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By R. Gordon, A. J. Power, Ross Street

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This proves that p is a sum of two squares. If x0 and y0 are 47 also rational integers with p = x20 + y02 , then (x0 + iy0 )(x0 − iy0 ) = (x + iy)(x − iy). By a previous homework assignment, the norm of each of these four factors being a rational prime implies they are each irreducible. By Theorem 58, the units in R are just ±1 and ±i. Since R is a UFD, we deduce that x0 + iy0 is (x ± iy) for some in {1, −1, i, −i}. In any case, {x20 , y02 } = {x2 , y 2 }, and the theorem follows. • When can a positive integer be written as a sum of two squares?

N−1 ) is called a Van der Monde determinant, and it will follow by our first lemma below that it is non-zero. This then will imply that ∆(ω (1) , . . , ω (n) ) is non-zero for some algebraic numbers ω (1) , . . , ω (n) in Q(α). Given two bases (not necessarily integral), say {ω (1) , . . , ω (n) } and {ω (1) , . . , ω (n) }, the values of ∆(ω (1) , . . , ω (n) ) and ∆(ω (1) , . . , ω (n) ) differ by the square of a rational number. If the numbers ω (1) , . . , ω (n) , ω (1) , . . , ω (n) are algebraic integers and {ω (1) , .

N) ∈ Q(α), then ∆(β (1) , . . , β (n) ) = det T rQ(α) (β (i) β (j) ) . Proof. See the proof of Theorem 40. Theorem 44. Consider the basis 1, α, . . , αn−1 for Q(α) over Q. If f (x) is the minimal polynomial for α, then ∆(1, α, . . , αn−1 ) = (−1)n(n−1)/2 NQ(α) f (α) . Lemma. Let x1 , . . , xn be n variables. Then  1 1  det  .  .. x1 x2 .. 1 xn x21 x22 .. x2n ··· ··· .. 2 xn−1 1  xn−1 2  n(n−1)/2 ..  = (−1)  . ··· xn−1 n (xi − xj ). 1≤i≤n 1≤j≤n i=j Proof. It is clear that the determinant on the left-hand side is 0 if xi = xj for any distinct i and j.

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Coherence for Tricategories by R. Gordon, A. J. Power, Ross Street


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