Let \(k\) be an algebraically closed field and let \[ X\subseteq\AA^n_k \] be Zariski closed. Show that there is a unique finite decomposition \[ X=X_1\union\cdots\union X_r \] into irreducible Zariski-closed subsets such that no \(X_i\) is contained in a different \(X_j\). These subsets are the irreducible components of \(X\).
Solution
Uniqueness means uniqueness of the collection of components, without an ordering. The empty set has the empty decomposition.
1. The affine space \(\AA^n_k\) is a Noetherian topological space.
Let \[ Y_1\supseteq Y_2\supseteq\cdots \] be a descending chain of Zariski-closed subsets of \(\AA^n_k\). Their vanishing ideals form an ascending chain \[ I(Y_1)\subseteq I(Y_2)\subseteq\cdots \] in \[ k[x_1,\ldots,x_n]. \] By the Hilbert basis theorem, this polynomial ring is Noetherian, so the chain of ideals stabilizes. Since \[ V(I(Y_i))=Y_i \] for every closed \(Y_i\), the original chain of closed subsets also stabilizes. Thus \(\AA^n_k\) is Noetherian.
2. Every Zariski-closed subset of \(\AA^n_k\) is a finite union ofirreducible Zariski-closed subsets.
1. Call it \(Y\).
The set \(Y\) is nonempty, because the empty set is the empty finite union. It is not irreducible, because otherwise \(Y\) itself would be a one-term decomposition. Hence \[ Y=A\union B \] for proper closed subsets \[ A,B\subsetneq Y. \] By minimality of \(Y\), both \(A\) and \(B\) are finite unions of irreducible closed subsets. Combining those two finite decompositions gives one for \(Y\), a contradiction.
Therefore every closed subset admits a finite irreducible decomposition.
3. Every finite irreducible decomposition can be reduced to one in whichno member is contained in another.
Starting from a finite decomposition \[ X=Y_1\union\cdots\union Y_s, \] delete any repeated member and any member \(Y_i\) contained in another member \(Y_j\). Such a deletion does not change the union. Since there are only finitely many members, this process terminates and leaves a finite decomposition in which no distinct member contains another.
4. If an irreducible subset \(T\) is contained in a finite union of closedsets\[ C_1\union\cdots\union C_s, \]then\[ T\subseteq C_j \]for some \(j\).
Inside the irreducible space \(T\), \[ T=(T\intersect C_1)\union\cdots\union(T\intersect C_s) \] is a finite union of closed subsets. For \(s=2\), irreducibility says that one of the two closed subsets is all of \(T\). Induction on \(s\) gives the same conclusion for any finite number of closed subsets.
5. A decomposition satisfying the noncontainment condition is unique.
Suppose \[ X=X_1\union\cdots\union X_r = Y_1\union\cdots\union Y_s \] are two finite decompositions into irreducible closed subsets, with no member of either decomposition contained in a distinct member of the same decomposition.
Fix \(i\). Since \[ X_i\subseteq Y_1\union\cdots\union Y_s, \] step 4 gives \[ X_i\subseteq Y_j \] for some \(j\). Applying step 4 to \[ Y_j\subseteq X_1\union\cdots\union X_r \] gives \[ Y_j\subseteq X_h \] for some \(h\). Thus \[ X_i\subseteq X_h. \] The noncontainment condition in the first decomposition forces \(h=i\), so \[ X_i=Y_j. \]
Hence every \(X_i\) occurs among the \(Y_j\). By symmetry every \(Y_j\) occurs among the \(X_i\). The two decompositions therefore have exactly the same members.
6. The members of this unique decomposition are exactly the maximalirreducible closed subsets of \(X\).
Let \(X_i\) be a member and suppose \[ X_i\subseteq Z\subseteq X \] with \(Z\) irreducible and closed. Since \[ Z\subseteq X_1\union\cdots\union X_r, \] step 4 gives \[ Z\subseteq X_j \] for some \(j\). Then \[ X_i\subseteq X_j, \] so the noncontainment condition gives \(j=i\), and hence \(Z=X_i\). Thus every \(X_i\) is maximal irreducible.
Conversely, step 4 shows that every irreducible closed subset of \(X\) is contained in some \(X_i\), so every maximal irreducible closed subset is one of the \(X_i\).
7. Q.E.D.
Steps <1>1–<1>3 prove existence of a finite decomposition with the required noncontainment property, step 5 proves uniqueness, and step 6 identifies the members as the irreducible components.