The image of the natural homomorphism of Witt rings of orders in a global field (Q2851133)
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scientific article; zbMATH DE number 6214470
| Language | Label | Description | Also known as |
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| English | The image of the natural homomorphism of Witt rings of orders in a global field |
scientific article; zbMATH DE number 6214470 |
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The image of the natural homomorphism of Witt rings of orders in a global field (English)
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9 October 2013
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order
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Witt ring
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integral quadratic form
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field of fractions
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Given a Dedekind domain \(R\) whose field of fractions is a global field \(K\), and an order \(\mathcal{O} \subset R\), what quadratic forms over \(R\) are in fact defined over \(\mathcal{O}\)? It is known that the natural homomorphism of Witt rings \(WR \rightarrow WK\) is injective, so when studying the natural homomorphism \(\varphi : W\mathcal{O} \rightarrow WR\), the image can be inspected inside \(WK\).NEWLINENEWLINEWhen \(R\) is the ring of \(S\)-integers in \(K\) (for some set \(S\) of places including the Archimedean ones), it is known that the image of \(WR \rightarrow WK\) is spanned, as an additive group, by certain forms of dimensions \(1\) and \(2\) and by \(2\)-fold Pfister forms. The strategy is, therefore, to ascertain when those generators are defined over \(\mathcal{O}\) (i.e. in the image of \(\varphi\)).NEWLINENEWLINERecall that an order in \(R\) is a subring \(\mathcal{O} \subseteq R\) which is Noetherian and of Krull-dimension \(1\), such that \(R\) is a finite \(\mathcal{O}\)-module and equals to the integral closure of \(\mathcal{O}\) in \(K\). The conductor \(\mathfrak{f} = \{x \in R : x R \subseteq \mathcal{O}\}\) is an important invariant of the order. For example, it is shown that the one dimensional forms in the image of \(W\mathcal{O} \rightarrow WK\) are of the form \(\langle h \rangle\) where \(hR\) is the square of an ideal in \(R\), and \(hR + \mathfrak{f} = R\) (Thm~2.9). More generally if \(a_1,\dots,a_{\ell} \in \mathcal{O}\) and \(q = \langle a_1,\dots,a_\ell \rangle \in WK\) is defined over \(R\), and \(a_i R + \mathfrak{f} = R\) for each \(i\), then \(q\) is in fact defined over \(\mathcal{O}\) (Thm~5.6).NEWLINENEWLINEThe case when \(K\) is a quadratic extension of \(\mathbb{Q}\) or of a function field \(\mathbb{F}(X)\), \(\mathbb{F}\) finite field, is analyzed in detail, and we give here only a sample. Suppose \(K = \mathbb{Q}(\sqrt{D})\) and \(R\) is the ring of integers in \(K\). When \(D < 0\), the map \(W\mathcal{O}\rightarrow WR\) is surjective, unless \(D \equiv 3 \pmod{4}\) and \(\mathcal{O}\) has even conductor (Cor. 3.2 and 3.4).NEWLINENEWLINESuppose \(F = \mathbb{F}(X)\), \(K = F(\sqrt{D})\) where \(D\) is a nontrivial square-free polynomial, and \(R\) is the ring of integers with respect to all finite places (the infinite place having uniformizer \(1/X)\). If all scalar norms (i.e. norms in \(K/F\) which are in \(\mathbb{F}\)) are squares in \(\mathbb{F}\), the map \(W\mathcal{O} \rightarrow WR\) is surjective (Cor. 4.3 and Cor. 7.6). On the other hand if all scalars are norms, the conductor has odd-degree factors then the map is not surjective (Cor. 4.9); furthermore when \(K\) is real (i.e. the infinite valuation splits in \(K\)), the map is surjective if and only if all factors of the conductor have even degree (Cor. 7.4).
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