Assume that $F$ is an infinite subfield of a field $K$ such that its multiplicative group, $F^\times$, is divisible. Also, $a\in K$ and $[F(a):F]<\infty$. Can we conclude that the multiplicative group of $F(a)$ is a divisible group?

For example the root closure of $\mathbb Q$ in $\mathbb C$, is a divisible field. Can we obtain that any finite extension of this field is a divisible field?

  • $\begingroup$ "...such that its multiplicative group is divisible". Whose mult. group: $\;F\;$ or $\;K\;$ ?? $\endgroup$ – DonAntonio Dec 27 '16 at 18:56
  • $\begingroup$ $F^*$ is a divisible group. $\endgroup$ – Reza Fallah Moghaddam Dec 27 '16 at 19:04
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    $\begingroup$ I'm not sure, but may be we get a counterexample as follows. Let $k$ be an algebraic closure of $\Bbb{F}_p$, and let $F$ be the field of Puiseux series in indeterminate $T$ over $k$. I think that then $F^*$ is divisible. We can extract a root of $p$-power order termwise, and a root of order coprime to $p$ by the usual method. Then, if $a$ satisfies $a^p-a=1/T$, i.e. $a=T^{-1/p}+T^{-1/p^2}+T^{-1/p^3}+\cdots\notin F$, I wonder whether $a$ has roots of all orders in $K=F(a)$? $\endgroup$ – Jyrki Lahtonen Dec 27 '16 at 19:41
  • $\begingroup$ @JyrkiLahtonen I think the link you gave above might not be the right one, since I don't see the link with Puiseux series... That said, your example looks like it should work, but I wasn't able to flesh it out. $\endgroup$ – Pierre-Guy Plamondon Dec 27 '16 at 23:56
  • $\begingroup$ @Pierre-GuyPlamondon Oops. Copy/paste -error. Should be fixed now. $\endgroup$ – Jyrki Lahtonen Dec 28 '16 at 7:06

There is a finite extension of the root closure of $\mathbb{Q}$ whose multiplicative group is not divisible.

First, even though it is obvious, let us state it: a field $F$ has a divisible multiplicative group if and only if for every $n\in \mathbb{N}$, every element of $F$ admits an $n$-th root.

Recall the definition of the root closure $\mathbb{Q}^r$ of $\mathbb{Q}$. Define a root extension of $\mathbb{Q}$ to be a sequence of extensions of the form $$ \mathbb{Q}\subset \mathbb{Q}(a_1) \subset \mathbb{Q}(a_1, a_2) \subset \ldots \subset \mathbb{Q}(a_1, \ldots, a_r), $$ where a power of $a_1$ lies in $\mathbb{Q}$ and for each $i\in\{2,\ldots r\}$, a power of $a_i$ lies in $\mathbb{Q}(a_1, \ldots, a_{i-1})$.

Let $\mathbb{Q}^r$ be the union of all possible root extensions of $\mathbb{Q}$. It is not hard to see that $\mathbb{Q}^r$ is a subfield of $\mathbb{Q}^a$ (where $\mathbb{Q}^a$ is the algebraic closure of $\mathbb{Q}$). Moreover, since some polynomials over $\mathbb{Q}$ cannot be solved by radicals by the Abel-Ruffini Theorem, the field $\mathbb{Q}^r$ is strictly contained in $\mathbb{Q}^a$.

By construction, for every $n\in \mathbb{N}$, every element of $\mathbb{Q}^r$ admits exactly $n$ $n$-th roots in $\mathbb{Q}^r$. Thus $(\mathbb{Q}^r)^*$ is divisible.

Let $L$ be a finite extension of $\mathbb{Q}^r$ (such an extension exists, since $\mathbb{Q}^r \subsetneq \mathbb{Q}^a$). We can assume that $L/\mathbb{Q}^r$ is a Galois extension (if it isn't, replace $L$ by its Galois closure). Let $p$ be a prime number dividing the order of the Galois group $Gal(L/\mathbb{Q}^r)$. By Cauchy's theorem, there is a subgroup of $Gal(L/\mathbb{Q}^r)$ of order $p$. Hence, by the Galois correspondence, there is an intermediate field extension $$ \mathbb{Q}^r \subsetneq L' \subsetneq L $$ such that $[L:L']=p$.

Then the multiplicative group of $L'$ is not divisible. Indeed, assume that it is. Since $L'$ contains all roots of unity, then for any $n\in \mathbb{N}$, any element of $L'$ admits exactly $n$ $n$-th roots in $L'$. Now, the Galois group of $L/L'$ has order $p$, and is thus solvable; therefore, if $b\in L\setminus L'$, then the minimal polynomial $P_b$ of $b$ over $L'$ is solvable by radicals. But this implies that $b\in L'$, a contradiction.

Thus $(L')^*$ is not divisible, and there exists an $a\in L'$ and an $n\in \mathbb{N}$ such that $a$ does not admit any $n$-th root in $L'$.

Therefore, if we take $F=\mathbb{Q}^r$ and $a$ as above, then $F(a)^*$ is not divisible, even though $F^*$ is.

A finite example

The only finite field whose multiplicative group is divisible is $\mathbb{F}_2$. Thus, taking $F= \mathbb{F}_2$ sitting inside any finite extension $K$ gives us a counter-example.

  • $\begingroup$ Thaks for your response $\endgroup$ – Reza Fallah Moghaddam Dec 29 '16 at 9:31
  • $\begingroup$ IMO the useful/helpful answers are upvoted, and the answer that satisfies the asker is accepted. The OP's call, of course :-) $\endgroup$ – Jyrki Lahtonen Dec 29 '16 at 10:07
  • $\begingroup$ In your proof you said" The Galois group of $L/L'$ has order p, and is thus solvable;" why? $\endgroup$ – Reza Fallah Moghaddam Jan 1 '17 at 16:36
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    $\begingroup$ @RezaFallah-Moghaddam $L/L'$ is Galois because $L/\mathbb{Q}^r$ is Galois. Any group of order $p$ with $p$ prime is cyclic, and any cyclic group is solvable. If you think that the proof is wrong, it would be helpful if you explained why; I have read it again, and it seems all right to me. $\endgroup$ – Pierre-Guy Plamondon Jan 6 '17 at 3:09
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    $\begingroup$ @RezaFallah-Moghaddam There is a point of confusion with the fundamental theorem in what you say. If you have a Galois extension $L/K$, then the fundamental theorem gives a bijection between normal subgroups of $Gal(L/K)$ and subfields of $L$ that are Galois extensions of $K$. However, if $L'$ is a subfield of $L$ containing $K$, then the extension $L/L'$ is always Galois (even though $L'/K$ might not be). So in our situation, we do not need that the subgroup is normal. $\endgroup$ – Pierre-Guy Plamondon Jan 6 '17 at 12:50

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