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In the book "Commutative Algebra with a view toward Algebraic Geometry of David Eisenbud, he wrote about the Geometric interpretation of primary decomposition.

I summary as follows :

Let $I=\cap_{j}I_{j}$ be a minimal primary decomposition of $I$, where $I$ is an ideal in $k[x_{1},\cdots,x_{n}]$. Then $Z(I)=\cup_{j}Z(I_{j})$. So, if $I$ is a radical ideal then each of $I_{j}$ is a prime ideal minimal over $I$, and the primary decomposition give us the decomposition of $Z(I)$ as the union of irreducible variety $Z(I_{j})$.

My thinkings are follows:

If $I=\cap_{j}I_{j}$ is a primary decomposition of $I$ then each $I_j$ is a primary ideal then its radical $radI_{j}$ is a prime ideal $p_j$. Then $Z(I_{j})=Z(p_{i})$ and because $p_j$ are prime ideals, $Z(p_{j})$ are irreducible varieties. Thus, $Z(I)=\cup_{j}Z(I_{j})$ is a decomposition of $Z(I)$ into irreducible components.

So, what is the role of the radical property of $I$ that Eisenbud mentioned ?

Could anybody point out the geometric meaning of primary decomposition in a very concrete way?


Update: MattE has answered my second question(thank you for that), but I still have trouble in my first question. Eisenbud's argument have used the radical property of ideal $I$, and he conclude that $Z(I)$ can be decompose into union of irreducible component. However in my argument above, I have not used it and still get the same conclusion. So was I wrong in anywhere or we can ignore the radical properties of $I$ in Eisenbud's argument?

Please point it out for me. Thank you very much!

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Great question +1. I've never been forced to think about this until writing up an answer, but it's absolutely fundamental. –  Brett Frankel Feb 27 '12 at 5:22
Dear msnaber, I just noticed your edit (many months later), and you seem to have not understood my answer. I added an additional explanation, which should show that my answer completely answers your question. Regards, –  Matt E Nov 8 '12 at 23:50
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1 Answer

up vote 11 down vote accepted

The primary decomposition is more subtle than the decomposition into irreducible components. Namely, the various primes that appear are the associated primes of the quotient $R/I$ (thought of as an $R$-module).

That is, they are the prime ideals which appear as annihilators of some element of $R/I$.

Geometrically, we can think of $R/I$ as the global sections of the structure sheaf of Spec $R/I$, and the various $Z(p_i)$ are precisely those irreducible subsets of Spec $R/I$ which can be realized as the support of some particular element of $R/I$.

E.g. if $R = \mathbb C[x,y]$ and $I = (xy, x^2)$, then a primary decomposition of $I$ is $0 = (x,y)^2 \cap (x).$ Here $(x)$ appears because it is the radical of $I$: the quotient $R/I = \mathbb C[x,y]/(xy,x^2)$, although not a domain, becomes a domain after we quotient out by its nilradical, and so its Spec is irreducible. The other prime ideal that contributes is $(x,y)$: this appears because the element $x \in R/I$ is supported at the origin, i.e. at the point $(x,y)$. This is related to the fact that $x$ is nilpotent in $R/I$, although $R/I$ is generically reduced. We say that $(x,y)$ is an embedded point of Spec $R/I$.

Added: The OP has edited the question, remarking that this answer does not answer the first part of the question. I would just like to point out that in fact it does answer that part of the question.

If $I$ is radical, so that $R/I$ is reduced, then the associated primes of $R/I$ are just its minimal primes, and so (as Eisenbud notes) the primary decomposition of $I$ just corresponds to the union of Spec $R/I$ into its irred. comps.

However, if $I$ is not radical, so that $R/I$ is not reduced, then the primary decomposition of $I$ reflects the possible embedded components in Spec $R/I$, and so carries more subtle information than just the minimal primes of $I$ (or,equivalently, the irred. comps. of Spec $R/I$).

Concretely, if $\mathfrak p$ and $\mathfrak q$ are two primes corresponding to primary ideals in the primary decomposition of $I$, then it can happen that $\mathfrak p \subset \mathfrak q$, so that $Z(\mathfrak q) \subset Z(\mathfrak p)$. (See e.g. the explicit example above.) Hence $Z(\mathfrak q)$ will not be an irred. component of Spec $R/I$. (It is precisely an embedded component).

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