Let $k$ be a field such that its group of units $k^\times$ is finitely generated ; then is it true that $k$ is finite ? I can only show that $\text{char}(k) >0$ . Please help . Thanks in advance


Such a field would be a quotient of $R=\Bbb Z[X_1,\ldots,X_n]$. It is fairly well-known that all quotient fields of $R$ are finite. If you know the characteristic is $p$, then such a field is a quotient of $\Bbb F_p[X_1,\ldots,X_n]$ and so by the Nullstellensatz is a finite extension of $\Bbb F_p$.

(I omit the proof that the characteristic in nonzero; that is another application of the Nullstellensatz.)

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  • $\begingroup$ I can understand that $k^\times$ will be a homomorphic image of the additive group of $\mathbb Z[X_1,...,x_n]$ ; but why the field $k$ itself should be a homomorphic image ? And consequently because of this I cannot understand why should the subsequent argument work $\endgroup$ – user Jun 28 '17 at 13:08
  • $\begingroup$ Take a finite set of generators of $k^\times$ and their reciprocals and choose enough $X_i$ to map onto all of them. @users $\endgroup$ – Angina Seng Jun 28 '17 at 18:00

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