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 2


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 non-UFD $R$? $\endgroup$ Jan 5, 2010 at 7:08

Google "Kronecker function ring". This is the germ of the idea behind Kronecker's divisor theory. An elementary historically-minded introduction can be found in Harold Edward's book "Divisor Theory".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.