Is this function field extension a Galois extension ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:10:58Z http://mathoverflow.net/feeds/question/89098 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89098/is-this-function-field-extension-a-galois-extension Is this function field extension a Galois extension ? Lierre 2012-02-21T12:18:31Z 2012-02-22T23:16:41Z <p><strong>Setting and question</strong></p> <p>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 <code>$C^*$</code> the sheaf-theoric pull-back of $C$ in $X$. Assume that <code>$C^*$</code> is reduced, or even regular if you want.</p> <p>The function field of each irreducible component of <code>$C^*$</code> 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 ?</p> <hr> <p><strong>Example</strong></p> <p>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 <code>$C^*$</code> is given in $A'$ by the ideal $(y)$.</p> <p>Thus, the field extension is $k(\sqrt{z}) | k(z)$, which is Galois.</p> http://mathoverflow.net/questions/89098/is-this-function-field-extension-a-galois-extension/89235#89235 Answer by Jason Starr for Is this function field extension a Galois extension ? Jason Starr 2012-02-22T23:16:41Z 2012-02-22T23:16:41Z <p>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.</p>