Let $\mathbb{A}_{\mathbb{R}}^n$ be $\mathbb{R}^n$ endowed with the Zariski topology, where closed sets are algebraic sets (in $\mathbb{R}^n$) defined by real polynomials.

Suppose $V \subseteq \mathbb{A}_{\mathbb{R}}^n$ is an irreducible affine variety. Let $U$ be an open (with respect to the usual topology) ball $U$ around a non-singular point of $V$ and of small enough radius.

Does it then follow that the Zariski closure of $(V \cap U)$ is $V$? I thought it should be true (maybe not?), but I was wondering how I can show this. Any comments are appreciated. Thank you!


1 Answer 1


As noted in this answer to a previous question of yours, for any subvariety $W \subseteq \mathbb A_\mathbb R^n$, we have $$\dim_{\mathbb R} W(\mathbb R) \leq \dim W$$ (where $\dim W$ denotes the dimension in the sense of scheme theory, written there as $\dim_\mathbb C W(\mathbb C)$), with equality if $W$ has a smooth $\mathbb R$-point. In particular, if $W \subseteq V$ is the Zariski closure of $V \cap U$, then applying the above to both $V$ and $W$ gives $$\dim_\mathbb R W(\mathbb R) \leq \dim W \leq \dim V = \dim_\mathbb R V(\mathbb R).$$ But $W(\mathbb R)$ contains the full-dimensional subset $V(\mathbb R) \cap U$ of $V(\mathbb R)$, hence all dimensions must be equal. Since $V$ is irreducible and $\dim W = \dim V$, this forces $W = V$. $\square$

  • $\begingroup$ How do you define the dimension of $V(\mathbb R) \cap U$? $\endgroup$
    – LSpice
    Jul 8, 2018 at 23:20
  • $\begingroup$ @LSpice: this wasn't answered by the OP in the other question either. In my reading, a topological dimension was meant, so I was thinking of something like Lebesgue covering dimension (or inductive dimension, but they agree). $\endgroup$ Jul 9, 2018 at 0:24

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