8
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Matt D
    Commented May 20 at 14:25

1 Answer 1

5
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Minor variation: Use $$\frac1{1+x}+\frac1{1-x}=\frac2{1-x^2}$$ $\endgroup$
    – tkf
    Commented May 21 at 1:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .