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It's easy to embed the (cyclic) multiplicative group of a finite field into the multiplicative group of $\mathbb{C}$ (or other algebraically closed field of characteristic 0): assign to a generator of the finite cyclic group a corresponding primitive root of unity in $\mathbb{C}$ . Though such embeddings are usually not unique, a fixed one is needed for example to define the Brauer character of a $p$ -modular representation of a finite group.

More generally one can embed the multiplicative group $F^\times$ of an algebraic closure $F$ of $\mathbb{F}_p$ into $\mathbb{C}^\times$ . This is somewhat less elementary and certainly less familiar. I'd be curious to know about the historical origin of such embeddings, but my immediate question is just a reference request:

What is a convenient modern reference for this larger type of embedding?

P.S. To clarify what I'm asking for, the proof of the embedding statement is not the issue (it requires some version of Zorn's Lemma, as does the original proof of existence of an algebraic closure of a finite field). I'd expect to find an explicit statement in some book, even in the form of a structured exercise, but it's not immediately obvious where to look. Occasionally some use is made in research papers of an embedding (without further comment).

edited Apr 26 2012 at 11:50

asked Apr 25 2012 at 16:04

Jim Humphreys
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Isn't this just a recursive construction from the finite case? – Martin Brandenburg Apr 25 2012 at 16:11

See the comments to my (wrong) answer... – Igor Rivin Apr 26 2012 at 16:11

François Brunault · Answer 1 · 2012-04-25 16:42:29Z UTC

One way to construct such an embedding (although probably not the easiest) is to use Witt vectors.

For every perfect field $k$ of characteristic $p$, the ring $W(k)$ of Witt vectors over $k$ is a complete discrete valuation ring of characteristic zero with residue field $k$. It is equipped with a multiplicative map $[\cdot] : k \to W(k)$ which is a section of the residue map, so that $k^\times$ embeds in $W(k)^\times$. Now you can embed $W(\overline{\mathbf{F}_p})$ in $\mathbf{C}$ by purely formal arguments. Note this also works for other fields $k$, at least if the cardinality of $k$ is not too big.

anon · Answer 2 · 2012-04-25 18:13:15Z UTC

Choose a prime of lying over $p$ in the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Then the reduction map gives an isomorphism from the roots of 1 of order prime to $p$ onto the $F^{\times}$.

Added: the proofs only require the Axiom of Dependent Choice, which is strictly weaker than the Axiom of Choice. The same is true of the construction of $F$ (for example, you could define $F$ to be the residue field at the chosen prime).

Igor Rivin · Answer 3 · 2012-04-25 19:30:11Z UTC

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I might be confused, but isn't that group isomorphic to the $p$-primary part of $\mathbb{Q}/\mathbb{Z},$ which obviously does imbed in $\mathbb{C}^*?$

answered Apr 25 2012 at 19:30

Igor Rivin
31.1k●3●37●71

	The group is indeed isomorphic to a subgroup of $\varinjlim \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z}$ , but every element has order relatively prime to $p$. The point is that the direct limit is taken over all the $n′s$ of the form $p^k−1$. – Guillermo Mantilla Apr 26 2012 at 3:52
	Ah, yes, so I was confused. I guess there is no easy way to give a direct description of that group... – Igor Rivin Apr 26 2012 at 4:00
	The group is isomorphic to the sum of $q$-primary components of $\mathbb Q/\mathbb Z$ for $q\ne p$. That is, $\{a/b:(b,p)=1\}/\mathbb Z$. This group obviously embeds in $\mathbb C^$, however, I guess the problem is the isomorphism of $F^$ and the above mentioned subgroup of $\mathbb Q/\mathbb Z$ is not canonical. – Emil Jeřábek Apr 26 2012 at 12:06
	@Emil: is it obvious that that is the group? @Guillermo's comment does confirm that the existence of an embedding is soft/easy. – Igor Rivin Apr 26 2012 at 12:17
	Well, it’s a torsion group, and its $p$-primary component is trivial. For any $q\ne p$, the $q$-primary component has rank $1$ as there are only $q$ elements of order $q$, hence it is either the Prüfer group or $C(q^n)$ for some $n$. It must be the former, as the group has elements of order $q^n$ for arbitrary $n$. – Emil Jeřábek Apr 26 2012 at 12:37

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Source for embedding multiplicative group of an algebraic closure of a finite field?

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