**Setting and question**


Let $X$ be a variety over an algebraically closed field of null characteristic, and let $C$ be a (regular if you want) curve included in $X$.
Consider $X'$ the normalization of $X$ and $C^\*$ the sheaf-theoric pull-back of $C$ in $X$. Assume that $C^\*$ is reduced, or even regular if you want.


The function field of each irreducible component of $C^\*$ gives an extension of the function field of $C$. On all the examples that I've been able to compute, these extensions are Galois extension. How to prove it as a general fact ?




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**Example**


Let $X$ be the surface defined by $A = k[x,y,z]/(x^2-zy^2)$, and let $C$ be the curve given by $(x,y)$. Then $A'$ is $A[x/y]$, — that is to say $k[u,y,z]/(u^2-z)$, with $u=x/y$ —, and $C^\*$ is given in $A'$ by the ideal $(y)$.


Thus, the field extension is $k(\sqrt{z}) | k(z)$, which is Galois.



Is this function field extension a Galois extension ?
As requested by Francois: No, this is typically not correct. In affine space with coordinates $x$,$y$ and $z$, consider the variety cut out by the single equation $y^3−3yz^2−xz^3$. This is not normal; the normalization is obtained by adjoining the fraction $u=y/z$. The normalization is itself isomorphic to a hypersurface in the affine space with coordinates $x$, $u$, and $z$ with equation $u^3−3u−x$. The curve in the original variety cut out by $y=z=0$ pulls back to the smooth curve cut out by $z=0$. The map between the curves is degree 3 and not Galois.


