# Assistance with an exercise on field endomorphisms

Working through the problems in a book on field theory (Field Extensions and Galois Theory by Bastida). I came across one which I thought looked like a "routine" exercise, but has been particularly stubborn.

Suppose $$K$$ is a field with $$char(K) \neq 2$$ and $$u:K \rightarrow K$$ a map so that $$u(x+y) = u(x)+u(y)$$ for all $$x,y \in K$$, $$u(1) = 1$$ and $$u(x)u(1/x) = 1$$ for all $$x \in K^*$$. Show that $$u$$ is an endomorphism.

Seems straightforward, I just need to show $$u(xy) = u(x)u(y)$$. I figured out that I can reduce the problem to just verifying it for "squares." That is, if I can just show $$u(x^2) = u(x)^2$$ for all $$x$$.

In that case, we can say: $$u((x+y)^2) = u(x^2+ xy + xy + y^2) = u(x^2) + 2u(xy) + u(y^2)$$. But also $$u((x+y)^2) = u(x+y)u(x+y) = u(x)^2 + 2u(x)u(y) + u(y)^2$$. Then we have $$2u(xy) = 2u(x)u(y)$$ and since $$K$$ is a field not of characteristic 2, we can cancel the 2's and get the result.

Alas, proving the special case $$u(x^2) = u(x)^2$$ has resisted my efforts. Of course $$x = 0$$ is not a problem. Otherwise, I was trying to play around with $$u(x)u(1/x) = 1$$ and $$u(1) = 1$$ to get it. The closest I've managed to get by tinkering with these is $$u(x^2\cdot 1/x) = u(x)^2u(1/x)$$. A hint in the right direction would be appreciated, or, if this approach won't work, a nudge in another direction.

• It is from "Field Extensions and Galois Theory" by Julio Bastida, section 1.2. #7. It looks like I retyped the entire problem. Thanks for the interest in my question. It's the first one I've hit so far that caused me any grief. Commented May 20 at 14:25

I think you are going in the right direction. Your reduction to squares is a good start! It's a bit hard to give more of a hint without giving the entire game away. But roughly, I would re-frame the assumption $$u(x) u(1/x) = 1$$ as "$$u$$ commutes with taking multiplicative inverses", and then use this to deduce that $$u$$ commutes with "expressions built out of multiplicative inverses and addition/subtraction" (and play around a bit with such expressions). A complete solution is in the below spoiler tag:
We use the fact that $$x^2 = 1/(1/x - 1/(x + 1)) - x$$. Repeatedly applying the hypotheses, we end up with $$u(x^2) = u(x)^2$$. (Strictly speaking, you should observe that we never divide anything by zero. This is fine, because if $$x = 0$$ or $$x = -1$$, it's clear).
• Minor variation: Use $$\frac1{1+x}+\frac1{1-x}=\frac2{1-x^2}$$