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Extra info for Algebraic Number Theory [Lecture notes]
Example text
Since Trk/Q (1 · 1) = 2, Trk/Q (1 · −13) = 0 and Trk/Q ( −13 · −13) = −26, we get that dk = 2(−26) = 52. 1, we √ get that √ every ideal class contains an integral ideal of norm at most √12 52 = 26 < 6. So we need to look at ideals of norm 2, 3, 4, and 5. As before, the ideals of norm p = 2, 3, or 5 must be prime and they must be above the corresponding principal ideals (p). For p = 2, we get again that ok /(2) ∼ = Z2 [x]/((x + 1)2 ) and√so there is exactly one prime ideal above (2), namely the ideal I = (2, 1 + −13).
Thus, β ∈ p, and so p is nonzero. We conclude this section with a statement classifying non-Archimedean valuations, which is given without a proof. We need some notation. For a nontrivial non-Archimedean valuation || · ||, let p be the corresponding prime ideal in ok and for a nonzero fractional ideal F let vp (F ) be the exponent of p in the unique factorization of F . Note that the integer vp (F ) can be positive, negative, or zero, but it is nonnegative if F is an (integral) ideal. 10 Suppose || · || is a nontrivial non-Archimedean valuation of a number filed k and let p and vp be as above.
Let t be the order of GT . Then αt = 1 for all α ∈ GT , hence GT consists of all tth roots of unity from C. The group of all tth roots of unity is cyclic generated by the primitive tth root of unity. Therefore, GT is cyclic. √ As an example, let consider the group of units in ok , where k = Q( m), where m is a √ negative square-free integer. √ if m is nor equal to 1 modulo 4 then ok = Z[ m]. Since normk/Q (a + b m) = a2 − mb2 , we see that the norm is nonnegative. Furthermore, unless m = −1, there is only two units, 1 and −1.
