Sum of two prime ideals is prime ?

i was looking for a counter example.

and, i've not been able to think of any.

look at Z

mathwonk
Homework Helper
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the sum of two ideals is the ideal generated by their union right?

thus geometrically it is the ideal of the intersection of the two zero loci.

so look for a pair of irreducible algebraic sets whose intersection is reducible,

(like a quadric surface and a tangent plane.)

i.e. a prime ideal is one that has a (reduced and) irreducible zero set.

i did not understand the geometric bit.
well i considered the principal ideals <2> and <3>
their union includes 1 which is a unit in Z, so the ideal of the sum is nothing but Z itself right ?
and that can't be prime by definition ? (since an ideal P is prime => P /= R (the ring in consideration))

i did not understand the geometric bit.
well i considered the principal ideals <2> and <3>
their union includes 1 which is a unit in Z, so the ideal of the sum is nothing but Z itself right ?
and that can't be prime by definition ? (since an ideal P is prime => P /= R (the ring in consideration))

yea that's fine, just notice 1 = -1*2 + 1*3

another way to think about it is in terms of existence of gcd's. gcd(2, 3) = 1, so there are x, y in Z such that 2x + 3y = 1 and this is in <2> + <3>, but this is an ideal, so x = x*1 is in <2> + <3> for all x in Z, so yea Z = <2> + <3>, probably overkill but a useful observation

this can be generalized and is really useful, if R is a pid, then Ra + Rb = Rd where d = gcd(a, b)

when looking for counterexamples always think simple(doesn't always work but sometimes it does), like for example 2Z and 3Z are prime but 2Z n 3Z = 6Z is not, so the intersection of prime ideals isn't necessarily prime

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Hurkyl
Staff Emeritus