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Given a primary ideal I in a ring A, we can consider the subscheme V(I) of Spec(A). It is a nilpotentification (?) of the integral subscheme V(rad(I)) given by the radical rad(I) of I. My question is what kind of nilpotentifications you get that way. It is a vague question, but I'm asking because i am completely unable to understand the algebraic meaning of "primary ideal" (altho I learned it by heart)

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Geometrically, primary ideals (and most importantly, primary decomposition) gives you a way of visualizing, or at least identifying, embedded components in your variety/scheme. In fact, from the geometric point of view, it might be best to temporarily accept the definition of a primary ideal as slightly odd, make your way to the primary decomposition theorem, and reap the benefits of geometric intuition only after that point.

To steal an example I found online, here's a primary decomposition of an ideal corresponding to a intersection of three varieties:

$$ I=\langle xy,x^3-x^2,x^2y-xy\rangle=\langle x\rangle\cap \langle x-1,y\rangle \cap \langle x^2,y\rangle $$

so this intersection is seen to consist of the $y$-axis, the point $(0,0)$ ("embedded" on the $y$-axis), and the isolated point $(1,0)$ -- something not immediately discernible from the system of equations. Back to primary ideals, briefly: this nilpotentization you speak of is precisely the idea of giving extra fuzziness to this point $(0,0)$ as an embedded subscheme of the y-axis. So in some sense, it's just a slightly more nuanced version of the idea that, say, $(x,y)^2$ should correspond geometrically to a point of multiplicity 2, where you can now identify a point as a repeated point even though $(x,y)^2$ does not appear in the primary decomposition of this ideal.

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This answer is better than mine for what the op wanted, but if the op does want to gain some commutative algebra intuition for it, I still suggest he reads my post. =) – Harry Gindi Jan 28 '10 at 13:11

As Harry suggests in his answer, it is probably more intuitive to work with associated primes, rather than the slightly older language of primary decompositions.

If $I$ is an ideal in $A$, an associated prime of $A/I$ is a prime ideal of $A$ which is the full annihilator in $A$ of some element of $A/I$. A key fact is that for any element $x$ of $A/I$, the annihilator of $x$ in $A$ is contained in an associated prime.

The associated primes are precisely the primes that contribute to the primary decomposition of $I$. Geometrically, $\wp$ is an associated prime of $A/I$ if there is a section of the structure sheaf of Spec $A/I$ that is supported on the irreducible closed set $V(\wp)$. E.g. in the example given in Cam's answer, the function $x^2 - x$ is not identically zero on $X:=$ Spec ${\mathbb C}[x,y]/(x y, x^3-x^2, x^2 y - xy),$ but it is annihilated by $(x,y)$, and so is supported at the origin (if we restrict it to the complement of $(0,0)$ in $X$ then it becomes zero).

The non-minimal primes of $I$ that play a role in the primary decomposition of $I$ (i.e. appear as associated primes of $A/I$) are the generic points of the so-called embedded components of Spec $A/I$: they are irreducible closed subset of Spec $A/I$ that are not irreducible components, but which are the support of certain sections of the structure sheaf.

An important point is that if $I$ is radical, so that $A/I$ is reduced, then there are no embedded components: the only associated primes are the minimal primes (for the primary decomposition of $I$ is then very simple, as noted in the question: $I$ is just the intersection of its minimal primes).

There is a nice criterion for a Noetherian ring to be reduced: Noetherian $A$ is reduced if and only if $A$ satisfies $R_0$ and $S_1$, i.e. is generically reduced, and has no non-minimal associated primes. Geometrically, and applied to $A/I$ rather than $A$, this says that if $A/I$ is generically reduced, then the embedded components are precisely the irreducible closed subsets of Spec $A/I$ over which the nilpotent sections of the structure sheaf are supported. This may help with your ``nilpotentification'' mental image.

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I'm glad that someone was able to give an answer about commutative algebra that didn't immediately degenerate into geometry. (As you might be able to tell, I'm one of the few commutative algebra fans.) =) – Harry Gindi Jan 28 '10 at 20:41

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