Re: Product of Prime Ideals

On Thu, 6 Mar 2008 21:17:10 +0000 (UTC), magidin@xxxxxxxxxxxxxxxxx
(Arturo Magidin) wrote:

In article <e0da6af8-3672-4027-9b74-2ac8e981f3a1@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
hagman <google@xxxxxxxxxxxxx> wrote:
On 6 Mrz., 21:08, quasi <qu...@xxxxxxxx> wrote:
On Thu, 6 Mar 2008 19:53:10 +0000 (UTC), magi...@xxxxxxxxxxxxxxxxx

(Arturo Magidin) wrote:
In article <3gg0t3dq07fhdtdn73bmj7rm0u8l61m...@xxxxxxx>,
quasi <qu...@xxxxxxxx> wrote:

So then, how about an example of prime ideals P,Q in a commutative
unit ring R such that the product ideal P*Q is a proper subset of the
ideal (P intersect Q).

Probably there are easy examples, but thinking about it briefly, I
couldn't come up with one.

You mean, besides taking P=Q? (4) = (2)(2) =/= (2)/\(2) in Z.

Yes, of course, I meant to bar inclusion of either one into the other.

To restate the request ...

Give an example of prime ideals P,Q in a commutative ring R with 1,
where neither of P,Q is a subset of the other, and such that the
product ideal P*Q is a proper subset of the ideal (P intersect Q).

quasi

Since (P intersect Q) is in P, we must have P as a factor of (P
intersect Q).
Similarly, Q is a factor.
Since P and Q are different, they both appear in the prime ideal
decomposition
of (P intersect Q).
Hence P*Q subseteq (P intersect Q) = P * Q * ... subseteq P*Q

That's true if you are in a Dedekind domain, but need not be true in
arbitrary rings.

It is also true if P + Q = R. So what you want is some ring if Krull
dimension greater than 1, and pick P and Q distinct nonprincipal
primes contained in some larger ideal.

This suggests taking R[x,y,z] (R the reals, but really any integral
domain), and taking P=(x,y) (prime since the quotient is isomorphic to
R[z], which is an integral domain), and Q = (y,z) (prime since the the
quotient is R[x]).

Note that y in P/\Q.

On the other hand, PQ = (xy, y^2, xz, yz); note that y is not in
PQ.

Yes, that example works nicely.

Thanks.

quasi
.

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