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!