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Let $B$ be a noetherian integral domain and consider the polynomial ring $B[t]$. Let $\mathfrak{p}$ be a prime ideal in $B[t]$ of height $1$ such that $\mathfrak{p} \cap B = (0)$. Show that $B[t]_{\mathfrak{p}}$ is just $K[t]_{\mathfrak{m}}$ where $K$ is the field of fractions of $B$ and $\mathfrak{m}$ is some maximal ideal of $K[t]$.

Note this isn't a homework problem, it's actually a step in a proof in Hartshorne that I got stuck on, showing that the preimage of the generic point under the fibered product is a regular codimension $1$ point. For reference it's Proposition II.6.6.

It seems like this should be straight forward, but I'm having trouble making it rigourous. We have $B[t]_{\mathfrak{p}}$, and since $\mathfrak{p}$ doesn't contain any non-zero element in $B$, then $B[t]_{\mathfrak{p}}$ must at least contain $K[t]$. If $B$ was a UFD then it would be easy, since we could say that $\mathfrak{p}$ is principal. But say $\mathfrak{p}$ is generated by $\{ f_{1}, f_{2}, \ldots , f_{r} \}$. I can "see" intuitively that the result should be $K[t]_{\langle f_{1}, \ldots, f_{r}\rangle}$, and of course $\langle f_{1}, \ldots, f_{r} \rangle$ would be principal in $K[t]$. But I can't seem to be able to make this rigorous, which is worrying to me because I feel like this should be extremely straight forward. Any advice?

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  • $\begingroup$ Since it's just a step in a proof, are you sure all the hypotheses are necessary ? I seem to have a proof that doesn't use the height of $\mathfrak{p}$ $\endgroup$ – Maxime Ramzi May 5 '18 at 12:30
  • $\begingroup$ No I'm not sure that hypothesis is necessary for this particular claim. $\endgroup$ – Luke May 5 '18 at 13:19
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Let me give it a try: let $\mathfrak{m}$ be a maximal ideal of $K[t]$ containing $\mathfrak{p}$ ($\mathfrak{p}K[t]$ contains non nonzero constant element because $\mathfrak{p}$ doesn't, so it's a strict ideal: such an $\mathfrak{m}$ exists).

The idea of my proof is to show that $K[t]_\mathfrak{m}$ has the universal property of $B[t]_\mathfrak{p}$, which will allow us to prove the isomorphism between the two (and it's easy to check that it's "the right isomorphism")

Consider a morphism $f: B[t]\to C$, $C$ a ring such that for $x\notin \mathfrak{p}$, $f(x) \in C^\times$.

Compose with the inclusion $i: B\to B[t]$ this yields a morphism $f\circ i: B\to C$, and your hypothesis concerning $\mathfrak{p}\cap B$ + the hypothesis concerning $f$ yields that this extends to $g: K\to C$.

By the universal property of $K[t]$, sending $K$ to $C$ with $g$ and $t$ to $f(t)$ yields a morphism $s: K[t] \to C$, which corresponds to the following commutative diagram :

$\require{AMScd} \begin{CD} B @>>> K;\\ @VV^iV @VVV \\ B[t] @>>> K[t];\\ @VV^fV @VV^sV \\ C@>>^{id_C}>C \end{CD}$

Now let $A \in K[t], A\notin \mathfrak{m}$. For some $q\in B$, $qA\in B[t]$. $s(qA) = f(qA)$. Now if $qA \in \mathfrak{p}$ then $qA \in \mathfrak{m}$ hence (by maximality) $q\in \mathfrak{m}$ or $A\in \mathfrak{m}$. The second option is impossible by hypothesis, but so is the first since $q\in K^\times$. Hence $qA\notin \mathfrak{p}$ and so $f(qA)\in C^\times$. Hence $s(q)s(A)\in C^\times$, and so $s(A) \in C^\times$.

By the universal property of the localization at $\mathfrak{m}$, we get a morphism $e:K[t]_{\mathfrak{m}}\to C$ making the whole thing commute:

$\require{AMScd} \begin{CD} B @>>> K;\\ @VV^iV @VVV \\ B[t] @>>> K[t] @>>> K[t]_{\mathfrak{m}};\\ @VV^fV @VV^sV @VV^eV \\ C@>>^{id_C}>C @>>^{id_C}> C \end{CD}$

So apart from uniqueness which is left to prove, $K[t]_{\mathfrak{m}}$ satisfies the universal property of $B[t]_{\mathfrak{p}}$, so it is isomorphic to it.

Let's now prove the uniqueness part: assume the morphism $l: K[t]_{\mathfrak{m}}$ makes the traingle commute:

$\require{AMScd} \begin{CD} B[t] @>>> K[t]_{\mathfrak{m}};\\ @VV^fV @VV^lV \\ C@>>^{id_C}>C \end{CD}$

Then the induced map $K\to C$ is necessarily obtained by the universal property of $K$ with respect to $B$, because of the uniqueness there, similarly for the map $K[t]\to C$, and so the map $l:K[t]_\mathfrak{m}\to C$ is the one that is induced by the universal property with respect to $K[t]$ , that is $l=e$.

So uniqueness follows, and we have our proof.

(If anyone knows how to make diagonal arrows instead of putting $id_C$'s everywhere, please feel free to edit)

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Here's a proof with a more straightforward flavor (and no universal property reasoning). These very basic ideas come up again and again when working with polynomial domains and so-called uppers to zero, and once they're digested i feel the statement in question does become something of the one-liner-no-brainer that you hoped it would.

The only assumptions we need are that $B$ is an integral domain and $\mathfrak{p}$ is a nonzero prime ideal of $B[t]$ such that $\mathfrak{p} \cap B = 0$. (We could deduce that $\mathfrak{p}$ is nonzero from the much stronger assumption that it has height $1$).

We will just use two very elementary facts about polynomial localizations, which you can easily verify:

Given a domain $B$ and prime $\mathfrak{p} \subset B[t]$ such that $\mathfrak{p} \cap B = \mathfrak{p_0}$, it's always the case that

(1) $\mathfrak{p} B_{\mathfrak{p_0}}[t]$ is a prime ideal of $B_{\mathfrak{p_0}}[t]$.
(2) $B[t] \cap \mathfrak{p}B_\mathfrak{p_0}[t] = \mathfrak{p}$

In particular if $\mathfrak{p_0} = 0$ we get that

(1') $\mathfrak{p}K[t]$ is a maximal ideal of $K[t]$ (since primes are maximal in $K[t]$, it being a $PID$) and
(2') $B[t] \cap \mathfrak{p}K[t] = \mathfrak{p}$

By (1') we can take $\mathfrak{m} = \mathfrak{p}K[t]$.
By (2') we see that every element of $B[t] - \mathfrak{p}$ is also an element of $K[t] - \mathfrak{m}$, and thus $B[t]_\mathfrak{p}$ canonically embeds in $K[t]_\mathfrak{m}$: any fraction in $B[t]_\mathfrak{p}$ is a perfectly good fraction in $K[t]_\mathfrak{m}$.

The opposite inclusion is even simpler. Consider any element of $K[t]_\mathfrak{m}$. Write it as $\frac{g}{h}$ with $g \in K[t]$ and $h \in K[t] - \mathfrak{p}K[t]$. Multiply $g$ and $h$ by a common denominator $d \in B$ to get an equivalent fraction $\frac{g'}{h'}$ with $g',h' \in B[t]$. Since $h'$ is also not in $\mathfrak{p}K[t]$ (else $h = \frac{h'}{d}$ would have been too) we have that $h' \in B[t] - \mathfrak{p}$, and $\frac{g}{h} \sim \frac{g'}{h'}$ is a fraction in $B[t]_\mathfrak{p}$.

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