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I want to find the irreducible components of the algebraic set $Y$, in $\mathbb{A}^{3}$ given by the zero-locus of the equations $x^{2}-yz$ and $xz-z$. I also want to compute the dimension of $Y$.

Well if $xz-z=0$ then $x=1$ or $z=0$. If $z=0$ then $x=0$ so we obtain the $y-axis$. Now if $x=1$ then $yz-1=0$ hence:

$Y = V_{1}(x,z) \cup V_{2}(x-1,yz-1)$. (Here $V_{i}$ denotes the locus-set).

Now $k[x,y,z]/(x,z) \cong k[y]$ so $V_{1}(x,z)$ is irreducible because $k[y]$ is a integral domain.

I'm stuck in showing $V_{2}(x-1,yz-1)$ is irreducible.

Can you please help? (Also in general how do you compute the dimension?)

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We have that $k[V_2]=k[x,y,z]/(x-1,yz-1)\cong k[y,z]/(yz-1)$. You can show that the map from $V_2$ to $\mathbb{A}^1-\{0\}$ defined by $(x,y,z)\mapsto y$ is an isomorphism by showing that the corresponding morphism of coordinate rings $k[\mathbb{A}^1-\{0\}]=k[y,y^{-1}]\rightarrow k[y,z]/(yz-1)$ (i.e., the morphism defined by $y\mapsto y$, $y^{-1}\mapsto z$) is an isomorphism. Because $\mathbb{A}^1-\{0\}$ is a non-empty open set of the irreducible set $\mathbb{A}^1$, it is also irreducible, and hence $V_2$ is irreducible as well.

I feel like this may be a bit roundabout, but I think it works.

The dimension of $Y$ is 1, because a maximal-length chain of irreducible subsets of $Y$ will start at either of its irreducible components, each of which is dimension 1. The irreducible components are a line, which is clearly of dimension 1 (also: its coordinate ring is $k[y]$, which is a ring of dimension 1), and a hyperbola $yz-1$, which (as we showed above) is isomorphic to $\mathbb{A}^1-\{0\}$, and by Prop 1.10 in Hartshorne, $\dim(Y)=\dim(\overline{Y})$ for any quasi-affine variety $Y$.

I often find it helpful to try to visualize what's going on - here is the output from Mathematica. The two planes are the zero locus of $xz-z$, and the cone is the zero locus of $x^2-yz$ - their intersection is, as expected, the $y$-axis and a hyperbola which lies on the plane $x=1$.

enter image description here

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  • $\begingroup$ Sorry I meant $V_{1}=<x,z>$. $\endgroup$ – user6495 Feb 22 '11 at 2:23
  • $\begingroup$ @user6495, Ah, you are right. I have also added an explanation about the dimension. You can edit your question to correct the typo - there are buttons at the bottom left. $\endgroup$ – Zev Chonoles Feb 22 '11 at 2:37
  • $\begingroup$ @user6495 - No problem, glad I can help! It is good practice for me too, I just learned this stuff last semester :) $\endgroup$ – Zev Chonoles Feb 22 '11 at 3:42

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