Bourbaki

Exercise 2, § 8

Algebra, V, p. 153

Let KK be a field and AA a KK-algebra of finite degree nn. For each integer mm relatively prime to nn we have A∗m∩K∗=K∗mA^{*m} \cap K^* = K^{*m}; in other words, if x∈K∗x \in K^* and if there exists an element yy of AA such that x=ymx = y^m, then there exists an element zz in KK such that x=zmx = z^m.

Solution

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Transcribed from the printed volume. Not translated.