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I am trying to understand the Basic Argument from Chapter 9.1 of Washington's "cyclotomic fields", and I can understand all but one part. Under Assumption 1: $p \nmid h^{+}(\mathbb{Q}(\zeta_p))$ it says:

"Note that $\overline{B_0} = B_0$ and $(1-\zeta) \nmid B_0 $, so $B_0$ arises from $\mathbb{Z}[\lambda]$. Since $B_0^p$ is principal in $\mathbb{Z}[\lambda]$..."

We already know that $B_0$ is an ideal in $\mathbb{Z}[\zeta]$, and by definition $\lambda = (1-\zeta)(1-\zeta^{-1})$. We know $$(\omega + \theta) = (\lambda)^{m-\frac{p-1}{2}}B_0^p $$ with $\lambda, \omega, \theta \in \mathbb{Z}[\lambda]$ pairwise relatively prime.

But what does it mean for $B_0$ to "arise" from $\mathbb{Z}[\lambda]$ (I might not understand this part because of the language barrier) and how do we conclude that?

lewy
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1 Answers1

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that means that B0 is generated by elements in Z[λ]. In other words, it may be defined as the ideal produced by a few Z[λ] elements. It is concluded by the fact that B0 is principal and (1−ζ)∤B0, which means that (1−ζ) is not a factor of any element of B0. Since (1−ζ) is a factor of λ, and λ is in Z[λ], this means that B0 must be generated by elements in Z[λ] that do not involve (1−ζ). Therefore, B0 arises from Z[λ].

An example would be if B0 = <3, 1-ζ>, which is generated by the elements 3 and 1-ζ. Here, B0 is an ideal in Z[ζ], but it is generated by elements in Z[λ], which is the ring of integers of the extension of Q by a root of lambda, where lambda = (1-ζ)(1-ζ^-1). So, in this example, B0 "arises" from Z[λ].

Ahmed Izz Murtaja
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  • Thank you for the reply, I still don't understand thogh how we know B0 is principal in the first place and where do we use that the conjugate of every element of B0 is in B0? – lewy Jan 28 '23 at 23:35
  • "Arising from" in this context means that the ideal B0 is generated by elements of Z[λ], that is, it can be expressed as the set of all linear combinations of elements of Z[λ] with coefficients in Z. This is concluded from the fact that (1-ζ) is in B0 and (1-ζ-1) is in Z[λ], and B0 is generated by the product (1-ζ)(1-ζ-1)=λ. – Ahmed Izz Murtaja Jan 28 '23 at 23:39
  • Regarding B0 being principal, a principal ideal is an ideal that is generated by a single element, so it is a principal ideal if there exists a single element of the ring that generates it. In the case of B0, it is principal because it is generated by the element λ. The fact that B0 is principal in Z[λ] is important for the argument in the text because it means that Bp0 can be expressed as (λ)^(m-p-1/2) for some m, which is a key step in the argument. – Ahmed Izz Murtaja Jan 28 '23 at 23:40
  • Regarding the conjugate of every element of B0 being in B0, this is a property of the ideal B0 in Z[ζ]. The conjugate of an element in Z[ζ] is the element with the same coefficients but with each occurrence of ζ replaced by its complex conjugate ζ^p. Since ζ and its conjugate are roots of the same minimal polynomial, the conjugate of an element of B0 is also in B0, and this property is used in the argument to show that B0 is a subfield of Q(ζp) – Ahmed Izz Murtaja Jan 28 '23 at 23:41