2 characteristic zero

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.

1

Is this function field extension a Galois extension ?

Setting and question

Let $X$ be a variety over an algebraically closed field, 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.