So basically this question is from Serge Lang "Undergraduate Algebra" Third Edition, III, $\S$3, exercise 2.
Show that a ring-homomorphism of a field $K$ into a ring $R \neq {0}$ is an isomorphism of $K$ onto its image
My attempt:
let
$f : K \rightarrow R$ is a ring-homomorphism.
Need to show that $K \cong f(K)$.
It is enough to show that $f$ is bijective with respect to sets $K$ and $f(K)$.
(injectivity)
$$
\begin{align*}
x &\neq y \\
x – y &\neq 0 \\
f(x-y) &\neq 0 \\
f(x) – f(y) &\neq 0 && (*) \\
f(x) &\neq f(y)
\end{align*}
$$
(surjectivity)
From the definition $f(K)$ is an image of $K$ under $f$, so it must be surjective.
From this follows that $f: K \rightarrow f(K)$ is a bijection, and thus two sets are isomorphic.
(note) In $(*)$ we assume that $R \neq 0$, otherwise injectivity fails
Question:
I am not sure whether this solution is correct or not, because I did not use the fact that $K$ is a field. Could you please verify it?
Best Answer
Proof for injectivity:
You have showed that $\ker f=\{0\}$. We will continue from this observation.
Select arbitrary $x,y\in K$ s.t. $x\ne y$. Since $x-y\ne 0$, we yield $x-y\notin\ker f$. By definition of kernel, \begin{equation} a\in \ker f\iff f(a)=0\cdots(\star). \end{equation} Since ($A\iff B$) $\implies$ ($\neg A\iff\neg B$), we infer from ($\star$) that $$ a\notin \ker f\iff f(a)\ne 0 \cdots(\star\star).$$
Since $x-y\notin \ker f$, from ($\star\star$) we infer $$ f(x-y)\ne 0. $$ Therefore $f(x)-f(y)=f(x-y)$ is non-zero, i.e. $f(x)\ne f(y)$.