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Consider the morphism: $$ f: (\mathbb{P}^2 -\{(0:0:1),(0:1:0) \} )\to \mathbb{P}^3 $$ Given by $f((x:y:z))=(x^2:xy:xz:yz)$, my problem is to find the closure of the image of $f$, my argument was: in the image $x$ is not vanishing so we have in affine coordinates $(1:\frac{y}{x}:\frac{z}{x}:\frac{yz}{x^2})$ i.e. is given by $Z(W-YZ)$ then considering the projective closure I have to homogenize the polynomial. Then $\overline{Imf}=Z(XW-YZ)$ is it correct? Thanks!

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Yeah, that's pretty much right. Let's say $[x:y:z]$ are homogeneous coordinates on $\mathbb{P}^2$ and $[X:Y:Z:W]$ are homogeneous coordinates on $\mathbb{P}^3$. Then the affine chart $U_x=\{[x:y:z] : x\ne 0\} \cong \mathbb{A}^2$, which has coordinates $(\frac{y}{x}, \frac{z}{x})$, is mapped into the affine chart $U_X = \{[X:Y:Z:W] : X\ne 0\} \cong \mathbb{A}^3$, which has coordinates $(\frac{Y}{X}, \frac{Z}{X},\frac{W}{X})$, by the affine morphism $(\frac{y}{x},\frac{z}{x}) \mapsto (\frac{y}{x},\frac{z}{x},\frac{y}{x}\cdot\frac{z}{x})$, whose image in $\mathbb{A}^3$ is given by $\frac{Y}{X}\cdot\frac{Z}{X} - \frac{W}{X} = 0$. As you say, the projective closure of this set is $C = Z(XW-YZ)$. So we know the closure of the image of $f$ is at least as big as $C$. Now you can check that on the rest of the domain, e.g. on $U_y\cap U_z$ (which is also affine), the image of $f$ is contained in $C$, so the closure of the image is actually $C$.

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