Compute the Zariski closure of a set

Question

I have to compute the Zariski closure of the image of the following rational map: $f:{P}^2 \rightarrow \mathbb{P}^4$ $[x_0:x_1:x_2]\rightarrow [x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2]$

I have already proved that this map is a morphism from $U$ to $\mathbb{P}^4$, where $U=\mathbb{P}^2 \setminus \{[1:0:0]\}$ and that cannot be extended to a map defined on the whole $\mathbb{P}^2$.

Now, I don't know how to compute the Zariski closure of $f(U)$. I know the useful relation $\overline{S}=V(I(S))$ for a set $S$: does this mean that I have to compute the ideal of $f(U)$, or is there a better way?

Moreover, I am asked to say if this closure of $f(U)$ is irreducible and what is its dimension. I have already proved that the closure is irreducible if and only if $f(U)$ is irreducible, but I still do not know how to proceed.

Any help would be highly appreciated. Thanks.

Answer 1

One way to compute equations for the Zariski closure of the image is to find the kernel of the graded ring homomorphism $k[y_0,y_1,y_2,y_3,y_4]\to k[x_0,x_1,x_2]$ by $y_0\mapsto x_0x_1$, $y_1\mapsto x_0x_2$, $y_2\mapsto x_1^2$, $y_3\mapsto x_1x_2$, $y_4\mapsto x_2^2$. There are a few obvious relations coming from considering different products of $y_i$s which are equal after plugging in the $x_i$s, and it turns out that these obvious relations actually generate the kernel and cut out the closure (exercise: prove this).

You're almost there on the irreducibility - the only missing ingredient for you is that the image of an irreducible set is irreducible. For more details, see here on MSE.

To find the dimension, you can work locally: on the affine open $D(y_4)$, you have a map $\Bbb A^2\to\Bbb A^4$ given by $(\frac{x_0}{x_2},\frac{x_1}{x_2})\mapsto (\frac{x_0x_1}{x_2^2},\frac{x_0}{x_2},\frac{x_1^2}{x_2^2},\frac{x_1}{x_2})$, which shows that the dimension of the image is 2.