Geometric meaning of primary decomposition In the book "Commutative Algebra with a view toward Algebraic Geometry" of David Eisenbud, he wrote about the Geometric interpretation of primary decomposition.
I summarize 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 thinking is 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?
Thanks.
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!
 A: 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).
