$\mathbb{R}$ is not a variety when considered as a subset of $\mathbb{C}$

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, $$V(S)=\{a\in \mathbb{K}^n:f(a)=0\:\forall f\in S\}$$

• Polynomials in $x+iy$ have complex derivatives, but $\overline{z}$ doesn't have complex derivative. – conditionalMethod Oct 16 '19 at 9:25
• For 2), quote the definition of "variety" to be used. Without the definition, we can prove nothing about it. – GEdgar Oct 16 '19 at 9:50
• @GEdgar I have added the definition of variety being considered – cookiemonster Oct 16 '19 at 9:57
• 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. – GEdgar Oct 16 '19 at 10:02

1 Answer

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.