Let $A \subseteq B$ be integral domains and let $\phi:A \rightarrow \Omega$ be a homomorphism of $A$ into the infinite algebraically closed field $\Omega$. Let $x \in B$ and suppose that $x$ is transcendental over $A$, i.e. it is the root of no non-zero polynomial with coefficients in $A$.
I would like to see whether my understanding about extending $\phi$ to $A[x]$ is correct. Let $g(x) \in A[x]$ be a monic polynomial of positive degree. Then it will be non-zero because $x$ is transcendental over $A$. Denote by $g^{\phi}$ the polynomial of $\Omega[z]$ obtained by taking its coefficients to be the images of the coefficients of $g$ under $\phi$. Then $g^{\phi}$ will also be non-zero and it will have a finite number of roots. Since $\Omega$ is infinite, we can find a $\xi \in \Omega$ such that $g^{\phi}(\xi) \neq 0$. Then we can extend $\phi$ to $A[x]$ by sending $x$ to $\xi$. Is this argument correct?
Now, if $A$ is a field and we are interested in extending $\phi$ to $A(x)$, the above argument is not valid anymore, because we can not guarantee that a non-zero element of $A(x)$ will be sent to a non-zero element of $\Omega$. Is this argument correct?
Thanks.