# How logarithm is properly defined on a field?

Given a field $(F,+,\times)$, an exponential function is defined as a function $E:F\to F$ s.t. $E(x+y)=E(x)E(y)$ and $E(0)=1$ where $0$ is the additive identity and $1$ is the multiplicative identity.

I am curious how logarithm is properly defined for $F$ and how the connection with the exponential function is made? Is it defined as a $L:F \to F$ function s.t. $L(xy)=L(x)+L(y)$ and $L(1) = 0$? Any explanation and reference is welcome. Thank you!

• By using formal power series. Commented Sep 28, 2016 at 17:04
• @user26857 if $F$ has a topology ... Commented Sep 28, 2016 at 18:39
• @user251257 I have no idea why need a topology. Characteristic zero should be enough. Commented Sep 28, 2016 at 18:44
• @user26857 what do you do with that formal series? Commented Sep 28, 2016 at 18:46
• Logarithm has a Taylor expansion which can be thought of as a definition over fields of characteristic zero. Commented Sep 29, 2016 at 6:30

I suppose that the function defined by the functional equation is continuous, so to avoid ''wild'' solutions. In this case, to see when we can define an inverse (logarithm) function, define: $$A_0=\{x\in F : E(x)=1\}$$

Now we can prove that if $A_0=\{0\}$ that $E(x)$ is invertible.

Prove by contraposition:

If we have $x_1,x_2 \in F$ such that $E(x_1)=E(x_2)$ than $E(x_1-x_2)=E(x_1)E(-x_2)=E(x_1)E(x_2)^{-1}=E(x_1)E(x_1)^{-1}=1$, so $x_1-x_2 \in A_0$: contradiction.

In this case we can define an inverse of the $E$ function: $$L=E^{-1}:E(F)\to F \quad L(a)=x \quad \mbox{such that}\quad E(x)=a$$ and we can prove that $L(ab)=L(E(x)E(y))=L(E(x+y))=x+y=L(a)+L(b)$. This is the case if $F=\mathbb{R}$.

But, if there is $x_0\ne0 \in A_0$ than we can prove that $E(x)$ is a periodic function because: $E(x+nx_0)=E(x)E(nx_0)=E(x)E(x_0)^n=E(x)\cdot 1^n= E(x)$. So the function $E$ is not invertible and if we want define a ''logarithm'' we have to chose one period that fix a ''principal value'' for the inverse function. This is the case if $F=\mathbb{C}$.

• Where do you need continuity? Commented Sep 28, 2016 at 23:55
• If we don't require continuity ( or at least misurability) we can have many different functions that satisfies the functional equation Commented Sep 29, 2016 at 7:52
• yeah. But you haven't shown that there is at most one solution. So you don't really need that for what you have written in your answer, do you? Commented Sep 29, 2016 at 7:59