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Does anyone know an example of a curve $X$ over a perfect field $k$ such that if $\tilde{X}$ is its noramlisation, there exists a point $x \in X$ and a point $y \in \tilde{X}$ over $x$ such that $k(y) / k(x)$ is not trivial?

(If we remove the hypothesis that it is over a field, there is such an example: $\mathbb{Z}[11\sqrt{3}] \to \mathbb{Z}[\sqrt{3}]$ and the points $(11, 11\sqrt{3}), (11)$.

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Maybe it's not clear from the answers given, but one can get any finite extension whatsoever (of $k$ for simplicitly). For a finite extension $K$ of $k$ let $y$ be a closed point of $\mathbb{A}^1_k$ with $k(y) = K$. Let $A$ be the sub $k$-algebra of $k[X]$ generated by the maximal ideal $m$ correponding to $y$; if $m = (f)$ for an irreducible polynomial $f$ of degree $d$ then $A$ is generated by $f,Xf,\dots, X^{d-1}f$. The normalisation of $A$ is $k[X]$ and $m$ gives rise to a point $x$ of $X := Spec(A)$ lying below $y$ such that $k(x) = k$. – ulrich Aug 18 '11 at 11:16
Thanks ulrich. If you are interested in generalisations here is another question I've had kicking around for a while. Given a finite, flat, surjective morphism of varieties $f: Y \to X$ with $X$ normal, and a point $x \in X$ we obtain a morphism $\tilde{Y} \times_X x \to Y \times_X x$ of finite Artin algebras over $k(x)$. Which morphisms of finite Artin algebras over a field can be obtained in this way? – name Sep 1 '11 at 16:06
up vote 7 down vote accepted

Sure, consider $\text{Spec }\mathbb{R}[x, ix]$ which has normalization $\text{Spec }\mathbb{C}[x]$. Of course the real numbers $\mathbb{R}$ is a perfect field. Over the origin, the residue field changed from $\mathbb{R}$ to $\mathbb{C}$.

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In case you'd like an example which is geometrically integral (as Karl's isn't), here's one. Let $A$ be the subring of $\mathbb{R}[x]$ consisting of polynomials which take real values at $\pm i$. Explicitly, $A$ is generated by $u:=x^2+1$ and $v:=x(x^2+1)$, and can be written as $\mathbb{R}[u,v]/\langle v^2+u^2=u^3 \rangle$.

As you would expect, the ideal of polynomials vanishing at $\pm i$ (also known as the ideal $\langle u,v \rangle$) has residue field $\mathbb{R}$. The normalization is $\mathbb{R}[x]$ (in terms of $u$ and $v$, we have $x = v/u$), the preimage of $\langle u,v \rangle$ is $\langle x^2+1 \rangle$ and the corresponding residue field is $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.

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Correction: the equation is $v^2+u^2=u^3$. Ths is a plane cubic having the origin as an ordinary (real) double point with imaginary tangents. – Laurent Moret-Bailly Aug 18 '11 at 13:00
Corrected, thanks! – David Speyer Aug 18 '11 at 13:08

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