Let $\alpha: F \rightarrow F$ be a one to one field homomorphism, prove $\alpha$ fixes the prime subfield of $F$ elementwise

Question

Let $\alpha: F \rightarrow F$ be a one to one field homomorphism, prove $\alpha$ fixes the prime subfield of $F$ elementwise.

So I already proved that $\alpha$ fixes the identity element and used that to prove that $\alpha$ fixes the prime subfield of $F$ when $Char(F)=p$, which was easy because in this case the prime subfield is just $\mathbb{Z_p}$. I'm not trying to do this when $Char(F)=0$, so the prime subfield of $F$ is going to be isomorphic to $\mathbb{Q}$. I as going to proceed the same way, let $\frac{a}{b} \in \mathbb{Q}$ then $\alpha (\frac{a}{b})=\alpha(\frac{1+1+...+1_a}{1+1+1...+1_b})$.. but I got a little thrown off because i'm not used to using the multiplicative properties of a homomorphism when there is a denominator involved... I guess I could write the denomoniator $(1+1+1+....1_b)^{-1}$ and go from there but I just wanted to double check with yall, and besides that every time I ask something on here I get some new insight into something so I thought why not. Thanks in advance!

Yes, I also get some new insight form this site, reading so many great answers. I saw this answer, by the way. — Dietrich Burde, Oct 16 '18 at 16:24

score 1 · Answer 1 · answered Oct 16 '18 at 16:35

1

You appear to be comfortable showing that $\alpha(a)=a$ for $a\in\mathbb{Z}$. It is now enough to show that $$\displaystyle b\alpha\left(\frac{a}{b}\right)=a.$$ Well, \begin{align} b\alpha\left(\frac{a}{b}\right)&=\alpha(b)\alpha\left(\frac{a}{b}\right)\\ &=\alpha\left(b\frac{a}{b}\right)\\ &=\alpha(a)\\&=a. \end{align}

answered Oct 16 '18 at 16:35

David Hill

12,483

Also, using $\alpha(ab^{-1})=\alpha(a)\alpha(b)^{-1}$ is another fine proof. – David Hill Oct 16 '18 at 16:36

Let $\alpha: F \rightarrow F$ be a one to one field homomorphism, prove $\alpha$ fixes the prime subfield of $F$ elementwise

1 Answers1