i was looking for a counter example.
and, i've not been able to think of any.
look at Z
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))
more hints please.
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
Think of rings like R[x, y]. Algebraic curves (like the parabola y - x^2 = 0) correspond to ideals (like the ideal <y - x^2>). Sums of ideals relate to intersections of curves. Can you work out why? Do you see how a non-prime ideal corresponds, in some sense, into a curve that is the union of two or more other curves?
a prime ideal of a cone is (Z^2 -X^2-Y^2).
a prime ideal of a plane is (Y).
the sum of these ideals (Z^2 -X^2-Y^2, Y) = (Z^2 -X^2, Y), is the ideal of the intersection, which is the two lines Z=X, Z=-X, in the X,Z plane.
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