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7.2: Roots of 1 in fields

  • Page ID
    180022
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    As the last example indicates, the representations of a group over a field \(F\) depend on the roots of \(1\) in the field. The \(n\)th roots of \(1\) in a field \(F\) form a subgroup \(\mu_{n}(F)\) of \(F^{\times}\), which is cyclic (see [it20b]).

    If the characteristic of \(F\) divides \(n\), then \(|\mu_{n}(F)|<n\). Otherwise, \(X^{n}-1\) has distinct roots (a multiple root would have to be a root of its derivative \(nX^{n-1}\)), and we can always arrange that \(|\mu_{n}(F)|=n\) by extending \(F\), for example, by replacing a subfield \(F\) of \(\mathbb{C}{}\) with \(F[\zeta]\), where \(\zeta=e^{2\pi i/n}\), or by replacing \(F\) with \(F[X]/(g(X))\), where \(g(X)\) is an irreducible factor of \(X^{n}-1\) not dividing \(X^{m}-1\) for any proper divisor \(m\) of \(n.\)

    An element of order \(n\) in \(F^{\times}\) is called a primitive \(n\)th root of \(1\). To say that \(F\) contains a primitive \(n\)th root \(\zeta\) of \(1\) means that \(\mu_{n}(F)\) is a cyclic group of order \(n\) and that \(\zeta\) generates it (and it implies that either \(F\) has characteristic \(0\) or it has characteristic a prime not dividing \(n\)).


    This page titled 7.2: Roots of 1 in fields is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by James S. Milne.

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