[proofplan] We prove existence by contradiction: a variety with no finite irreducible decomposition would force an infinite strictly descending chain of subvarieties, equivalently (under the bijection $V \leftrightarrow I(V)$) a strictly ascending chain of ideals in $k[X_1, \ldots, X_n]$ — but this ring is Noetherian by the *Hilbert Basis Theorem*, contradicting the existence of such a chain. We then strengthen existence to an irredundant decomposition by removing components contained in others. Uniqueness up to reordering follows from an "irredundancy plus irreducibility" argument: each irreducible component on one side is contained in some component on the other, and irredundancy forces equality. [/proofplan]

[step:Reduce existence to a Noetherian-induction argument] We argue by contradiction. Let \begin{align*} \mathcal{S} = \{V \subset \mathbb{A}^n_k : V \text{ is a variety with no finite irreducible decomposition}\}, \end{align*} and suppose for contradiction that $\mathcal{S} \neq \varnothing$. We will build a strictly descending infinite chain of varieties starting from any $V_0 \in \mathcal{S}$. If $V_0 \in \mathcal{S}$, then $V_0$ itself is reducible (in the sense that it admits a decomposition into two proper closed subvarieties; this is the negation of irreducibility, which requires that $V_0$ cannot be written as a union of two proper closed subvarieties). Indeed, were $V_0$ irreducible, $V_0$ would be its own finite irreducible decomposition (a one-term union), contradicting $V_0 \in \mathcal{S}$. Hence we can write \begin{align*} V_0 = V_0' \cup V_0'' \end{align*} with $V_0', V_0'' \subsetneq V_0$ both proper closed subvarieties. Moreover, at least one of $V_0', V_0''$ must lie in $\mathcal{S}$: if both admitted finite irreducible decompositions, then concatenating those decompositions would give one for $V_0$, contradicting $V_0 \in \mathcal{S}$. Choose such a piece and call it $V_1$, so $V_1 \in \mathcal{S}$ and $V_1 \subsetneq V_0$. Iterating produces a strictly descending chain \begin{align*} V_0 \supsetneq V_1 \supsetneq V_2 \supsetneq \cdots, \end{align*} with each $V_i \in \mathcal{S}$. [/step]

[step:Translate the descending chain of varieties to an ascending chain of ideals]Apply the ideal-of-set operator $I$ to the descending chain. The map $I$ is order-reversing on subsets of $\mathbb{A}^n_k$: \begin{align*} W \subset W' \implies I(W') \subset I(W). \end{align*} Furthermore, $I$ is injective on closed subvarieties of $\mathbb{A}^n_k$: closed subvarieties are precisely the loci $V(J)$ of ideals, and the Zariski closure operation gives the recovery identity $V(I(W)) = W$ for any closed $W \subset \mathbb{A}^n_k$ (a direct unwinding of the definitions of $V$ and $I$ — see, e.g., the discussion of the [Ideal Operations](/theorems/2121)). So the strict inclusions $V_i \supsetneq V_{i+1}$ produce strict inclusions \begin{align*} I(V_0) \subsetneq I(V_1) \subsetneq I(V_2) \subsetneq \cdots \end{align*} in $k[X_1, \ldots, X_n]$.[/step]

[guided]We turn the geometric chain into an algebraic one and then invoke a structural property of the polynomial ring. The map \begin{align*} I : \{\text{closed subsets of } \mathbb{A}^n_k\} &\to \{\text{ideals of } k[X_1, \ldots, X_n]\} \\ W &\mapsto \{f \in k[X_1, \ldots, X_n] : f|_W = 0\} \end{align*} is order-reversing: a smaller variety has more polynomials vanishing on it, so $W \subsetneq W' \implies I(W) \supsetneq I(W')$. Why is the inclusion still \emph{strict}? Because closed subvarieties are recovered from their ideals: $V(I(W)) = W$ for any closed $W \subset \mathbb{A}^n_k$ (this is the closure operation for the Zariski topology). If we had $I(V_i) = I(V_{i+1})$, applying $V$ would give $V_i = V_{i+1}$, contradicting strict inclusion in the variety chain. So the ideal chain is strictly ascending: \begin{align*} I(V_0) \subsetneq I(V_1) \subsetneq I(V_2) \subsetneq \cdots \end{align*}[/guided]

[step:Derive a contradiction from the Noetherian property of $k[X_1, \ldots, X_n]$] Form the union \begin{align*} J = \bigcup_{i=0}^{\infty} I(V_i). \end{align*} As a directed union (the chain is totally ordered) of ideals, $J$ is itself an ideal of $k[X_1, \ldots, X_n]$. By the *Hilbert Basis Theorem* (starting from the Noetherian ring $k$ and iterating the Basis Theorem $n$ times), $k[X_1, \ldots, X_n]$ is Noetherian, so every ideal — in particular $J$ — is finitely generated. Let $f_1, \ldots, f_m$ be generators of $J$. Each $f_j$ lies in some $I(V_{N_j})$, and taking $N = \max_j N_j$, every generator lies in $I(V_N)$. Hence \begin{align*} J \subset I(V_N) \subset J, \end{align*} so $J = I(V_N)$, which forces $I(V_N) = I(V_{N+1})$. This contradicts the strict inclusion $I(V_N) \subsetneq I(V_{N+1})$ established in Step 2. Hence $\mathcal{S} = \varnothing$, and every variety admits a finite irreducible decomposition.[/step]

[guided]We have a strictly ascending chain of ideals in $k[X_1, \ldots, X_n]$ and we want to derive a contradiction. The general principle is: in a Noetherian ring, every ascending chain of ideals stabilises (the \emph{ascending chain condition}, ACC). The polynomial ring $k[X_1, \ldots, X_n]$ is Noetherian by the *Hilbert Basis Theorem* — which states that if $R$ is Noetherian then $R[X]$ is also Noetherian. Starting from the field $k$ (Noetherian because its only ideals are $(0)$ and $k$) and iterating the Basis Theorem $n$ times gives that $k[X_1, \ldots, X_n]$ is Noetherian. Why does ACC follow from "every ideal is finitely generated"? Form the union $J = \bigcup_i I(V_i)$ of the ascending chain. We must check $J$ is an ideal: \begin{itemize} \item Closure under addition: if $f \in I(V_i)$ and $g \in I(V_j)$, take $\ell = \max(i, j)$ so both $f, g \in I(V_\ell)$, hence $f + g \in I(V_\ell) \subset J$. \item Closure under scalar multiplication: $f \in I(V_i)$ and $r \in k[X_1, \ldots, X_n]$ give $r f \in I(V_i) \subset J$. \end{itemize} So $J$ is an ideal. Being finitely generated by, say, $f_1, \ldots, f_m$, each $f_j$ already lives in some level $I(V_{N_j})$ of the chain. Setting $N = \max_j N_j$, all $f_j$ lie in $I(V_N)$, which therefore contains the entire generating set of $J$ and hence (being closed under the ideal operations) contains $J$. The opposite inclusion $I(V_N) \subset J$ is automatic. So $J = I(V_N)$, and the chain has stabilised by index $N$: \begin{align*} I(V_N) = I(V_{N+1}) = I(V_{N+2}) = \cdots. \end{align*} But the chain was strictly ascending — contradiction. The original assumption that $\mathcal{S} \neq \varnothing$ is therefore false, and every variety admits a finite irreducible decomposition.[/guided]

[step:Refine to an irredundant decomposition] Given $V = W_1 \cup \cdots \cup W_r$ with each $W_i$ irreducible, we may pass to an \emph{irredundant} decomposition (no $W_i$ is contained in another) as follows: while there exist indices $i \neq j$ with $W_i \subset W_j$, drop $W_i$ from the list. Each removal preserves the union (since $W_i \cup W_j = W_j$ when $W_i \subset W_j$) and decreases $r$ by one, so the process terminates after finitely many steps. The resulting decomposition $V = V_1 \cup \cdots \cup V_s$ has each $V_a$ irreducible and no $V_a$ contained in any other $V_b$. [/step]

[step:Prove uniqueness of the irredundant decomposition up to reordering]Suppose \begin{align*} V = V_1 \cup \cdots \cup V_r = V_1' \cup \cdots \cup V_s' \end{align*} are two irredundant irreducible decompositions. Fix $i \in \{1, \ldots, r\}$. Intersecting both sides with $V_i$, \begin{align*} V_i = V_i \cap V = V_i \cap (V_1' \cup \cdots \cup V_s') = \bigcup_{j=1}^s (V_i \cap V_j'). \end{align*} Each $V_i \cap V_j'$ is closed in $V_i$ (intersection of closed sets), and $V_i$ is irreducible, so the decomposition $V_i = \bigcup_j (V_i \cap V_j')$ as a finite union of closed subsets forces $V_i = V_i \cap V_j'$ for some $j$ — equivalently, $V_i \subset V_j'$. Applying the same reasoning with the roles reversed: there exists $\ell \in \{1, \ldots, r\}$ with $V_j' \subset V_\ell$. Combining, \begin{align*} V_i \subset V_j' \subset V_\ell. \end{align*} By irredundancy of the first decomposition — no $V_i$ is contained in $V_a$ for any $a \neq i$ — the containment $V_i \subset V_\ell$ forces $\ell = i$. Hence the chain collapses to equalities: \begin{align*} V_i = V_j' = V_\ell. \end{align*} This produces, for each $i \in \{1, \ldots, r\}$, at least one $j \in \{1, \ldots, s\}$ with $V_i = V_j'$; moreover this $j$ is unique, because if $V_i = V_j' = V_{j'}'$ with $j \neq j'$ then the second decomposition would contain two equal components, contradicting irredundancy (a decomposition with two equal components contains one component inside another). We therefore obtain a well-defined map \begin{align*} \sigma : \{1, \ldots, r\} \to \{1, \ldots, s\}, \qquad V_i = V_{\sigma(i)}'. \end{align*} \emph{Injectivity of $\sigma$.} If $\sigma(i_1) = \sigma(i_2) = j$, then $V_{i_1} = V_j' = V_{i_2}$. By the irredundancy of the first decomposition (applied as above to exclude equal components), $i_1 = i_2$. \emph{Surjectivity of $\sigma$.} Running the same existence argument from the second decomposition's side produces $\tau : \{1, \ldots, s\} \to \{1, \ldots, r\}$ with $V_j' = V_{\tau(j)}$. Then for any $j$, $V_j' = V_{\tau(j)} = V_{\sigma(\tau(j))}'$, so $\sigma(\tau(j)) = j$ (again by the uniqueness of the index in $\sigma$), which exhibits $j$ as a value of $\sigma$. Hence $\sigma$ is a bijection, $r = s$, and the two decompositions agree up to reordering. Combining with the existence statement of Step 4, this completes the proof.[/step]

[guided]We have two ways of writing $V$ as an irredundant union of irreducibles, and we want to show they are the same list (up to permutation). \textbf{Step A — every component on one side is contained in some component on the other.} Fix an arbitrary $V_i$ from the first decomposition. Intersecting with $V$ written via the second decomposition, \begin{align*} V_i = V_i \cap V = V_i \cap (V_1' \cup \cdots \cup V_s') = (V_i \cap V_1') \cup \cdots \cup (V_i \cap V_s'). \end{align*} This expresses the irreducible variety $V_i$ as a finite union of closed subsets of itself. The defining property of an irreducible variety is precisely that whenever it is written as a finite union of closed subsets, one of those subsets must equal the whole. So there exists some $j$ with $V_i \cap V_j' = V_i$, equivalently $V_i \subset V_j'$. \textbf{Step B — the containment chain closes via irredundancy.} By symmetry (running the same argument on $V_j'$ using the first decomposition), there exists some $\ell \in \{1, \ldots, r\}$ with $V_j' \subset V_\ell$. Concatenating the two containments: \begin{align*} V_i \subset V_j' \subset V_\ell, \end{align*} so $V_i \subset V_\ell$. The decomposition $V = V_1 \cup \cdots \cup V_r$ is irredundant: no $V_i$ is contained in any other $V_a$ with $a \neq i$. Hence we are forced to $\ell = i$, and the inclusions collapse to equalities: \begin{align*} V_i = V_j' = V_\ell. \end{align*} \textbf{Step C — indices are determined uniquely by their components.} We pause to verify a subsidiary fact: within an irredundant decomposition, no two distinct indices point to the same component. Suppose $V_{i_1} = V_{i_2}$ with $i_1 \neq i_2$ in the first decomposition. Then $V_{i_1} \subset V_{i_2}$, contradicting irredundancy (the decomposition is irredundant precisely when no component is contained in another with a different index). Hence $V_i = V_a$ forces $i = a$; the map "component $\mapsto$ index" is well-defined within each decomposition. \textbf{Step D — a bijection of indices.} Define $\sigma : \{1, \ldots, r\} \to \{1, \ldots, s\}$ by $\sigma(i) = j$, where $j$ is the index given by Step B satisfying $V_i = V_j'$. Uniqueness of $j$ within the second decomposition (Step C applied there) makes $\sigma$ well-defined. Similarly, running the same argument from the second decomposition's side gives $\tau : \{1, \ldots, s\} \to \{1, \ldots, r\}$ with $V_j' = V_{\tau(j)}$. \emph{We show $\sigma \circ \tau = \mathrm{id}_{\{1, \ldots, s\}}$.} Fix $j \in \{1, \ldots, s\}$. Then $V_j' = V_{\tau(j)}$, and by definition of $\sigma$, $V_{\tau(j)} = V_{\sigma(\tau(j))}'$. Combining: $V_j' = V_{\sigma(\tau(j))}'$. Step C applied to the second decomposition then forces $j = \sigma(\tau(j))$. The symmetric argument shows $\tau \circ \sigma = \mathrm{id}_{\{1, \ldots, r\}}$. Therefore $\sigma$ is a bijection, $r = s$, and the lists are permutations of each other. \textbf{Why irredundancy is essential.} If we allowed $V_1 \subset V_2$ in the first decomposition, then $V_1 \cup V_2 = V_2$ and we could shorten the decomposition. The uniqueness statement only holds for irredundant decompositions; in the general statement of the theorem, we mean "irreducible decomposition" in the irredundant sense, which Step 4 produced for us.[/guided]