2
$\begingroup$

Suppose $V$ is an affine algebraic set defined by real polynomials. Let $\mathbb{A}_{\mathbb{R}}^n$ be $\mathbb{R}^n$ endowed with Zariski topology where the closed sets are algebraic sets (in $\mathbb{R}^n$) defined by real polynomials. Let $\mathbb{A}_{\mathbb{C}}^n$ be the usual affine $n$ space. Suppose $V(\mathbb{R})$ has at least one non-singular point. Then I think it should be that $$\dim_{\mathbb{A}_{\mathbb{R}}^n} V(\mathbb{R}) = \dim_{\mathbb{A}_{\mathbb{C}}^n} V(\mathbb{C}).$$

I was wondering how I can prove this... Any comments are appreciated. Thank you very much.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

The upper bound on the real Dimension needs no assumption on the non-singular point. This frees us to perform induction on $n$. With finitely many exceptions, the fibers of the projection to the first coordinate have complex dimension one less than $\dim V$, and the remainder have dimension $\dim V$. Hence by induction their real dimensions are bounded by the same. Whatever your definition of dimension, it should be easy to conclude.

The lower bound follows from considering the tangent space at the non-singular point.

Improve this answer
$\endgroup$
2
  • $\begingroup$ "The lower bound follows from considering the tangent space at the non-singular point." Can you explain it in more details? Thanks $\endgroup$
    – Pew
    Commented Sep 30, 2022 at 20:33
  • $\begingroup$ @Pew The tangent space at the non-singular point is an $n$-dimensional complex vector space, defined over $\mathbb R$, thus contains an $n$-dimensional real subspace. Consider a projection to $\mathbb R^n$ that is injective on this vector space. By the implicit function theorem, using the nonsingularity of $V$, this projection is locally an isomorphism $V(\mathbb R) \to \mathbb R^n$. $\endgroup$
    – Will Sawin
    Commented Sep 30, 2022 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .