# Can every algebraic set be written as a COUNTABLE union of irreducible algebraic sets?

Let $k$ be an algebraically closed field, and $\mathbb{A}^n(k)$ the affine space corresponding to $k[x_1, \dots, x_n]$.

In proving that every algebraic set can be written as the finite union of irreducible algebraic sets, the proof in Garrity, Belshoff et al, Algebraic Geometry: A Problem-Solving Approach, p. 215, Exercise 4.7.13, proceeds by contradiction, assuming that the algebraic set cannot be written as a finite union of irreducible algebraic sets.

In their proof, they seem to assume that any algebraic set, which cannot be written as a finite union of irreducible algebraic sets, can be written as a countable union of an increasing sequence of irreducible algebraic sets. This is equivalent to being the countable union of irreducible algebraic sets (e.g. see here: https://math.stackexchange.com/a/1321943/327486).

But if all we know is that $V$ can't be written as the finite union of irreducible algebraic sets, how do we know that it can be written as the countable union of irreducible algebraic sets? Why can we exclude the possibility that it can only be written as the uncountable union of irreducible algebraic sets?

By the Weak Nullstellensatz, every point in $\mathbb{A}^n$ is an irreducible algebraic set. So every non-empty subset of $\mathbb{A}^n$ can be written as the arbitrary (usually uncountable) union of irreducible algebraic sets. Why then can algebraic sets be written as the countable union of irreducible algebraic sets?

I understand everything past this point, i.e. how the Noetherian/ascending chain condition leads to a contradiction if we have that the set is equal only to an increasing countable union of irreducible algebraic sets. However, I don't think the Noetherian condition applies to an uncountable union of irreducible algebraic sets -- the whole point of Noetherian rings is to enable induction upon ideals, but induction requires countability. How can we infer that the countability condition holds?

Context:
The problem is based on a proposition and corollary in Hartshorne, Algebraic Geometry, 1977, specifically Proposition I.1.5 and Corollary I.1.6 on p.5. I looked at Hartshorne's proof, and it seems like his method sidesteps entirely the claim implicit in Garrity et al about any algebraic set equaling the countable union of an increasing chain of irreducible algebraic sets.

Exercise 4.7.13. Let $V$ be an algebraic set. Assume that $V$ cannot be written as the union of a finite number of irreducible algebraic sets.
(1) Show that there is an infinite descending chain of algebraic sets in $\mathbb{A}^n$: $$V \supset V_1 \supset V_2 \supset \dots$$ [Doesn't this chain as written have to be countable? Doesn't it have to be countable to apply the ascending chain condition for $I(V) \subset I(V_1) \subset \dots$?]

• @William As the proof eventually shows, every algebraic set in $\mathbb{A}^n_k$ is the union of finitely many irreducible algebraic sets, but at this point of the proof we don't know that. – Fan Zheng Oct 11 '16 at 19:02
We get the sequence essentially by induction. $V$ is reducible, so $V = V_1 \cup V_2$ for proper subsets. Because $V$ can't be written as a finite union of irreducible algebraic sets, neither can $V_1$ or $V_2$. They are also algebraic, and thus they are reducible, so e.g. $V_1 = V_{11} \cup V_{12}$ and the same argument applies, ad (countable) infinitum, i.e. $V \supset V_1 \supset V_{11} \supset V_{111} \supset \dots$, each additional set in the chain being generated from the definition of reducible algebraic set, and the chain never terminating because if it terminated then one could construct a finite union of irreducible algebraic sets.