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If $f : \mathbb{R} \to \mathbb{R}$ is a univariate irreducible polynomial, Galois theory says that all roots are equivalent up to field automorphism (specifically, an automorphism of the field extension fixing the base field).

Can anything similar be said for the multivariate case? Specifically, if $f : \mathbb{R}^d \to \mathbb{R}$ is an irreducible multivariate polynomial, is the local geometry of the zero set $V = f^{-1}(0)$ similar at different points in some sense?

Clearly this isn't true at all points of $V$; the easiest counterexample is the lemniscate (http://mathworld.wolfram.com/Lemniscate.html), where $V$ is manifold except at one point. However, $V$ does have some self similarity except at this lower dimensional set: in particular, the curvature elsewhere is nonzero. As another example, I would expect that $V$ is locally a ruled surface either almost everywhere or almost nowhere. Is there a general statement that would encompass these and related similarities between different points of $V$?

A hint at a possible answer: I believe the statement is true for any geometric property that can be expressed algebraically in terms of the differential structure near $x \in V$, since otherwise adding these equations would separate $V$ into multiple components, contradicting the irreducibility of $V$.

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  • $\begingroup$ Irreducible varieties over non-algebraically closed fields are somewhat subtle though. For instance, $V (x^2 + 1) \subseteq \mathbb{A}^1_{\mathbb{R}}$ is irreducible but becomes reducible after base change to $\mathbb{C}$. $\endgroup$ – Zhen Lin May 28 '13 at 7:09
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In general, it is not true that points on an algebraic variety "look the same". If $V$ is a smooth variety, and $p, q \in V$, it is not true that there are necessarily neighborhoods $U,V$ of $p,q$ which are isomorphic by an isomorphism taking $p$ to $q$ (a property which is obvious in differential geometry). If $V$ is smooth and projective, such an isomorphism would actually extend to an automorphism of $V$ taking $p$ to $q$. So in fact, on a sufficiently generic variety, no two points look the same!

What I've said amounts to the following thing: for different points, the local rings $\mathcal O_p$ are generally different. However, the process of completion allows us to make sense of the intuition from differential geometry. The local ring $\mathcal O_p$ is a local ring with maximal ideal $\mathfrak m_p$. Its dimension is equal to the dimension of $V$. We can take the formal completion $\hat{\mathcal O_p}$ along the maximal ideal $\mathfrak m_p$; this has essentially the effect of replacing rational functions with power series. Taking Taylor expansion of a rational function allows us to embed the ring $\mathcal O_p$ into $\hat{\mathcal O_p}$. If $V$ is smooth at $p$, the local ring $\mathcal O_p$ is a regular local ring (this is what it means to be smooth); and we have the following theorem characterizing the completion of a regular local ring of dimension $n$ over a field:

Theorem: If $(A, \mathfrak m)$ is a regular local $k$-algebra of dimension $n$, then $\hat A \cong k[[x_1, \dots, x_n]]$, the ring of formal power series in $n$ indeterminates.

Thus, all of the complete local rings $\hat{\mathcal O_p}$ are isomorphic, if $V$ is smooth. However, there is generally no preferred isomorphism from $\hat{\mathcal O_p}$ to $\hat{\mathcal O_q}$.

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  • $\begingroup$ Thanks, that answers a strong form of the question (exact similarity of finite neighborhoods is unlikely), but not a much weaker (and vaguer!) form of the question: what infinitesimal geometric properties are different points likely to share? $\endgroup$ – Geoffrey Irving May 28 '13 at 16:41
  • $\begingroup$ @GeoffreyIrving: You are welcome. I have added a bit of meat to my answer. Best regards, $\endgroup$ – Bruno Joyal May 28 '13 at 18:58
  • $\begingroup$ @Bruno, I promised an upvote. :) But I wanted to comment that there's certainly no preferred homeomorphism or diffeomorphism in the case of manifolds. What's more, since you said differential geometry, it's very unlikely that there will be isometric neighborhoods. $\endgroup$ – Ted Shifrin May 30 '13 at 3:26

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