Visualizing embedded primes:
In P^2, a one dimensional scheme cannot have embedded points unless its ideal has more than one generator, by the unmixedness theorem of Macaulay. So imagine we have two polynomials that define a one dimensional scheme in P^2. We will imagine this scheme as a limit of zero dimensional schemes. First take two quadratic polynomials, one of which is a product of two linear factors, i.e. take one pair of lines meeting at p, and another irreducible conic. In general the irreducible conic C meets each of the lines twice, away from p. Thus the two qudratic polynomials define a zero dimensional scheme of 4 points.
Now hold fixed the two intersections of C with one of the lines L, and let the two intersections of C with the other line M approach p, i.e. let C become tangent to M at p. When this occurs, the conic C now contains three distinct points of L, hence C has become reducible and contains L. Now the scheme defined by intersecting L+M with C has become one dimensional, reducible, and consists set theoretically only of the line L. I claim the point p is an embedded point of the component L of the scheme defined by L+M and C.
This is easy algebraically, since the ideal of the given scheme is (xy,(x(x-y)) = (x^2, xy) which is the intersection of the primary ideals (x) and (x^2, xy, y^2), with associated primes (x) and (x,y). Hence (x,y) is an embedded prime. I.e. the origin is an embedded point on the y axis for this scheme. This also helps explain the apparent failure of Bezout's theorem for this intersection of two conics apparently not having degree 4.
In general, in P^n, a scheme S with embedded subschemes must be defined by intersecting more hypersurfaces than the codimension of S. Thus such an S can always be viewed as a limit of lower dimensional schemes. It seems to me that embedded subschemes should arise when these lower dimensional schemes are reducible and some lower dimensional component comes to lie on a larger dimensional component of the limit. I do not know if this intuition is the only possibility, and since the world is wide, probably not.
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