Extending $\phi: A \rightarrow \Omega$ to $A[x] \rightarrow \Omega$ where $A$ is integral domain and $x$ transcendental over $A$ 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.
 A: Gathering up some of the comments, including P-Y Gaillard's and Th. Andrews':
First, we might demonstrate that a "transcendental" behaves just like an "indeterminate". The polynomial ring $A[X]$ in an indeterminate $X$ over ring $A$ is really the free $A$-algebra on one generator, meaning that, given any ring hom $f:A\rightarrow B$ and given $b\in B$, there is a unique extension $F:A[X]\rightarrow B$ such that $F(X)=b$. (Note, no "solving" of equations is required: $X$ can go anywhere, $0$ or otherwise.)
Thus, letting $x$ be the "transcendental", the identity map $A\rightarrow A$ extends uniquely to $A[X]\rightarrow A[x]$ extending $X$ to $x$. Now observe that the "transcendental" condition is exactly equivalent to the kernel being trivial, so $A[X]$ is in natural bijection with $A[x]$.
Thus, given a ring hom $A\rightarrow B$, and given any $b\in B$, there is a (unique) extension $A[x]\rightarrow B$ sending $x$ to $b$. 
Such maps need not extend to the field of fractions $A(x)$, however, as obvious examples show: for example, with $B=A$ and the identity map on $A$, $x$ can be sent anywhere at all, say to $b\in B=A$, but then $x-b$ maps to $0$, so this map cannot be extended to the fraction $1/(x-b)$. 
