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As I know there is a theorem, which says the following:

The prime ideals of $B[y]$, where $B$ is a PID, are $(0), (f)$, for irreducible $f \in B[y]$, and all maximal ideals. Moreover, each maximal ideal is of the form $m = (p,q)$, where $p$ is an irreducible element in $B$ and $q$ is an irreducible element in $(B/(p))[y]$.

I managed to prove, that any prime non-principal ideal $m \subset B[y]$ is maximal. This gives the first part of the theorem. Now I want to prove, that $m = (p,q)$. I can show, that $m \cap B \neq (0)$. Thus $m$ contains a non-zero element $a$ from B. Since $B$ is PID and $m$ is prime, $m$ must contain an irreducible element from $B$ (we can take an irreducible factor $p$ of $a$).

As I understand, now I need to show, that $m$ contains an element $q$ (see the theorem above). Then it follows that $(p,q) \subseteq m$. And after that I have to prove, that $m \subseteq (p,q)$. How do I make these last steps?

Yes, I have seen similar questions with $B = k[x]$ or $\mathbb{Z}$, but I still cannot prove the last part.

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Since $\mathfrak m\cap B\neq0$ it is a prime ideal of $B$, generated by an irreducible element$p$. By the 3rd isomorphism theorem, $\mathfrak m$ corresponds to the maximal ideal $\mathfrak m/pB[x]$ of $B[x]/pB[x]\simeq B/(p)[x]$.

However, $B/(p)$ is a field and $B/(p)[x]$ is principal, hence $\mathfrak m/pB[x]$ is generated by a polynomial $\bar q(x)$, irreducible modulo $p$, hence $\mathfrak m =\bigr((p, q(x)\bigl)$. Furthermore $q(x)$ is irreducible in $B[x]$ since it is irreducible modulo $p$.

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  • $\begingroup$ Thanks, now I see this $1-1$ correspondence between $m$ and $m/pB[x]$. I have one more silly off topic question. If $m = pB[x]$, then $B[x]/m$ is not a field (which is a contradiction, since $m$ is maximal), so we can find such $\overline{q}$. But how one can see quicker, that $m \neq pB[x]$? $\endgroup$ – user128245 Sep 10 '16 at 0:57
  • $\begingroup$ Maybe you can say $(B/pB)[x]$ has Krull dimension $1$ (being a polynomial ring over a field), while $B[x]/\mathfrak m$ has dimension $0$, but it's not fundamentally different from the other argument – which isn't very long, really. $\endgroup$ – Bernard Sep 10 '16 at 1:17
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Let $p$ be a generator of $m\cap B$, so that $p$ is prime, and let $k=B/p$. Since $\mathfrak{m}$ is maximal in $B[X]$, $\mathfrak{m}/p$ is maximal in $B[X]/p \cong k[X]$.

Since $k$ is a field, maximal ideals of $k[X]$ are exactly those generated by irreducible polynomials, there is some polynomial $q\in B[X]$ whose image in $k[X]$ is irreducible and generates $\mathfrak{m}/p$. Then $\mathfrak{m}=(p,q)$.

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