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Mar 23, 2011 at 18:22 comment added Ramsey @Lubin. Thanks - that's the formula I had in the back of my mind, but didn't think it through. I suppose I should have been a little more specific and a little less lazy...
Mar 23, 2011 at 16:38 comment added Lubin "Grows exponentially": sure, in the $p=2$ case, $K^*/(K^*)^2$ has dimension $n+2$ over $\mathbb{Z}/(2)$, so there are $2^{n+2}-1$ quadratic extensions over a field of degree $n$ over $\mathbb{Q}_2$
Mar 23, 2011 at 13:49 history answered Ramsey CC BY-SA 2.5