I'm having trouble trying to prove the following theorem:

Proposition: Let $R$ be a PID. The prime ideals of $R[y]$ are precisely the ideals of the following form:

  • $(0)$,
  • $(f(y))$ where $f$ is an irreducible polynomial in $R[Y]$
  • $(p, f(y))$ where $p \in R$ is prime and $f(y)$ is irreducible in $R[y]$ and its image is irreducible in $(R/p)[y]$.

The proof essentially goes as follows: (Sketch) Suppose $P$ is a non-principal prime ideal that contains a prime ideal $p'=(p_0)$ of $R$ where $p_0$ is prime in $R$. Then the image of $P$ under the canonical map $\phi:R[Y]\rightarrow (R/p')[Y]$ is non-zero and prime which implies that $\phi(P)=(f')$ where $f'\in (R/p')[Y]$ is irreducible. Now, I can show that $P=(p_0, f)$ where $f$ is some element in the pre-image of $f'$. However, I need $f$ to be irreducible in $R[Y]$ as well.

  1. How do I choose $f$ to be an irreducible element in $R[Y]$
  2. Why are prime ideals of the third form maximal ideals in $R[Y]$.

As suggested by user26857:

We should consider the following map: $R[Y]\rightarrow (R/p)[Y]\rightarrow (R/p)[Y]/(\bar{f})$. The kernel of this surjective map is $(p,f(y))$ and $(R/p)[Y]/(\bar{f})$ is a field which implies that $(p,f(y))$ is maximal.

Any ideas or suggestions? (For the first question)

Jhon Doe
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  • Try this: $R[y]/(p,f(y))\simeq (R/p)[y]/(\bar f(y))$. – user26857 Feb 28 '21 at 22:19
  • Thanks. I think that solves the 2nd question but do you have any ideas about the first question? – Jhon Doe Mar 01 '21 at 02:27
  • Duplicate of [What do prime ideals in $k[x,y]$ look like?](https://math.stackexchange.com/questions/56916/what-do-prime-ideals-in-kx-y-look-like) – KReiser Mar 01 '21 at 02:34
  • @KReiser Im proving the theorem suggested by Qiaochu Yuan in that link. But in his theorem the polynomial $f\in R[Y]$ is just required to have a irreducible image in $(R/p)[Y]$ whereas in my case I need $f$ to be irreducible in $R[Y]$ as well. So I do not know how to retrieve such a irreducible $f$. – Jhon Doe Mar 01 '21 at 02:37
  • @KReiser The link you suggested does not answer question (1) – Jhon Doe Mar 01 '21 at 02:47

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