Re: A most tiresome "ideal" problem... can one feed this to a computer algebra program?

On 09.10.2006 19:19, Jannick Asmus wrote:
On 09.10.2006 16:04, Jose Capco wrote:

Let Z be the whole numbers
let Z[T] be the polynomial ring over it
define two prime ideals in this ring by

p=<t^2-2>=<p>
q=<t^2-3>=<q>

define the multiplicative set
S=Z[T]/{p \cup q}

Define the ring (a qoutient ring of Z[T])
A=Z[T]S^{-1}

Define an ideal of A
I = pA \cap qA

and another ring
B={(j,k) in AxA : j-k in I}
(note: this can be expressed as a fiber product)

Now, Spec B (or the prime ideals of B) can easily be enumerated.

There are 4 ideals of B

p1 = {(j,0) : j is in I}, p2 = {(0,j) : j is in I}
p3= {(j,j+k) : j in pA, k in I}
p4= {(j,j+k): j in qA, k in I}

Also one sees that
p1 + p2 = p3 \cap p4

(but p1+p2 =/= p3 or p4)

Now define
C=B_p3
i.e. B localized on the prime ideal p3...

the claim which is also the problem/question is that

p1C + p2C = p3C ?

(Note that C has now only 3 prime ideals.. p1C, p2C and p3C)

You could use the following lemma (where R=B, p3=p and p4=q):

Let R be a unitary commutative ring with prime ideals p and q one of
which is not contained in the other. Then (p n q)R_p=pR_p.

Of course, the hypothesis on the inclusion of p and q can be weakened to
that q is not contained in p.

Hint: Localization commutes with intersections and qR_p=R_p.

J.
.

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