Posted by: hilbertthm90 | March 8, 2009

Lying Over and Going Up Part II

I realized there was one more result I probably should have included last time. Oh well. Here goes:

Let R^*/R be integral, \frak{p} a prime ideal in R and \frak{p}^*, \frak{q}^* prime ideals in R^* lying over \frak{p}. If \frak{p}^*\subset \frak{q}^*, then \frak{p}^*=\frak{q}^*.

Proof: Recall that S^{-1}R^* is integral over S^{-1}R by last time for any multiplicative set, and also that prime ideals are preserved in rings of fractions. Thus the hypotheses still hold if we localize at \frak{p}. Thus R_\frak{p}^* is integral over R_\frak{p}, and \frak{p}^*R_\frak{p}^*\subset \frak{q}^*R_\frak{p}^* are prime ideals. Thus we can WLOG replace R^* and R by their localizations and hence assume they are local. So now \frak{p} is a maximal ideal in R. Thus by last time \frak{p}^* is maximal. Since \frak{p}^*\subset \frak{q}^*, we have \frak{p}^*=\frak{q}^*.

Now we are ready for the two big theorems. Here is the “Lying Over” Theorem. Let R^*/R be an integral extension. If \frak{p} is a prime ideal in R, then there is a prime ideal \frak{p}^* in R^* lying over \frak{p}, i.e. \frak{p}^*\cap R=\frak{p}.

Proof: First note that R \stackrel{i}{\longrightarrow} R^* \stackrel{h^*}{\longrightarrow} S^{-1}R^* and R \stackrel{h}{\longrightarrow} R_\frak{p} \stackrel{j}{\longrightarrow} S^{-1}R^* form the two sides of a commutative diagram. By last time S^{-1}R^* is integral over R_\frak{p}. Choose a maximal ideal \frak{m}^* in S^{-1}R^*. Thus \frak{m}^*\cap R_\frak{p} is maximal in R_\frak{p}. But R_{\frak{p}} is local with unique max ideal \frak{p}R_\frak{p}, so \frak{m}^*\cap R_\frak{p}=\frak{p} R_\frak{p}. But the preimage of a prime ideal is prime, so \frak{p}^8=(h^*)^{-1}(\frak{m}^*) is a prime ideal in R^*.

Now we just diagram chase: (h^*i)^{-1}(\frak{m}^*)=i^{-1}(h^*)^{-1}(\frak{m}^*)=i^{-1}(\frak{p}^*)=\frak{p}^*\cap R. And also: (jh)^{-1}(\frak{m}^*)=h^{-1}j^{-1}(\frak{m}^*)=h^{-1}(\frak{m}^*\cap R_\frak{p})=h^{-1}(\frak{p}R_\frak{p})=\frak{p}.

Thus \frak{p}^* lies over \frak{p}.

Our other big theorem is the one about “Going Up”: If R^*/R is an integral extension and \frak{p}\subset \frak{q} are prime in R, and \frak{p}^* lies over \frak{p}, then there is a prime ideal \frak{q}^* lying over \frak{q} with \frak{p}^*\subset \frak{q}^*.

Proof: By last time (R^*/\frak{p}^*)/(R/\frak{p}) is an integral extension where R/\frak{p} is embedded in R^*/\frak{p}^* as (R+\frak{p}^*)/\frak{p}^*. Now we just replace R^* and R by these rings so that both \frak{p}^* and \frak{p} are \{0\}. Now we just apply the Lying Over Theorem to get our result.

So as we see here integral extensions behave extremely nicely. These theorems guarantee that se always have prime ideals lying over ones in the lower field. This has some important applications to the Krull dimension that we’ll start looking at next time.

Posted in algebra | Tags: , , , , ,

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Responses

  1. [...] Given a prime , there is a prime lying above . I hinted at the proof in the previous post, but to save time and avoid too much redundancy I’ll refer interested readers to this post. [...]

    By: e, f, and the remainder theorem « Delta Epsilons on September 12, 2009
    at 11:57 am


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