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Jun 12, 2012 at 7:32 comment added Olivier Benoist I edited my answer ; it should be easier to read. The equality L'=L\otimes_K K' comes by base change from the cartesian square involving \mathbb{A}^1_K, C,\mathbb{A}^1_{K'} and C_{K'}$.
Jun 12, 2012 at 7:26 history edited Olivier Benoist CC BY-SA 3.0
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Jun 12, 2012 at 1:01 comment added David Zureick-Brown @Olivier: Thanks for the response! One question: why is L' = L \otimes_K K'? It seems like for K = Q, K' = Q(i), C = x^2 + y^2 + 1 = 0, and p' = (0,i), that L = L'.
Jun 11, 2012 at 23:35 history answered Olivier Benoist CC BY-SA 3.0