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Consider the following projective algebraic set

$$\mathcal{V}(x_0^2+x_1x_2,x_0^2-x_1x_2+2x_3^2)\subset\mathbb{P}^3_{\mathbb C}.$$

I need to find the irreducible components. My approach has been the following.

If we set $x_0=0$ then we obtain $x_1x_2=0$ and $x_3=0$, so we get two points at infinite $P_1=[0:1:0:0]$ and $P_2=[0:0:1:0]$.

Otherwise, setting $x_0=1$ and adding both equations we get that $1+x_3^2=0$; and $1+x_1x_2=0$. But the first one can be decomposed as $(1+ix_3)(1-ix_3)$. So we get

$$ \mathcal{V}(x_0^2+x_1x_2,x_0^2-x_1x_2+2x_3)=\mathcal{V}(1+x_1x_2,1+ix_3)\cup\mathcal{V}(1+x_1x_2,1-ix_3)\cup P_1\cup P_2.$$

Is this correct? I feel like something is off.

Thanks in advance.

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1 Answer 1

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Your calculations in the affine patch look correct to me. You can simplify it a bit more by doing all the computations in projective space rather than passing to an affine patch (which is what you're doing when you set $x_0 = 1$). Indeed, adding the two defining equations yields $x_0^2 + x_3^2 = 0$, which can be factored as $(x_0 + ix_3)(x_0 - ix_3) = 0$, so we get $$\mathcal{V}(x_0^2+x_1x_2,x_0^2-x_1x_2+2x_3^2)=\mathcal{V}(x_0^2+x_1x_2,x_0+ix_3)\cup\mathcal{V}(x_0^2+x_1x_2,x_0-ix_3),$$ with no need to treat the two points where $x_0 = 0$ separately. You also need to verify that these two components are not contained in each other and are themselves irreducible, but that should be pretty straightforward.

Note that the points you denote $P_1$ and $P_2$ are not irreducible components; in fact, they're the intersection of the two irreducible components. So for your work to give the decomposition into irreducible components, you need to re-homogenize the polynomials in $x_0$ so they're closed subvarieties in projective space, and then recognize that $P_1$ and $P_2$ are contained in these and hence aren't irreducible components themselves.

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  • $\begingroup$ Thank you so much for your answer! I have a question - the number of irreducible components of an algebraic set is not unique? Or the way I expressed the algebraic set is not valid as a decomposition in irreducible components? $\endgroup$ Oct 12, 2022 at 16:39
  • $\begingroup$ The decomposition into irreducible components is unique. The points $P_1$ and $P_2$ aren't irreducible components, they're the points in the intersection of the two irreducible components. I've edited my answer to clarify. $\endgroup$ Oct 12, 2022 at 19:40
  • $\begingroup$ That was very clarifying, thank you kindly! $\endgroup$ Oct 12, 2022 at 20:28

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