Let $\mathbb{R}=\{z:z=\overline{z}\}$

"$\mathbb{R}$ is not a variety when considered as a subset of $\mathbb{C}$, since complex conjugation is not a polynomial operation"

I don't understand this particular statement.

1) Is it possible to show that there is no way to get from $x+iy$ to $x-iy$ using polynomial addition and multiplication?

2) Even if I consider the latter half of the statement, how does it prevent $\mathbb{R}$ from being a variety? Is it because then $\mathbb{R}$ will not form an ideal?

Edit: The definition of variety being considered,

Let $\mathbb{K}$ be a field. Let $S\subseteq \mathbb{K}[x_1,\cdots,x_n]$ be a set of polynomials. The variety defined by S is the set, $V(S)=\{a\in \mathbb{K}^n:f(a)=0\:\forall f\in S\}$

  • 3
    $\begingroup$ Polynomials in $x+iy$ have complex derivatives, but $\overline{z}$ doesn't have complex derivative. $\endgroup$ – conditionalMethod Oct 16 '19 at 9:25
  • $\begingroup$ For 2), quote the definition of "variety" to be used. Without the definition, we can prove nothing about it. $\endgroup$ – GEdgar Oct 16 '19 at 9:50
  • $\begingroup$ @GEdgar I have added the definition of variety being considered $\endgroup$ – cookiemonster Oct 16 '19 at 9:57
  • 2
    $\begingroup$ Okay. Elements of $\mathbb C[x_1]$ are complex polynomials (in one complex variable). For 2) we need to know: the zero-set of such a polynomial is either a finite set or $\mathbb C$. Then note that $\mathbb R$ is not the intersection of any collection of these. $\endgroup$ – GEdgar Oct 16 '19 at 10:02

The key thing to understand here is: which ground field am I working with ?

So I guess you want to know if $\mathbb{R}$ is a subvariety over $\mathbb{C}$ of the variety $\mathbb{C}$. So you know that a subvariety of $\mathbb{C}$ is given as the zero locus of a finite set $I\subset \mathbb{C}[X]$ of polynomials in one variable. Now, you know that such polynomials have finitely many zeros. Thus the zero locus of polynomials in $I$ must be finite (or empty, or $\mathbb{C}$ if $I=\{ 0 \}$). In particular, it cannot be $\mathbb{R}$.

In algebraic geometry, when thinking of variety over $\mathbb{C}$, you must forget the $\mathbb{R}^2$ interpretation of $\mathbb{C}$, but think $\mathbb{C}$ just as a algebraically closed field.


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