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Let $Y:z=x^{2}+y^{2}$ and $Z:z=-x^{2}-y^{2}$ be two affine varieties in $\mathbb{R}^{3}$, then how does it satisfy the Affine Dimension Theorem here (on Page 3)?

Grigory M
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celilia
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  • Thanks a lot for any helpful answers! – celilia Dec 18 '13 at 07:45
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    It's hard to tell with certainty, but in algebraic geometry - in particular when dealing with varieties as sets of points - one often assumes the ground field to be algebraically closed. This may or may not have been explained in an earlier chapter. Here in $\Bbb{C}^3$ the intersection is the union of two lines: $z=0,x=iy$ and $z=0, x=-iy$. In a more scheme theoretic language we would see something like the spectrum of $\Bbb{R}[x,y]/(x^2+y^2)$ in the intersection, but unfortunately I don't remember/know how intersection is defined in the language of schemes, so that's fuzzy. – Jyrki Lahtonen Dec 18 '13 at 08:05
  • @JyrkiLahtonen: Dear Jyrki, thank you for your comment, but is there a similar but different Affine Dimension Theorem for non-algebraic-closed ground field? Thanks a lot! – celilia Dec 18 '13 at 08:10
  • I think there is. If we consider the field of fractions $F$ of the domain $\Bbb{R}[x,y]/(x^2+y^2)$, we see that $y/x$ is a square root of $-1$ in there. So $F\cong\Bbb{C}(x)$, and that has transcendence degree one as promised. – Jyrki Lahtonen Dec 18 '13 at 08:24
  • It is just that over a field that's not algebraically closed, we need to redefine many things. The coordinate ring survives, but a point becomes (roughly) a maximal ideal of the coordinate ring. This leads to things as follows: the pair of conjugate points $(x,y)=(1,\pm i)$ coalesce to form "a point of degree two over the reals". This is particularly crucial to doing algebraic geometry over finite fields. – Jyrki Lahtonen Dec 18 '13 at 08:28
  • You get into all sorts of problems when dealing with non-algebraically-closed fields. Take for instance $((y^2 - 2)^2 + (x^2 - 3)^2 + z^2)\in \Bbb R^2[x, y, z]$. The polynomial is irreducible, but the corresponding solution set consists of four points, and is thus reducible of dimension 0 (instead of irreducible and of dimension $2$, as it "should"). – Arthur Dec 18 '13 at 08:32
  • See for example this answer for more discussion. – Jyrki Lahtonen Dec 23 '13 at 22:06

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