Suppose $X=V(F_1,\cdots, F_r) \subseteq \mathbb P^n $ is an (irreducible) projective variety and $\varphi:X\rightarrow \mathbb P^m $ a regular map $\varphi=[\varphi_0:\cdots :\varphi_n]$ with $\varphi_i\in k[x_0,\cdots, x_n]$ all homogeneous. The image of $\varphi$ will be a closed subset of $\mathbb P^n$: how does one go about finding the equations defining the image $\varphi(X)$?
Here's what I am thinking. Let $x_0,\cdots, x_m$ be the coordinates on $X$ and $y_0,\cdots, y_n$ the coordinates on $\mathbb P^n$. It suffices to find the generators of the $I(\varphi(X))$. The kernel of the induced map $\varphi^*:k[\mathbb P^m]\rightarrow k[X]$ should be $I(\varphi(X))$ since $$\varphi^*(f)=0 \iff f\circ \varphi=0 \iff f|_{\varphi(X)}=0.$$ So we only need to determine this kernel...?
I seem to remember reading somewhere (can't recall the source, so this is just me piecing things together...) that, for example, if $X=V(F)\subseteq \mathbb P^2 $ and $\varphi:X\rightarrow \mathbb P^2$ is defined by $y_i=\varphi_i(x_0,x_1,x_2)$ then the equations defining $\varphi(X)$ can be found by looking at the ideal $$ \big(y_0-\varphi_0(x_0,x_1,x_2),y_1 -\varphi_1(x_0,x_1,x_2), y_2-\varphi_2(x_0,x_1,x_2), F\big)\subseteq k[y_0,y_1,y_2,x_0,x_1,x_2], $$ the generators of which will be the equations defining the image. Why can the generators of this ideal be written purely in terms of the $y_i$? Why would these generators define the image?