# Thread: Ring homomorphisms - statement

1. ## Ring homomorphisms - statement

If for all r $\in$ R there exists some $x \in R$ such that $x^2 = r$ then the same statement is true in $\varphi$ [R].

This shows that there is no homomorphism that maps the Complex numbers onto the real numbers (surjective)

i'm not quite sure how it can be used to disprove some onto homomorphisms' existences like the one above

2. ## Re: Ring homomorphisms - statement

Suppose we have a surjective homomorphism $\varphi:R\to R'$ and suppose

$\forall r\exists x\;x^2=r$ (*)

holds in R. Then (*) holds in R' as well. Indeed, fix any $r'\in R'$. Since $\varphi$ is surjective, there exists an $r\in R$ such that $\varphi(r)=r$. By (*), $x^2=r$ for some $x\in R$, so $\varphi(x)^2=\varphi(r)=r'$, as required.

Since (*) is true in $\mathbb{C}$ but false in $\mathbb{R}$, there is no surjective homomorphism from $\mathbb{C}$ to $\mathbb{R}$.

3. ## Re: Ring homomorphisms - statement

thanks for the reply. I'm nearly there with my question (the question is different to the one above).

just quickly: is it allowed to do this for any given homomorphism:

Given that $\varphi (\sqrt{a}) \in Z$
then $\sqrt{\varphi(a)} \in Z$ also ?

4. ## Re: Ring homomorphisms - statement

Well, at least when $\sqrt{a}$ and $\sqrt{\varphi(a)}$ are unique, then $\varphi(\sqrt{a})=\sqrt{\varphi(a)}$. Suppose $b^2=a$, $\varphi(a)=a'$ and $\varphi(b)=b'$. Then $a'=\varphi(a)=\varphi(b^2)=\varphi(b)^2=b'^2$, so $\sqrt{\varphi(a)}=b'=\varphi(\sqrt{a})$.