Definition. A complex algebraic curve is the zero zet of a polynomial $P:\mathbb{C}^2 \to \mathbb{C}$.
If I am given a parametrisation of such a zero set, e.g. $P(t^2, t^3+1)=0, \forall t \in \mathbb{C}$, is there a constructive method to find a polynomial $P(x,y)$ that fulfills this condition for $x=t^2$ and $y=t^3+1$?
Here is a standard way to get an equation for the image of a plane parametrized curve:
$$x = \phi(t) \\ y = \psi(t)$$
From the first equality obtain a polynomial equation for $t$ with coefficients dependent on $x$, and similarly from the second:
$$P(t, x) = 0\\ Q(t, y) = 0$$
Now, impose the condition on the polynomials $P(t)$, $Q(t)$ in $t$ to have a common root $t$. This involves the resultant:
$$\operatorname{Res}(P, Q) = 0$$
The above will be an equation in $x$, $y$.
Let's take a simple example
$$x = \frac{1- t^2}{1+t^2} \\ y = \frac{2 t}{1+t^2}$$
Then
$$P(t, x) = (1+t^2) x - (1-t^2)\\ Q(t,x) = (1+t^2) y - 2 t$$ With WA we get
$$\operatorname{Res}(P(t), Q(t)) = 4(x^2 + y^2-1)$$
Yes, this can be done using elimination theory. See the reference to Cox, Little, and O'Shea's book I've given here or here.
We can find the desired polynomial by computing a Gröbner basis using SageMath. For the given $x$ and $y$, we find that they satisfy the polynomial $P(x,y) = x^3 - y^2 + 2 y - 1 = x^3 - (y-1)^2$.
