The base field $\mathbb F$ is probably not important, but I'm using the rational expressions over the binary field, $\mathbb F=\mathbb F_2(x)$.

One subfield of $\mathbb G$ consists of ratios of polynomials in roots of $x$, with coefficients in $\mathbb F_2$, such as


and such expressions are added and multiplied using the usual rules for rational functions, and $x^ax^b=x^{a+b}$ for $a,b\in\mathbb Q$.

Squaring and square roots are linear over $\mathbb F_2$, so we can simply halve the exponents on $x$ to get a square root of the expression. Thus, any $2^n$th root always exists. For example,


But this doesn't work for other $n$th roots; it's not possible to write


where $p,q$ are polynomials and $m\in\mathbb N$. So it's necessary to formally adjoin $\sqrt[3]{x+1}$ to the field, or instead $\sqrt[3]{x^{1/2}+1}$, etc.

Is there a unique, well-defined field $\mathbb G$ of such algebraic expressions over $\mathbb F$?

Note that I don't want $n$ different $n$th roots of each element, just a single root (unless $\mathbb F$ already has roots of unity; but I chose $\mathbb F_2$ to avoid this).

Given the algebraic closure $\mathbb A\supseteq\mathbb F$, we might just take the intersection $\mathbb G\overset?=\bigcap\{\mathbb B\}$ of all intermediate fields $\mathbb A\supseteq\mathbb B\supseteq\mathbb F$ with the property $\forall n\in\mathbb N,\,\forall a\in\mathbb B,\,\exists b\in\mathbb B,\,a=b^n$. But this doesn't work because different fields have different roots of $a$, so their intersection contains no root of $a$. Presumably there's some way to use the axiom of choice to construct $\mathbb G$, either through $\mathbb A$, or directly from $\mathbb F$. Is this the case? Can the Wiki proof of existence (I haven't followed it in detail) be modified to give $n$th roots of everything without introducing new roots of unity? And what about uniqueness?

Is there a simpler construction of $\mathbb G$ for the special case of $\mathbb F_2(x)$, that doesn't use the axiom of choice? Here I don't require uniqueness. See for example this answer; we would be using polynomials of the form $x^p-a$ which are irreducible over the field defined by the previous polynomials.

Having several $n$th roots of $a\neq0$ is equivalent to having an $n$th root of unity: If $x_1^n=x_2^n=a$ and $x_1\neq x_2$, then $(x_1/x_2)^n=1$ and $(x_1/x_2)\neq1$. Conversely, if $\omega^n=1$ and $\omega\neq1$, and $x_1^n=a$, then $(\omega x_1)^n=a$ and $x_1\neq\omega x_1$.

If $\mathbb F$ has a primitive $mn$th root of unity, then it also has a primitive $n$th root of unity; so we need only consider prime numbers. Fix two primes $p\neq q$. If $\mathbb F$ has a primitive $p$th root of unity $\omega_1$, then $\mathbb G$ should have a primitive $p^n$th root of unity $\omega_n$ for all $n$, since non-primitive $p^n$th roots of unity can never reach $\omega_{n-1}$ as a $p$th power. If $\mathbb F$ doesn't have a primitive $q$th root of unity, then $\mathbb G$ shouldn't either, since we already have $1^q=1$ and $(\omega_n^r)^q=\omega_n$ where $r=q^{-1}\bmod p^n$.

  • $\begingroup$ The existence of an algebraic closure for $\mathbb{F}_2(x)$ (or more generally, for any well-orderable field) does not need the axiom of choice. $\endgroup$ Sep 6, 2020 at 20:43
  • $\begingroup$ @EricWofsey - That was evident from the linked answer. What's your point? $\endgroup$
    – mr_e_man
    Sep 6, 2020 at 20:48
  • $\begingroup$ Well, you were asking for a simpler construction "that doesn't use the axiom of choice", but nothing you have discussed does use the axiom of choice in any essential way. $\endgroup$ Sep 6, 2020 at 20:50
  • $\begingroup$ @EricWofsey - The proof of existence of $\mathbb A$ given on Wikipedia uses Zorn's lemma which is equivalent to the axiom of choice. I was asking for a construction, not of $\mathbb A$, but of $\mathbb G$, with or without the axiom of choice, but preferably without. $\endgroup$
    – mr_e_man
    Sep 6, 2020 at 20:54
  • $\begingroup$ The existence of multiple roots is equivalent to existence of roots of unity in the field. A good candidate could then be taking $\mathbb{G}'$ to be the field in which you adjoin all roots of all elements, and then set $\mathbb{G}$ to be $\mathbb{G}'$ stabilised by $Gal(\bar{\mathbb{F}}_2 /\mathbb{F}_2) $. There are two issues with this approach: first, does this Galois group acts on $\mathbb{G}'$? Second, does there always exist a root of an element which is stabilised by such action? The answer to 1 is probably yes, while 2 seems dubious. $\endgroup$ Sep 12, 2020 at 11:03

3 Answers 3


Well, from my thrashing-about in the comments, it’s probably clear that I haven’t fully understood the depth of this problem. But let me make some remarks anyhow, first about the very special case $\mathbf F=\Bbb F_2(x)$. It happens, and some highly-experienced people seem not to know this, that when $\mathbf F$ is of transcendence degree one over a perfect field of characteristic $p$ and not already perfect, there is exactly one radicial (=purely inseparable, that’s French radiciel) extension of each possible degree $p^m$. I think this should clarify your thinking on a square-root-closed extension of $\Bbb F_2(x)$.

Second, let me just point out the difficulty of describing any construction for your field $\mathbf F=\Bbb F_2(x)$: for $d$ odd, the extensions $\mathbf F(\sqrt[d]x\,)$ and $\mathbf F(\sqrt[d]{x+1}\,)$ have nothing to do with each other: their intersection is the ground field $\mathbf F$. Stick any $\Bbb F_2$-irreducible polynomial under the radical sign and get another totally unrelated extension. So conceptually anyway, this is getting to be a mess; you need to worry about rational expressions, too.

I guess that you were thinking of specifying at the outset that in every case, the $n$-th root of $1$ that you choose is $1$ itself. Even if you do that, it’s not clear to me that in your choosing lots of unspecified $n$-th roots of other elements, you may inadvertently induce the presence of other roots of unity than $1$ itself. This would actually simplify matters, but I do think you’re making things hard for yourself. It seems to me that if you agree at the outset that all roots of unity should be added (this would make the constant field $\overline{\Bbb F_2}\,$, algebraically closed), then the existence of your field is now easily seen, though an explicit construction would remain still something of a problem.

  • 1
    $\begingroup$ I haven't studied these subjects enough to know the significance of separability or perfectness. Anyway, for me, the goal was to find the smallest field (other than the trivial $\mathbb F_2$) whose multiplicative group is isomorphic to a $\mathbb Q$ vector space, thus providing a different answer to this question. $\endgroup$
    – mr_e_man
    Sep 10, 2020 at 22:06
  • $\begingroup$ Aha, @mr_e_man . If I had known your ultimate aim, I would have written a very different answer; in fact I would have had few suggestions for you. But good luck: you should delve more deeply into all this material, since it has many beautiful aspects. $\endgroup$
    – Lubin
    Sep 11, 2020 at 1:37

Here is a good benchmark. If $\Bbb F$ is a field, and $\Bbb K$ is a field extension such that:

  1. $\Bbb{K/F}$ is infinite dimensional, and
  2. for every irreducible $p\in\Bbb F[x]$, if $p$ has at least two roots in $\Bbb K$, then there is an automorphism of $\Bbb K$ (fixing $\Bbb F$) which is a complete derangement of the roots of $p$ in $\Bbb K$.

In this case, there is a model of $\sf ZF$ in which there is a field which is "morally isomorphic to $\Bbb K$, but not internally isomorphic to it". That is to say, we add a new copy of $\Bbb K$, but we remove the isomorphism, and indeed any bijection, while preserving the field structure, and every field extension of $\Bbb F$ embedding to both will be finite dimensional.

It's not hard to see that your "smallest field" will satisfy these properties, or at the very least, we can find such a field which will ensure no "smallest" exists.

  • $\begingroup$ To be clear: You're saying that, denying choice in a certain way, the answer is "No" for any $\mathbb F$? $\endgroup$
    – mr_e_man
    Sep 14, 2020 at 15:46
  • $\begingroup$ No, because algebraically closed fields still exist, and real closed fields still exist too. But in principle, that should be the only restriction. $\endgroup$
    – Asaf Karagila
    Sep 14, 2020 at 15:50
  • $\begingroup$ I guess I was thinking that $\mathbb G/\mathbb F$ should always be infinite-dimensional, which is obviously wrong.... $\endgroup$
    – mr_e_man
    Sep 14, 2020 at 15:58
  • $\begingroup$ Yes. In any case, a related result will appear in a paper I am currently revising. $\endgroup$
    – Asaf Karagila
    Sep 14, 2020 at 16:00
  • $\begingroup$ Sorry, it is hard for me to see property 2 for my "smallest field". What's wrong with the existence of an irreducible polynomial with several roots in $\mathbb G$, where every automorphism fixes some root? Does that imply that some polynomial of the form $x^n-a$ has several roots in $\mathbb G$? $\endgroup$
    – mr_e_man
    Sep 14, 2020 at 19:55

Instead of constructing a radical closure $\Bbb{F}^{rad}$ by intersection, why not build it up by a process of union? Start with your base field $\Bbb{F}$ (in your particular case, the base field is $\Bbb{F}_2(x)$) which, by assumption, is not already radically closed (or this process will terminate at the very first step, lol). Your next field $\Bbb{F}_1 \supset \Bbb{F}$ is the union of all radical extensions of $\Bbb{F}$, $$\Bbb{F}_1 := \bigcup_{a \in \Bbb{F}} \left( \bigcup_{n \geq 2} \Bbb{F}(\sqrt[n]{a}) \right);$$ the next field $\Bbb{F}_2 \supset \Bbb{F}_1$ is the union of all radical extensions of $\Bbb{F}_1$, $$\Bbb{F}_2 := \bigcup_{a \in \Bbb{F}_1} \left( \bigcup_{n \geq 2} \Bbb{F}_1(\sqrt[n]{a}) \right);$$ and continuing in this way, you acquire a sequence of radical extensions $$... \supset \Bbb{F}_n \supset ... \supset \Bbb{F}_2 \supset \Bbb{F}_1 \supset \Bbb{F}.$$ Clearly the extension $\Bbb{F}_n/\Bbb{F}_{n-1}$ will be unique up to isomorphism at each stage, and so the entire chain is as well. Therefore $\Bbb{F}^{rad} := \bigcup_{n \geq 1} \Bbb{F}_n$ is unique up to isomorphism and every element of $\Bbb{F}^{rad}$ has an $n$th root in $\Bbb{F}^{rad}$ for all $n \geq 2$.

Edit: This chain method still works even if, as in OP's case, we just want a root for each element of the field. Start with the base field $F := \Bbb{F}_2(x)$. Construct $F_1 \supset F$ via $$F_1 := \bigcup_{a \in F^\times} \bigcup_{q \in \Bbb{Q}} F(a^q)$$ where to avoid Axiom of Choice we regard $a^q$, $q \in \Bbb{Q} \setminus \Bbb{Z}$, to be purely formal "powers" of the elements $a$ under appropriate equivalence relations, such as: $$(a^n)^q \sim (a^q)^n \text{ for all } q \in \Bbb{Q}, n \in \Bbb{Z}, a \in F^\times,$$ $$a^q a^r \sim a^{q+r} \text{ for all } q, r \in \Bbb{Q}, a \in F^\times,$$ or $$(ab)^q \sim a^q b^q \text{ for all } a, b \in F^\times, q \in \Bbb{Q}.$$ In most fields, root-of-unity issues would make setting up these equivalence relations fraught with difficulties, but since the only root of unity in $F = \Bbb{F}_2(x)$ is $1$ itself, these equivalence relations end up behaving the way we want them to, without much fuss. (Edit: We should also specify that $x \sim 0$ if $x^n = 0$ for some $n$.)

Keep iterating this construction to get a new field $F_{k+1}$ from the previous field $F_k$: $$F_{k+1} := \bigcup_{a \in F_k^\times} \bigcup_{q \in \Bbb{Q}} F(a^q),$$ where we also add that our equivalence relations should obey $(a^q)^r \sim (a^r)^q \sim a^{rq}$ for all $a \in F_{k-1}$ and all $q, r \in \Bbb{Q}$. Then as before we get the desired field of radicals via $\bar{F} := \bigcup_{k \geq 1} F_k$, and any other field which is closed under radicals and contains the base field $F$ must contain a field isomorphic to $\bar{F}$, as $\bar{F}$ by construction contains all possible radical expressions which could be made out of elements of $F = \Bbb{F}_2(x)$ (or, at any rate, elements equivalent to same).

  • $\begingroup$ You are implicitly using the axiom of choice here. $\endgroup$
    – Asaf Karagila
    Sep 14, 2020 at 8:20
  • $\begingroup$ @AsafKaragila You would think so, but no! :) It seems like I am at first, because it looks like I have to make an infinite number of choices of $\sqrt[n]{a}$ at each stage in the construction. But regardless of $\Bbb{F}$'s characteristic, at the very first stage, $\Bbb{F}_1$ contains every possible root of unity, because we adjoin all radicals of all roots of unity in $\Bbb{F}$. This means $\Bbb{F}_1$ is independent of our "choices" of $\sqrt[n]{a}$ for $a \in \Bbb{F}$. By the same token, $\Bbb{F}_k$ is independent of our "choices" of $\sqrt[n]{a}$ for $a \in \Bbb{F}_{k-1}$ for all $k$. $\endgroup$ Sep 14, 2020 at 8:55
  • $\begingroup$ You know that I'm an expert on the axiom of choice, right, and I don't make these claims lightly? Unless you only included a single root at each step, you are using choice to argue that the limit is the smallest, and if you are adding one root at a time, then you're using the axiom of choice to choose said root. $\endgroup$
    – Asaf Karagila
    Sep 14, 2020 at 8:56
  • $\begingroup$ @AsafKaragila Perhaps the confusion is thinking that my construction involves "picking" an element of the algebraic closure $\Bbb{A} = \bar{\Bbb{F}}$. Actually, at each stage, I'm constructing the new field $\Bbb{F}_k$ by adjoining the formal symbol $\sqrt[n]{a}$ for each $a \in \Bbb{F}_{k-1}$, which I don't believe involves any choice. It then turns out, since $\Bbb{F}_1$ already had all possible roots of unity, that $\Bbb{F}_k$, "interpreted" as a subfield of $\Bbb{A}$, contains the same elements regardless of my individual "interpretations" of the $\sqrt[n]{a}$'s as elements of $\Bbb{A}$. $\endgroup$ Sep 14, 2020 at 9:07
  • 1
    $\begingroup$ @mr_e_man It’s not “just union”. Let me clarify that construction. I’m talking about adjoining the formal elements $a^q, a \in F_k, q \in \Bbb{Q},$ to $F_k$ at each stage of the construction with the indicated properties. $F_{k+1}$ is then the field of fractions of the vector space of linear combinations of such expressions with coefficients in $F_k$, where multiplication is defined in the obvious way. $\endgroup$ Sep 14, 2020 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.