1
$\begingroup$

I'm filling the gaps in a proof and I'm stuck in this part:

Suppose $R$ is a UFD and $Q$ is a prime ideal of $R[x]$, if $F$ is the quotient field of $R$ and $R\cap Q=\{0\}$, then $QF[x]\cap R[x]=Q$.

I've been dealing with this problem for some days and every idea I have never works. I consider now is the time to come here, so I hope that some of you could give me a hand with this.

I don't post what I've done in order to solve the problem because, as I have told, anything of it worked.

Hints will suffice. Thanks a lot.

$\endgroup$
  • $\begingroup$ It seems you missed the following: if $P$ is a prime ideal of a commutative ring $R$, $S\subset R$ a multiplicative set, and $P\cap S=\emptyset$, then $S^{-1}P\cap R=P$. This is very easy to prove and I suggest you to try it. (Btw, this has nothing to do with $R$ being an UFD.) $\endgroup$ – user26857 Aug 31 '14 at 20:50
  • $\begingroup$ Ok, i'm really thankful. I've proved what you suggested me, now I want to apply it to my original problem, I think this is what I need to do: Your $R$ is my $R[x]$, your $P$ is my $Q$ and $S$ is my original $R$ minus $\{0\}$, I'd only need to show that $QF[x]=S^{-1}Q$ and that's all. Am i right? $\endgroup$ – Daniel Aug 31 '14 at 21:15
  • $\begingroup$ Yes, you are right. $\endgroup$ – user26857 Aug 31 '14 at 21:23
1
$\begingroup$

Your claim is wrong!

Take $R=\mathbb Z$ and $Q=2\mathbb Z[X]$. Then $Q\mathbb Q[X]=\mathbb Q[X]$.

However, if you add to the hypothesis $Q\cap R=(0)$, then the equality holds since then the extension of $Q$ to the ring of fractions $F[X]=S^{-1}R[X]$, where $S=R-\{0\}$, is $S^{-1}Q\ne F[X]$.

$\endgroup$
  • $\begingroup$ This answer was posted before the edit (which was actually suggested in the answer) of the question. $\endgroup$ – user26857 Aug 31 '14 at 20:52
  • $\begingroup$ I'm sorry, I should have clarified that. I was about to thank you, but the other answer appeared and I began to analyze it. Thanks again. $\endgroup$ – Daniel Aug 31 '14 at 21:17
  • $\begingroup$ @Devilathor Actually there are no two answers: in my comment under your question I wanted to clarify you what exactly I've meant in the second part of my answer. $\endgroup$ – user26857 Aug 31 '14 at 21:25
  • $\begingroup$ Oh, I'm sorry, I didn't even realized it was you the one in the comment under my question. Thanks again, you took me out from a jam. I only have one last question if you don't mind: $S^{-1}Q$ is defined as $\{q/s:q\in Q, s\in S\}$ and $QF[x]$ is the set of all finite sums of elements of the form $qf(x)$ with $q\in Q$ and $f(x)\in F[x]$, isn't it? Can I define $S^{-1}Q$ as the set of all finite sums of elements of the form $q/s$ with $q\in Q$ and $s\in S$? $\endgroup$ – Daniel Aug 31 '14 at 21:38
  • $\begingroup$ The answer to both questions is yes, but why define the last ideal in that way? If you add all that fractions you get only one with numerator in $Q$ and denominator in $S$. $\endgroup$ – user26857 Aug 31 '14 at 21:40

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.