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I'm trying to work out the following past paper question and I've got stuck.

$R$ is an integral domain and $S = R[t]$, the polynomial ring in one variable over $R$. We have that $Q$ is a prime ideal strictly contained in another prime ideal $P$ with $$ P \cap R = Q \cap R = P_0. $$ We also have $Y = R\setminus P_0$.

I have shown that $P_0S$ is a prime ideal, and that after localisation we have $$ \frac{SY^{-1}}{P_0SY^{-1}} \simeq F[t] $$ where $$ F = \frac{RY^{-1}}{P_0Y^{-1}}. $$ This is a field since $P_0 = R\setminus Y$, so $P_0Y^{-1}$ is a maximal ideal of $RY^{-1}$.

What I now have to show is that $P_0S = Q$, and the question says that i should do this using the above isomorphism of polynomial rings, but I can't figure it out.

A hint or solution would be great, thanks!

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The property you want to prove says that there are no chains of three prime ideals in $R[X]$ lying over the same prime ideal in $R$ (and this doesn't require $R$ an integral domain).

You have $Q\subsetneq P$ prime ideals in $R[X]$ with $P\cap R=Q\cap R=\mathfrak p$. It follows that $\mathfrak p[X]\subseteq Q$, and want to prove the last containment can't be strict. Suppose the contrary. Then look at the factor ring $R[X]/\mathfrak p[X]\simeq(R/\mathfrak p)[X]$. Here you have a chain of three prime ideals lying over $(0)$. Thus you can assume that $R$ is an integral domain (replacing $R/\mathfrak p$ by $R$), and there is a chain of there primes in $R[X]$ lying over $(0)$. Now invert all non-zero elements in $R$, and get a chain of three primes in $K[X]$, where $K$ is the field of fractions of $R$, which is a contradiction.

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    $\begingroup$ Sorry, I know this is a stupid question, but what exactly do you mean by the ideals are `lying over $(0)$'? Do you mean for each of $\mathfrak{p}[X],Q$ and $P$ they all get mapped to the zero ideal after quotienting out by $\mathfrak{p}[X]$? $\endgroup$ – CameronJWhitehead May 27 '15 at 11:50

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