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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.

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The notation is a bit of a mess, because we usually mean $A[x]$ to be a ring of polynomials with an indeterminate $x$. As it turns out, since $x$ is transcendental, your $A[x]$ is isomorphic to that ring of polynomials over $A$, however, so the existence of extensions of $\phi$ follows from universal properties of $A[x]$ –  Thomas Andrews Aug 13 '12 at 14:53
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For your last paragraph: if a nonzero element of $A(x)$ is sent to $0\in\Omega$, what can you say about the kernel of $\phi$? (Remember that $A(x)$ is itself a field.) –  M Turgeon Aug 13 '12 at 14:56
    
@MTurgeon: The kernel of any field homomorphism must be the zero ideal. –  Manos Aug 13 '12 at 14:58
    
@Manos Then what can you conclude about your last paragraph? –  M Turgeon Aug 13 '12 at 14:59
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Dear Manos: If $\phi:A\to B$ is a commutative ring morphism, $X$ an indeterminate, and $b$ an element of $B$, then there is a unique morphism $A[X]\to B$ which coincides with $\phi$ on $A$ and maps $X$ to $b$. –  Pierre-Yves Gaillard Aug 13 '12 at 15:54

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up vote 3 down vote accepted

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 $1/(x-b)$.

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Thank you for your nice answer :-) –  Manos Aug 14 '12 at 13:45

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