0
$\begingroup$

Suppose we have a ring R. We can define E as the class of functions from R to R such that

$$ \forall f \in E. \, x y \in R. \, f (x \cdot y) = f (x) \cdot f(y) $$ $$ \forall f \in E. \, x \in R. \, n \in \mathbb{N} . \, f (x^n) = (f (x))^n $$

Where exponention by naturals is defined as repeated multiplication.

We can then define

$$ E_0 (x) = R_1 $$ $$ E_1 (x) = x $$ $$ \forall f g \in E. \, x \in R. \, (f \cdot g)(x) = f(g(x)) $$ $$ \forall f g \in E. \, x \in R. \, (f + g)(x) = f(x) \cdot g(x) $$

Where $R_1$ is the identity element.

$$ E_0 (x) \cdot E_0 (y) = R_1 \cdot R_1 = R_1 = E_0 (x \cdot y) $$ $$ (E_0 (x))^n = (R_1)^n = R_1 = E_0 (x^n) $$

So $E_0$ is in $E$.

$$ E_1 (x) \cdot E_1 (y) = x \cdot y = E_1 (x \cdot y) $$ $$ (E_1 (x))^n = x^n = E_1 (x^n) $$

So $E_1$ is in $E$.

$$ (f \cdot g) (x) \cdot (f \cdot g) (y) = f (g (x)) \cdot f (g (y) = f (g (x) \cdot g(y)) = f(g(x \cdot y)) = (f \cdot g) (x \cdot y)$$ $$ ((f \cdot g) (x))^n = f (g(x))^n = f (g(x)^n) = f(g(x^n)) = (f \cdot g) (x ^n) $$

So $(f \cdot g)$ takes two $E$ to $E$.

$$ (f + g) (x) \cdot (f + g) (y) = f (x) \cdot g(x) \cdot f (y) \cdot g (y) = f (x \cdot y) \cdot g(x \cdot y) = (f + g)(x \cdot y)$$ $$ ((f + g) (x))^n = f (x)^n \cdot g(x)^n = f (x^n) \cdot g(x^n) = (f + g)(x^n) $$

So $(f + g)$ takes two $E$ to $E$.

I think $E$ is isomorphic to the reals when the ring is reals and just functions $\lambda x. x^a$.

Is $E$ always isomorphic to $R$? What would $E$ be for more complicated rings such as the ring of square matrices?

I think it is important to note that

$$ \ln f (e^{x + y}) = \ln f (e^x) + \ln f(e^y) $$ $$ \ln f (e^{nx}) = n \ln f (e^x) $$

so

$$ g(x) = \ln f(e^x) $$

is a linear map if appropriate logarithm and exponential functions are defined.

1 Answers 1

0

First of all, note that the property $f(x^n) = f(x)^n$ for all $x \in R$ is already implied by $f(x \cdot y) = f(x) f(y)$ for all $x, y \in R$. So we are really dealing with maps which just respect the multiplication of $R$.

Note, however, that we are not addressing the additive structure of $R$ in any way to define $E$, so it seems unreasonable that you should get back this structure via this construction.

For the real numbers you seem to hope that every element of $E$ is of the form $x \mapsto x^a$ for some $a \in \mathbb{R}$. The absolute value $\mathbb{R} \to \mathbb{R}, x \mapsto |x|$ is an element of $E$ which is not of this form. Also note that maps of the form $x \mapsto x^a$ are not defined on all of $\mathbb{R}$ for arbitrary $a \in \mathbb{R}$.

In any way, $E$ cannot be isomorphic to $\mathbb{R}$, since it is not even a ring. The addition does not give rise to an abelian group, for example not even $E_1$ has an additive inverse.