2

Let $R$ be a PID. Show that if $\mathfrak p \subset R[x]$ is a prime ideal, $(r) = \left\{h(0) \colon h(x) \in \mathfrak p \right\}$, and $$\mathfrak p = (r, f(x), g(x)),$$ where $f(x), g(x) \in R[x]$ are nonconstant irreducible polynomials, then $f(x)$ and $g(x)$ are associates. So $\mathfrak p = (r, f(x))$.

user26857
  • 52,094
pseudoname123456
  • 101

2 Answers2

3

After seeing the comments below I see that

1)[most importantly] my answer was clearly wrong
2) the OP question cannot be solved as the counterexample of user26857 shows: $(2,x,x+2)=(x,2)=\{h(x) \in \mathbb Z[x] \mid h(0)=2\}$, so $r=2$ and $x$ and $x+2$ are not associated.

Since the OP aims seems to be following

I am trying to solve an exercise to show that every prime ideal in R[x] is of the form (r,f(x))

as amend I'll give a sketch of a possible solution to the problem.

The idea is to consider for a generic prime $p$ in $R[x]$ the ideal $p'=p \cap R$ which is prime in $R$. Since $R$ is a PID we have two possibilities

  • $p'=(0)$: in this case one can consider an irreducible polynomial $f$ of minimal degree in $p$ (this polynomial exists as a consequence of the primality of $p$). The one can consider a generic $g \in p \setminus \{0\}$ and the ideal $(f,g)$ in $K(R)[x]$ proving that $(f,g)=(f)$ in $K(R)[x]$ and then using Gauss Lemma to conclude that $p=(f)$.
  • $p'=(r)$ for some $r \ne 0$: in this case $R[x]/p \cong (R[x]/(r))/(p/(r))$. Up to the isomorphism $R[x]/(r) \cong R/(r)[x]$ we have that $R/(r)$ is a field (because $(r)$ is a not null prima ideal of a PID, hence it is maximal) so $p/(r)$ is principal ($R/(r)[x]$ is a PID, indeed an Euclid ring) and so it is generated by the image of a polynomial $f \in R[x]$ under the homomorphism $R[x] \to R/(r)[x]$. Since $p/(r)=(\bar f)$ by the theorem on ideals of quotient rings $p=(r,f)$, of course the case $f=0$ is contemplated.

The hardest part to work out is the first one ($p'=(0)$) because the details are a little boring but it is an easy exercise, feel free to ask if something does not add up.

user26857
  • 52,094
Giorgio Mossa
  • 18,150
  • Very clear, thanks. – pseudoname123456 Jun 29 '15 at 15:10
  • Upon going over this again, I think there might be a problem with your answer. You say that $d(x)$ must be associated with $\underline{f(x)}$ and $\underline{g(x)}$, but $d(x)$ could be a constant; for example, if $f(x)$ and $g(x)$ are not associates, then $d(x) = 1$. Notice that your solution doesn't use the hypothesis that the ideal $p$ is prime. – pseudoname123456 Aug 02 '15 at 19:18
  • I have to edit the question to add that $r$ is defined as the generator of $\left{f(0) \colon f(x) \in p \right}$. – pseudoname123456 Aug 02 '15 at 19:56
  • Notice as well that the answer, as written, appears to imply that any two nonconstant, irreducible polynomials are associates, which is not true. – pseudoname123456 Aug 02 '15 at 23:03
  • Well, I had originally accepted it, but then I went through it again and saw this potential error, and so I was waiting to have my observation validated before unaccepting the answer. – pseudoname123456 Aug 05 '15 at 17:33
  • @pseudoname123456 The bad news is that (not only this answer, but) the question is wrong as well. – user26857 Aug 06 '15 at 06:01
  • Yea. I am trying to solve an exercise to show that every prime ideal in $R[x]$ is of the form $(r, f(x))$ and I came up with this (false) claim that I asked above. I know a prime $\mathfrak{p}$ is of the form $(r, f_1(x), ..., f_n(x))$ since $R[x]$ is Noetherian, and I was trying to go from there. – pseudoname123456 Aug 06 '15 at 14:12
  • Note that we may use the primality of $\mathfrak{p}$ to say that the $f_i(x)$ are nonconstant irreducible. I want to proceed by induction, showing $f_n(x) \in (r, f_1(x), ..., f_{n-1}(x))$ (trying to generalize your example above). Say $\mathfrak{p} = (r, f(x), g(x))$. I want to show $g(x) \in (r, f(x))$. If $f(x)$ and $g(x)$ are not associates, then, passing to the field of fractions, there are $a(x)$, $b(x)$ in $R[x]$ and $s \in R$ such that $a(x)f(x) + b(x)g(x) = rs$. I cannot figure out how to proceed from here. – pseudoname123456 Aug 06 '15 at 15:24
  • Also, I think I defined $r$ incorrectly. I think it is supposed to be the generator of the set of constants in $\mathfrak{p}$, contrary to what I wrote above. – pseudoname123456 Aug 06 '15 at 16:11
  • Unfortunately I've been for a long time far away from an internet connection. I see the problems with the answer, thanks for pointing out. – Giorgio Mossa Aug 23 '15 at 18:38
  • I've made some changes to the answer, with some additional data that I hope may help you to solve the problem you were addressing. – Giorgio Mossa Aug 23 '15 at 19:57
  • Actually the proof for $R=\mathbb Z$ works as well for $R$ a PID. See http://math.stackexchange.com/questions/174595/classification-of-prime-ideals-of-mathbbzx – user26857 Aug 23 '15 at 20:20
  • @user26857 indeed my proof is based on an adaptation of the proof for $\mathbb Z$ :) – Giorgio Mossa Aug 23 '15 at 21:08
1

As stated, the claim being made in the post is false, as user26857 makes clear in the comments. To reiterate, a counterexample is $$\mathfrak{p} = (2, x, x+2) \subset \mathbf{Z}[x].$$

pseudoname123456
  • 101