Timeline for How to show $\dim_{\mathbb{A}_{\mathbb{R}}^n} V= \dim_{\mathbb{A}_{\mathbb{C}}^n} V$?
Current License: CC BY-SA 3.0
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| Sep 30, 2022 at 21:04 | comment | added | Will Sawin | @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 $V(\mathbb R) \to \mathbb R^n$. | |
| Sep 30, 2022 at 20:33 | comment | added | Pew | "The lower bound follows from considering the tangent space at the non-singular point." Can you explain it in more details? Thanks | |
| Apr 19, 2018 at 9:15 | vote | accept | Johnny T. | ||
| Apr 4, 2018 at 18:02 | history | answered | Will Sawin | CC BY-SA 3.0 |