I am trying to find solutions of functional equation $f(xy)=f(x)+f(y)$ for all $x,y\in\mathbb Q_p\setminus\{0\}=\mathbb Q_p^\star$. Where $f:\mathbb Q_p^\star\to\mathbb R$. I know some solutions:

1) $f(x)\equiv 0$.

2) $f(x)=\log |x|_p$

Here, $p$ is prime and $\mathbb Q_p$ is the field of all $p-$adic numbers.

My question is: Does there exist any other solution which of cause not similar to these solutions as above? (ie.. I mean $f(x)=\log |x|^n_p$ is "similar" to 2).)

  • $\begingroup$ $f(0\cdot 0)=f(0)+f(0)$ implies $f(0)=0$. Then $f(x\cdot 0)=f(x)+f(0)$ implies $f(x)=0$. $\endgroup$ – Hagen von Eitzen Apr 6 '13 at 16:23
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    $\begingroup$ My question might be silly, how does the $p$-adic field differ from the rational number field? $\endgroup$ – Easy Apr 6 '13 at 16:29
  • $\begingroup$ @Easy: $\mathbb{Q}_p$ is uncountable, complete with respect to the $p$-adic metric, etc. $\endgroup$ – Martin Brandenburg Apr 6 '13 at 16:38
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    $\begingroup$ @Hai: Probably you want $\mathbb{Q}_p^*$ instead of $\mathbb{Q}_p$. $\endgroup$ – Martin Brandenburg Apr 6 '13 at 16:39
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    $\begingroup$ I don't think that constant function $\equiv 1$ works. Also, probably you meant $f(x)=n\log |x|_p$. That is, a multiple rather than a power of the logarithm of the $p$-adic absolute value $\endgroup$ – Jyrki Lahtonen Apr 7 '13 at 7:23

You are asking for homomorphisms from the multiplicative group $\mathbb{Q}_p^*$ to the additive group of real numbers.

It is known that $\mathbb{Q}_p^*$ is a direct product of three subgroups. The infinite cyclic subgroup generated by $p$, the cyclic subgroup of order $p-1$ consisting of non-zero Teichmüller elements (generated by $\zeta$, a primitive root of unity of order $p-1$), the group $1+p\mathbb{Z}_p$ of $p$-adic integers congruent to $1$.

A homomorphism from a direct product of groups is determined, if we know its restriction to the constituent groups. The infinite cyclic group is easy to handle, as we can select the image of $p$ to by any real number $a$ we want, and define $f_a(p^n)=na$. The finite cyclic group is also easy to handle: the additive groups has darn few elements of a finite order, so such a homomorphism has to be trivial.

The last factor, $U:=1+p\mathbb{Z}_p$, is more interesting. It is actually a bit scary to think about classifying the homomorphisms from $U$ to $\mathbb{R}$. For one, I can't write down a single non-trivial one. Yet the following argument shows that they exist. Consider the subgroup $K=\langle 1+p\rangle \le U$. This is an infinite cyclic group, so we get a homomorphism $\phi_b$ from $K$ to $\mathbb{R}$ simply by sending $1+p$ to an arbitrary real number $b$, and extending that. Because the additive group of $\mathbb{R}$ is divisible, it is an injective object in the category of abelian groups (see e.g. Hilton-Stammbach). Therefore there is a homomorphism $g_b:U\to\mathbb{R}$ such that the restriction of $g_b$ to $K$ is equal to $\phi_b$. The proof of this involves a typical use of Zorn's lemma, so if you belong to the church that does not believe in the axiom of choice, then I cannot convince you that such a homomorphism exists.

Given the above there will other such functions. Namely those of the form $$ f(p^n\zeta^j u)= na + g_b(u). $$ If $g_b$ is the trivial homomorphism, then you get a scalar multiple of the logarithm of the $p$-adic value. Otherwise you get something else.

  • $\begingroup$ A very nice answer, thanks $\endgroup$ – Jie Fan Apr 7 '13 at 7:57
  • $\begingroup$ Note that the multiplicative unit group $U_1:=1+p{\bf Z}_p$ is isomorphic to ${\bf Z}_p$ under addition, which is clearly a ${\bf Z}_{(p)}$-vector space of uncountable dimension (where ${\bf Z}_{(p)}:={\bf Z}_p\cap{\bf Q}$). This allows a full characterization of $\hom({\bf Q}_p,{\bf R})$ as ${\bf R}^{\oplus{\frak c}}$. (For the background facts you mention, I plug this answer of mine.) Or we could have just noticed $\dim_{\bf Q}{\bf Q}_p={\frak c}$ to begin with.. $\endgroup$ – anon Apr 7 '13 at 8:01
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    $\begingroup$ @anon: The use of the isomorphism of $U_1$ and the additive group of $\mathbf{Z}_p$ is a nice idea. But why is $\mathbf{Z}_p$ a free $\mathbf{Z}_{(p)}$-module? Assuming that's what the rest of your argument is based on? $\endgroup$ – Jyrki Lahtonen Apr 7 '13 at 8:08
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    $\begingroup$ Erm, I forgot ${\bf Z}_{(p)}$ is not a field. Now it's not as clear to me if ${\bf Z}_p$ is a free ${\bf Z}_{(p)}$-module. $\endgroup$ – anon Apr 7 '13 at 8:12
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    $\begingroup$ @anon: No problem, obviously. I was hoping that you had found a way of avoiding the use of injectivity of $\mathbf{R}$. I didn't, so I didn't bother using the $\mathbf{Z}_p$-module structure of $U_1$. $\endgroup$ – Jyrki Lahtonen Apr 7 '13 at 8:17

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