)$ we're taking $A = \mathbb{R}[x,z]$ and dividing with respect to the $y$ variable. This gives a remainder $g$ which has $y$-degree zero, so this remainder $g \in A = \mathbb{R}[x,z]$. We then divide $g$ by the monic polynomial $z - x^4$ where we take $A = \mathbb{R}[x]$.