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Jan 10, 2017 at 20:26 comment added Neil Epstein What if $k(Y)$ is only algebraically closed in $k(Z)$ (without the pure transcendentality condition)? In that case, the normalization of $Y$ in $k(Z)$ is generically just $Y$, right? I'd like to know whether the fiber product is integral in that case too.
Apr 26, 2015 at 8:36 vote accept Ron
Apr 25, 2015 at 15:43 history answered Will Sawin CC BY-SA 3.0