# Is this function field extension a Galois extension ?

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 ?

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.

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Minor remark : you might want to assume $\mathrm{car}(k)=0$ in your question (in your example the field extension is not separable if $\mathrm{car}(k)=2$). –  François Brunault Feb 21 '12 at 12:38
Yes I do ! Thank you for your remark. –  Lierre Feb 21 '12 at 12:40
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. –  Jason Starr Feb 21 '12 at 13:16
Thank you ! I'll study your counter-example. –  Lierre Feb 21 '12 at 13:22
@Jason : Maybe you could write this as an answer, so the question doesn't remain in the "unanswered" category. –  François Brunault Feb 22 '12 at 11:59
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.