2
$\begingroup$

In a proof in my algebra book they state that it is sufficient to show that: if L/K is an algebraic extension and if $\varphi: L \rightarrow L$ is a surjective $K$-homomorphism, then $\varphi \in Aut_KL$.

I know this is true if $L/K$ is a finite algebraic extension. This because $L$ considered as vector space over $K$ has a finite dimension, and then surjectivity of $\varphi$ implies injectivity.

But how about infinite algebraic extensions? Could anyone point me into the right direction?

  • 0
    Well, it has to be $\{0\}$, but in the finite algebraic extension case, you can use the dimension formula to prove this. How would one approach this when the dimension is infinite?2011-12-14

3 Answers 3

3

Isn't it just because $\ker \varphi$ is an ideal of the field $L$ and $\varphi$ is certainly not the zero map because it fixes $K$? (I assume that "K-homomorphism" means "homomorphism of $K$-algebras".)

  • 0
    Yes, indeed and a field only has trivial ideals, thus the kernel has to be $\{ 0 \}$ thanks! I'll accept your answer as soon as it lets me.2011-12-14
1

Nontrivial ring homomorphisms between fields are necessarily injective.

HINT: The kernel of such a homomorphism is an ideal in $L$.

  • 0
    Oh right, a field only has trivial ideals. Thus it has to be $\{0\}$ that i didn't think of that -.-. Thanks!2011-12-14
0

The endomorphism need not be surjective. Injectivity is a property all homomorphisms that fix K share. Hint: a K-endomorphism maps any root of a given polynomial to a root of the same polynomial.