The real dimension of any real algebraic set equals the complex dimension of its complexification

Question

I want to prove the following statement. Please help!

Given any semialgebraic set $A$, consider its real Zariski closure $V_{\mathbb{R}}$ (which always has the same real dimension of $A$). Now consider the complex ideal $I_{\mathbb{C}}(V_{\mathbb{R}})$, and its complex solution set $V_{\mathbb{C}}$. Note that $V_{\mathbb{R}}$ is the restriction to $\mathbb{R}^{n}$ of $V_{\mathbb{C}}$ (which we can assume irreducible). Prove that the complex dimension of $V_{\mathbb{C}}$ equals the real dimension of $V_{\mathbb{R}}$.

One avenue is to emulate Tao's solution to Every real variety contains non-singular points and prove that the complex dimension of $V_{\mathbb{C}}$ equals the real dimension of $V_{\mathbb{R}}$ by working with smooth points, as follows.

If $V_{\mathbb{C}}$ is not invariant by complex conjugation, then one may intersect this variety with its complex conjugate, dropping the dimension of this variety without affecting $V_{\mathbb{R}}$ as a set. Thus, by induction on dimension, one may assume that $V_{\mathbb{C}}$ is invariant by complex conjugation, that is is definable over the reals.

Now, we're left by induction with some smooth real point of $V_{\mathbb{C}}$. At such point of $V_{\mathbb{C}}$, the complex tangent space is invariant by complex conjugation, and is thus the complexification of a real space of the same dimension.

However, from here I can't conclude that $V_{\mathbb{R}}$ is smooth at the point, unless by using the equality between dimension which I want to prove in the first place!

Also check out the comments under https://mathoverflow.net/a/262737/501777

Answer 1

Let $x$ be a real point of a complex variety $V_\mathbb C \subseteq \mathbb C^n$ that is smooth of dimension $\dim(V_\mathbb C)=m$ and defined by polynomials over the real numbers. Then we can always choose exactly $n-m$ elements $f_1,\dotsc, f_{n-m}$ of the ideal of $V_\mathbb C$ whose derivatives together generate the conormal space to $V_{\mathbb C}$ at $x$. The complex vanishing set of $f_1,\dotsc, f_{n-m}$ is smooth of dimension $m$ at $x$ by the complex intermediate value theorem, and the real vanishing set of $f_1,\dotsc, f_{n-m}$ is smooth of dimension $m$ in a neighborhood of $x$ by the real intermediate value theorem.

If $V_{\mathbb R}$ is not smooth of dimension $m$ at $x$, then it is a proper subset of the real vanishing set of $f_1,\dotsc, f_{n-m}$ in every neighborhood of $x$, so $V_\mathbb C$ is a proper subset of the complex vanishing set of $f_1,\dotsc, f_{n-m}$ in every neighborhood of $x$, contradicting that $V_\mathbb C$ is smooth of dimension $m$ at $x$.

Answer 2

I think this topic is covered in Whitney, Hassler. "Elementary structure of real algebraic varieties." Annals of Mathematics Second Series, Vol. 66, No. 3 (Nov., 1957), pp. 545-556 (link). See esp. Lemmas 6, 8, and 9.