Algebraic Number Theory [Lecture notes] by Sergey Shpectorov PDF

By Sergey Shpectorov

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Additional resources for Algebraic Number Theory [Lecture notes]

Example text

So C = ||n|| log n is independent of the value of n > 1. 1 Finally, we can set c = log C and note that ||n|| log n = ec implies ||n|| = (ec )log n = (elog n )c = nc for all n > 1. Now for an arbitrary integer n we clearly have ||n|| = |n|c , as claimed. c = |a| = | ab |c . So the claim holds. If ab ∈ Q with a, b ∈ Z then || ab || = ||a|| ||b|| |b|c The following more general statement (currently without proof) shows that up to equivalence all Archimedean valuations of a number field k come from the embeddings of k into C.

Since −5 maps to the coset √ x + (x1 ), we see that the √ two ideals above (3) are the ideals J1 = (3, 1 + −5) and J2 = (3, 1 − −5). We claim that both these ideals are contained in the√ same ideal √ class as I = (2, 1 √+ −5). Indeed, if we multiply I with 1− 2 −5 then we √ √ √ −5) get (1 − −5, (1+ −5)(1− ) = (1 − −5, 62 ) = J2 . Also, noticing that 2 √ √ √ I = (2, 1 + −5) = √(2, 1 −√ −5), we obtain that I multiplied with 1+ 2 −5 √ √ −5) ) = (1 + equals (1 + −5, (1− −5)(1+ −5, 26 ) = J1 . So all three ideals, I, 2 J1 , and J2 belong to the same ideal class.

Now we turn to the ideals above (3). Similarly to the above, ok /(3) ∼ = 2 2 Z3 [x]/(x − 1). Since x − 1 =√(x − 1)(x + 1), there exist exactly two proper 2 ideals in ok above (3). Since −5 maps to the coset √ x + (x1 ), we see that the √ two ideals above (3) are the ideals J1 = (3, 1 + −5) and J2 = (3, 1 − −5). We claim that both these ideals are contained in the√ same ideal √ class as I = (2, 1 √+ −5). Indeed, if we multiply I with 1− 2 −5 then we √ √ √ −5) get (1 − −5, (1+ −5)(1− ) = (1 − −5, 62 ) = J2 .

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Algebraic Number Theory [Lecture notes] by Sergey Shpectorov


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