Minimal primes and ring extensions?
Suppose R is a subring of S and suppose p is a minimal prime ideal of R.
Prove or disprove:
p is contracted.
i.e. There is some prime ideal q of S such that q intersected with R is p.
Suppose R is a subring of S and suppose p is a minimal prime ideal of R.
Prove or disprove:
p is contracted.
i.e. There is some prime ideal q of S such that q intersected with R is p.
?
Favorite Answer
I think yes
First, let M=R-p
Since p is prime, M is a multiplicative set in S that does not contain 0.
Let N={I ideal in S| I intersected M=empty set}
N is nonempty (it contains 0} and satisfies the condition of Zorn's Lemma
So it has a maximal element q; which can be proved is prime
(take ab in q, a,b not in Q; so (q+Sa) intersected M and
(q+Sb) intersected M nonempty...
(this appears also in Matsumara, Commutative algebra, page 5)
Now q intersected R is a prime ideal in R that by construction of q is included in p.
Since p is minimal, the intersection is p.
no comments?!
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Have you tested this on the ring of the integers?