# Can completely multiplicative functions be extended to $\overline{\mathbb{Q}}$ or further?

I'm looking for a subject of study that handles the following question. I'm not the most familiar with algebra; I have a strong working knowledge and that's about it, but I've been considering researching more into it. I'm curious if the following exists somewhere in the literature.

Consider a completely multiplicative function $$q :\mathbb{Q} \to \mathbb{Q}$$ which is also an automorphism of the multiplicative group $$\{\mathbb{Q}^{\times},\cdot\}$$ (a bijective completely multiplicative function), and $$q\big{|}_{\mathbb{N}} = \mathbb{N}$$ (so for instance $$q(x) \neq \frac{1}{x}$$). These functions are essentially "prime permutations" on the unique factorization of elements of $$\mathbb{Q}$$. We can use a permutation notation, because of this. For instance, $$(1\, 2)$$ swaps the first prime number with the second; i.e $$2$$ for $$3$$. Therein if $$x$$ is factored, only finitely many $$e_i$$ are non-zero

$$x = \prod_{i=1}^\infty p_i^{e_i}$$

then if $$q = (1\, 2)$$

$$q(x) = \prod_{i=1}^\infty p_{q(i)}^{e_i}$$

For example $$q(2) = 3$$, $$q(3) = 2$$, $$q(2^4 3^2 5^3 7^{-6}) = 3^4 2^2 5^3 7^{-6}$$, so on and so forth across all elements of $$\mathbb{Q}$$. These can be infinite permutations as well, so perhaps the $$2n-1$$'th prime is switched with the $$2n$$'th prime for all $$n\ge 1$$, or something of the sort. We also make the convention $$q(0) = 0$$, $$q(1) = 1$$ and that $$q$$ is bijective so that $$q(-1) = -1$$.

Now there also exists a natural extension to $$\mathbb{Q}$$ when equipped with $$n$$'th roots. That in $$q(x^{1/n}) = q(x)^{1/n}$$ where we make the convention that positive values are sent to positive values. This quite frankly gives the natural result that $$q(\sqrt{2}) = \sqrt{3}$$.

Does there exist a natural extension $$\tilde{q}$$ to $$\{\overline{\mathbb{Q}}, \cdot\}$$--where $$\tilde{q}$$ is still completely multiplicative, bijective and $$\tilde{q}\Big{|}_{\mathbb{Q}} = q\Big{|}_{\mathbb{Q}}$$? I.e: can this be made to work on all algebraic numbers?

The reasoning I have for thinking of this is because of field extensions. By defining the field $$F = \{\mathbb{Q},\cdot,\oplus\}$$ where $$x \oplus y = q(q^{-1}(x) + q^{-1}(y))$$. We can take the polynomial ring $$F[X]$$ and talk about $$F[X]/p$$ for some irreducible polynomial $$p$$--which is isomorphic to some field extension $$F(\alpha)$$.

Now let's make the convention $$p(\alpha) = 0$$, then this can be thought of as

$$0 = p(\alpha) = a_n\alpha^n \oplus a_{n-1}\alpha^{n-1} \oplus...\oplus a_0$$ $$p(\alpha) = q(q^{-1}(a_1)q^{-1}(\alpha)^n + q^{-1}(a_{n-1})q^{-1}(\alpha)^{n-1} +...+q^{-1}(a_0))$$

so that

$$0 = q^{-1}(a_1)q^{-1}(\alpha)^n + q^{-1}(a_{n-1})q^{-1}(\alpha)^{n-1} +...+q^{-1}(a_0)$$

Therefore if $$q^{-1}(\alpha) = \beta$$ and $$b_n = q^{-1}(a_n)$$, then $$\alpha = q(\beta)$$ where $$\beta$$ satisfies

$$b_n\beta^n + b_{n-1}\beta^{n-1} + ... + b_0 = 0$$

This definition is consistent with the taking roots version. If $$\beta$$ satisfies $$\beta^2 - 2 = 0$$, then $$q(\beta) = \alpha$$ where $$\alpha^2 \ominus 3 = 0$$, aka $$q(\sqrt{2}) = \sqrt{3}$$. (Where the meaning of $$\ominus$$ should be self evident).

However, as you may have noticed, this does not "choose" which root of the polynomial corresponds to $$\beta$$. As in $$q(\sqrt{2}) = \pm\sqrt{3}$$. Nor does it explicitly give a manner of producing said $$\alpha$$. $$\alpha$$ just exists in the "isomorphic algebraic numbers" generated by the algebraic closure of $$\{\mathbb{Q},\cdot,\oplus\}$$ which coincides with roots. And for higher-order polynomials this becomes a serious problem. This led me to two different questions.

Can we distinguish elements belonging to the algebraic closure of $$\{\mathbb{Q},\cdot,\oplus\}$$ and $$\{\mathbb{Q}, \cdot, +\}$$? I.e, can we make sense of the correspondence, so that $$q(1 +\sqrt{2}) = > \alpha \in \overline{\mathbb{Q}}$$ and not just in some isomorphic copy of $$\overline{\mathbb{Q}}$$?

Secondly, if we consider $$q:\mathbb{Q}\to\mathbb{Q}$$ and we wish to extend it to $$\mathbb{Q}(\alpha)$$ then if $$q:\mathbb{Q}(\alpha) \to F(\beta)$$ there should exist multiple extensions (i.e: a $$q_{-}$$ and $$q_{+}$$ where $$q_{+}(\sqrt{2}) = \sqrt{3}$$ and $$q_{-}(\sqrt{2}) = -\sqrt{3}$$). If we were to add multiple field extensions, the intuition would say we would compound the amount of extensions of $$q$$.

For instance, if $$q = (1\,2)(3\,4)$$ (swap 2 and 3, and swap 5 and 7) then the field extension $$Q(\sqrt{2},\sqrt{7})$$ has four possible $$q$$'s extending its definition $$q_{ij}(\sqrt{2}) = (-1)^{i}\sqrt{3}$$ and $$q_{ij}(\sqrt{7}) = (-1)^j\sqrt{5}$$ for $$i,j = 0\,\text{or}\,1$$. But are these definitions compatible when adding multiple field extensions? I was only able to really construct it with a Galois extension (I think I mean a Galois extension).

Would this still continue to work if say, we took $$\overline{\mathbb{Q}}$$, we added all algebraic numbers to $$\mathbb{Q}$$. Therein, my gut says to ask:

Could there be an uncountable number of $$\tilde{q}:\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}$$'s which would extend $$q:\mathbb{Q}\to\mathbb{Q}$$? (i.e: $$\tilde{q}\Big{|}_{\mathbb{Q}} = q\Big{|}_{\mathbb{Q}}$$)

But I'm doubting this because I think some extensions may be equivalent when adding another element. So, my gut also says to ask:

Is there only a finite number of $$\tilde{q}$$, perhaps just one? Does the number of solutions depend on our choice of $$q$$?

And the question I hope is answered no

Is there no such $$\tilde{q}$$?

Any help or comments or suggestions is greatly appreciated. This one has had me stumped for a long time and I'm stuck not knowing what tools to use to approach the problem. If something like this has already been considered in the literature, I'd be very happy to read up on it. Does it already have a name, and a well founded approach? Is it meaningless? Am I garbling out nonsense? Anything is welcome.

Thanks!

EDIT: I should mention this came up as I was thinking about the absolute Galois group. Clearly elements of the absolute Galois group are automorphisms of $$\{\overline{\mathbb{Q}},\cdot\}$$--however these fix the base field $$\mathbb{Q}$$. What if we, instead, allow it to permute the basefield $$\mathbb{Q}$$; does this still make sense.

• Could you give a precise formulation of the properties that you want the extension to have? – Christian Remling Feb 9 '17 at 22:24
• Essentially, that it is completely multiplicative. I guess I should've mentioned that, obviously there are infinite extensions otherwise. I'm having trouble expressing exactly what I mean; I better edit it over again. – user78249 Feb 9 '17 at 23:07
• Oh and bijective, otherwise again there are trivially many extensions. – user78249 Feb 9 '17 at 23:29
• You mentioned $q(\sqrt{2})$ but not $q(1+\sqrt{2})$ ? And you didn't mentioned the unique factorization of ideals in $\mathcal{O}_K$ – reuns Feb 10 '17 at 13:07
• @user1952009 Quite frankly I don't know what $q(1+\sqrt{2})$ would be, I can only make sense of it in the isomorphic field to $\overline{\mathbb{Q}}$, the algebraic closure of $\{\mathbb{Q},\cdot,\oplus\}$. Would the unique factorization of ideals in $\mathcal{O}_K$ have significance in the context of my question? If so, how? – user78249 Feb 10 '17 at 16:25

Let $$H \subset G$$ with $$G$$ divisible, and $$q \in \operatorname{Aut}(H)$$. Then one shows that a pair $$(K, \tilde q)$$ with $$H \subset K$$ and $$\tilde q \in \operatorname{Aut}(K)$$ extending $$q$$, which is maximal for the obvious ordering, necessarily satisfies $$K = G$$.
So for example, there exists an extension of $$q \in \operatorname{Aut}(\mathbb Q^\times)$$ which swaps $$\alpha = 1 + \sqrt 2$$ with $$\beta = 4 + \sqrt{13}$$, because we may first extend $$q$$ to the group $$\mathbb Q^\times \oplus \alpha^{\mathbb Z} \oplus \beta^{\mathbb Z}$$ in any way we like, and then to $$\overline{\mathbb Q}^\times$$.
And of course, one can always compose $$\tilde q$$ with a $$\sigma \in \operatorname{Gal}(\overline{\mathbb Q} / \mathbb Q)$$.