Proof verification : every algebraic set is the union of finitely many irreducible algebraic subsets I have found various proofs of the result but I have come up with something very different and I wonder whether it is a valid argument:
Let $W$ be an algebraic set. Let $I=\mathcal{I}(W)$. We have $I=rad(I)=P_1 \cap ... \cap P_m$ an intersection of minimal prime ideals of $I$. Set $V_i=\mathcal{V}(P_i)$. Then $W$ is the union of the $V_i$'s which are irreducible algebraic sets. 
I feel as though something must be going wrong here, it seems to me that I would have found this proof somewhere if it were to be right... Can you give me your opinion here? Thank you so much!
 A: There is no garantee that the $V_i$'s are irreducible.
A: I see no flaw in your proof. Perhaps the reason why you haven't found it, is because you're trying to prove an elementary result from algebraic geometry using a less elementary result from commutative algebra, namely that in a Noetherian ring, each radical ideal is the finite intersection of the minimal prime ideals above it?
A: You can think of primary decomposition as a generalisation of the decomposition of an algebraic set in irreducible subsets, so it isn't a good idea to use it for a proof. Your proof is also a well known fact, for example read the introduction in
https://en.wikipedia.org/wiki/Primary_decomposition
A: The classic argument to show this is to use topology: Every noetherian topological set is the  finite union of irreducible components, and an irreducible algebraic set is of the form $V(P)$ where $P$ is a prime.
Your argument is true, in a ring, the intersection of all the prime $\cup_{P, prime}P$ is the nilradical. The nilradical of $A/r(I)$ is trivial, so $r(I)=\cap_{r(I)\subset P, P prime}P$. The nil-radical is an intersection of all prime ideals proof
So $V(r(I)) =\cup V(P), r(I)\subset P$.
