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**Sample text**

Then the behavior of pOK is determined by the factorization of the polynomial x2 −d in Fp [x].

2 These roots√are not in R, so R is not integrally closed in K. On the other hand, R = Z[ −1+2 −3 ] is integrally closed in K, as we will show shortly. We have therefore found a way to distinguish between these two choices for special subring of K. Note that as promised the property of being integrally closed corresponds to R being “large enough” in K; that is, R can not leave out any elements of K which are integral over R. What we are looking for, then, is a ring R which has K as its field of fractions, which is integrally closed in K, and which is as small as possible given the first two conditions.

If p does divide j, then ζ j = 1, so it has only the one conjugate 1, and TrK/Q (ζ j ) = p − 1. 48 2. RINGS OF INTEGERS By linearity of the trace, we find that TrK/Q (1 − ζ) = TrK/Q (1 − ζ 2 ) = · · · = TrK/Q (1 − ζ p−1 ) = p. We also need to compute the norm of 1 − ζ. For this, we use the factorization xp−1 + xp−2 + · · · + 1 = Φp (x) = (x − ζ)(x − ζ 2 ) · · · (x − ζ p−1 ); plugging in x = 1 shows that p = (1 − ζ)(1 − ζ 2 ) · · · (1 − ζ p−1 ). Since the 1 − ζ j are the conjugates of 1 − ζ, this shows that NK/Q (1 − ζ) = p.