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Oct 30, 2013 at 12:44 comment added Cantlog One can also take $A=\mathbb Q$, $B=\mathbb Q(t)$ and $B'=\overline{\mathbb Q}$, in which case there is no morphism between $B$ and $B'$ from either sides.
Oct 28, 2013 at 15:22 history edited Julian Rosen CC BY-SA 3.0
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Oct 28, 2013 at 15:16 history edited Julian Rosen CC BY-SA 3.0
added 17 characters in body
Oct 28, 2013 at 15:11 history answered Julian Rosen CC BY-SA 3.0