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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
up vote 15 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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