For any UFD $R$, the concept of a primitive polynomial (gcd of the coefficients is 1) makes sense in $R[x]$. The product of two primitive polynomials is primitive (Gauss's Lemma), and certainly 1 is a primitive polynomial, so the primitive polynomials form a multiplicative subset $S$ of $R[x]$  hence we can form the ring $S^{1}R[x]$. What can we say about it? What does this look like geometrically?
2 Answers
A prime ideal of $S^{1}R[X]$ is the extension of a unique prime ideal of $R$, so that the morphism $Spec(S^{1}R[X])\to Spec(R)$ is a bijection, and even an homeomorphism. All the extensions of residual fields induced are pure transcendental of transcendence degre $1$.
As an example, if you look at the case $R=\mathbb{Z}$, the morphism of schemes you get "puts in family" the extensions of fields $\mathbb{F}_p\hookrightarrow\mathbb{F}_p(X)$.

$\begingroup$ In the first sentence, I think you mean "A prime ideal of S^{1} R[X] is generated by elements of R". In fact, you could say that every ideal of S^{1} R[X] is the extension of a unique ideal of R. This would show you that the map between spectra is a homeomorphism. $\endgroup$ Jan 5, 2010 at 1:40

$\begingroup$ Yes of course ! Thanks for pointing out the inaccuracy ! $\endgroup$ Jan 5, 2010 at 1:42

$\begingroup$ Thanks for the explanation! I'm still a novice with schemes, but I think I get the gist of your answer. Let me see if I understand the argument: Let $\phi:R[x]\rightarrow S^{1}R[x]$ be the canonical map (injective since $R[x]$ is an integral domain). Because $R[x]$ is a UFD, every polynomial factors uniquely into a product of irreducibles of $R[x]$, which are either irreducibles of $R$ or irreducibles $f\in R[x]$, $f\notin R$. Each irreducible $f\in R[x]$, $f\notin R$ is necessarily primitive, so that $\phi(f)$ is a unit in $S^{1}R[x]$. $\endgroup$ Jan 5, 2010 at 7:07

$\begingroup$ Thus, for $I\subset R[x]$ an ideal, $S^{1}I$ is generated in $S^{1}R[x]$ by $\phi(I)\cap R$ (where we have identified $R$ with $\phi(R)$). Thus, $S^{1}I$ is the is the extension of the ideal generated in $R$ by $\phi(I)\cap R$, and the injectivity of $\phi$ shows that this is the only ideal of $R$ which $S^{1}I$ is an extension of. Finally, the fact that preimages of prime ideals are prime and that extension of a prime ideal is its image under $\phi$, establishes the bijection. $\endgroup$ Jan 5, 2010 at 7:07

$\begingroup$ So, this answer definitely depends heavily on $R$ (and hence $R[x]$) being a UFD, especially with "each irreducible $f\in R[x]$, $f\notin R$ is necessarily primitive". Out of curiosity, any ideas on the case of nonUFD $R$? $\endgroup$ Jan 5, 2010 at 7:08
Google "Kronecker function ring". This is the germ of the idea behind Kronecker's divisor theory. An elementary historicallyminded introduction can be found in Harold Edward's book "Divisor Theory".