# Prime ideal in a polynomial ring over an integrally closed domain

Let $R$ be an integrally closed domain. Then $R[x]$ is also integrally closed. Let $P$ be a prime ideal of $R[x]$ with the property that $P\cap R=\left\{ 0\right\}$ and $P$ contains a monic polynomial. Prove that $P$ is principal.

Help me a hint to prove.

Thank for any insight.

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To prove $P$ is principal, you need to show some element of $P$ divides every other element of $P$. What are the candidates for this job? – hardmath Nov 20 '13 at 12:33
Can consider the minimal polynomial of $P$? – user109584 Nov 20 '13 at 12:37
Hint: $P$ corresponds to a prime ideal in $Q(R)[x]$. – Martin Brandenburg Nov 20 '13 at 13:15
What is $Q(R)[x]$? – user109584 Nov 20 '13 at 13:22
Is it the quotiens field of $R[x]$? – user109584 Nov 20 '13 at 13:24

Suppose $P$ is a prime ideal of $R[x]$ that contains a monic polynomial and whose intersection with the base ring, integrally closed domain $R$, is trivial, $P \cap R = \{0\}$.

We are asked to prove that $P$ is principal. In other words we need to show $P$ contains an element that divides all the polynomials in $P$. What are the candidates?

We know that $P$ contains a monic polynomial, so let's start there. While $P$ will certainly contain more than one monic polynomial, we know that $P$ is a prime ideal. This suggests that we want to get at irreducibility of the monic polynomial.

One way to do this is by picking a monic polynomial of minimal degree in $P$. Note that since $P$ has trivial intersection with $R$, the degree of any polynomial in $P$ must be at least 1.

Now use the primality of $P$ to show that any monic polynomial $f$ of minimal degree in $P$ must be irreducible. Primality of $P$ means that if $f = gh$, either $g \in P$ or $h \in P$. If $f = gh$ is monic, then both $g,h$ are monic (up to multiplication by units). Thus $P$ would contain a monic polynomial of degree less than the degree of $f$; contradiction.

Finally, use some ideas from the division algorithm to show such $f$ divides every element of $P$.

At this point the fact $R$ is integrally closed comes into play. The Question connects this to $R[x]$ also being integrally closed. Though we are not asked to prove that, it bears noting it is an easy consequence of this:

Lemma Let domain $R$ be integrally closed in its field of quotients $k$. If $g,h \in k[x]$ are monic polynomials such that $gh \in R[x]$, then $g,h \in R[x]$.

If $R$ were a UFD, this could be considered a corollary of Gauss's lemma on polynomial irreducibility (see Prop. 1 here), but using the weaker assumption of being integrally closed is strengthened by assuming monicity of $gh$ over mere primitivity. A proof is sketched at the link above, but its brevity bears a few words of elaboration.

Proof of Lemma: In the algebraic closure $\overline{k}$ of $k$, we can fully factor $g = \Pi (x-a_i), h = \Pi (x-b_j)$, so that the coefficients of $g$ are (elementary symmetric) polynomial expressions in roots $a_i$, and similarly the coefficients of $h$ in roots $b_j$. Since the roots $a_i,b_j$ are also roots of monic polynomial $gh \in R[x]$, they are integral over $R$. Dedekind (1877) showed the integral elements over a domain form a ring (this is covered in J. Milne's Algebraic Number Theory, Chapt. 2), so the coefficients of $g$ (resp. of $h$) are also integral over $R$. But these coefficients are in $k$, and $R$ is integrally closed in $k$, so the polynomials $g,h$ are in $R[x]$. QED

Since prime ideal $P \subseteq R[x]$ contains no nonzero constant, by a standard argument $kP$ is a proper ideal of $k[x]$ and $kP \cap R[x] = P$. As $k[x]$ is a Euclidean domain (division algorithm, degree argument), there $kP$ is a principal ideal, say generated by monic polynomial $g$. Because $f \in P \subseteq kP$, $f = gh$ for some monic $h \in k[x]$. Then, according to the Lemma, $g \in kP \cap R[x] = P$ and by irreducibility of $f$, $f = g$.

Now apply the observation @user makes in a Comment above: Since $f$ is monic, any element of $P$ can be expressed as $qf + r \in R[x]$ "by division algorithm" where $\deg(r) \lt deg(f)$. However the same is true in Euclidean domain $k[x]$ where we see by uniqueness that $r=0$ (since $f=g$ generates $kP$). Thus $f$ divides every element of $P$ over $R[x]$, as we wished to show.

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I need more details on the proof that any polynomial $f$ of minimal degree in $P$ must be irreducible. Can you explain more precisely to me? – user109584 Nov 20 '13 at 13:51
Is that enough detail? – hardmath Nov 20 '13 at 13:57
Yes it's enough. – user109584 Nov 20 '13 at 14:04
For the proof that such $f$ divides every element of $P$, I don't understand why we can prove. The last time, can you explain more precisely to me? – user109584 Nov 20 '13 at 14:11
I think you can use the fact that $R[x]$ is an integrally closed domain. – chuyenvien94 Nov 20 '13 at 14:18

Take $S=R-\{0\}$ and consider $S^{-1}(R[x])=(S^{-1}R)[x]$. (In the following denote by $K$ the field of fractions of $R$, that is $K=S^{-1}R$.) Then $S^{-1}P$ is prime in $K[x]$, so there exists $p\in K[x]$ monic such that $S^{-1}P=pK[x]$. Since $P$ contains a monic polynomial $f$ we get $f=pg$, $g\in K[x]$. Now use Gauss' lemma to get $p\in R[x]$ and prove that $P=pR[x]$.

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Why $S^{-1}R$ is prime in $K[x]$? – chuyenvien94 Nov 20 '13 at 13:45
Since $P$ is prime in $R[x]$, $S^{-1}P$ is prime in $K[x]=S^{-1}R[x]$ – user109584 Nov 20 '13 at 13:54
@user: I need more details on the proof that $P=pR[x]$. Can you give me more details on it? – user109584 Nov 20 '13 at 14:26
An element of $P$ is a multiple of $p$ in $K[x]$ and in the same time it's of the form $pq+r$ in $R[x]$ by division algorithm (this works since $p$ is monic). Now use the uniqueness of $q$ and $r$ in $K[x]$ and find $r=0$. – user89712 Nov 20 '13 at 16:17