By definition, a set which is dense in $X$ is a set which meets every non-empty open set, hence (a) and (b) are equivalent. It is immediate that (c) implies (a), for if $U_1$ and $U_2$ are disjoint non-empty open sets, $U_1 \cup U_2$ is a disconnected open set. Finally let us show that (a) implies (c): if $U$ is a disconnected open set, it is the union of two disjoint non-empty sets $U', U''$ which are open in $U$ and hence also open in $X$, which implies that $X$ is not irreducible.

A Hausdorff space is irreducible only if it consists of a single point.

A subset $E$ of a topological space $X$ is called an irreducible set if the subspace $E$ of $X$ is irreducible. For this to be so, it is necessary and sufficient that, for every pair of sets $U, V$ which are open in $X$ and meet $E$, $U \cap V$ also meet $E$, or (what amounts to the same thing) that, for every pair of sets $F, G$ which are closed in $X$ and satisfy $E \subset F \cup G$, $E \subset F$ or $E \subset G$. By induction on $n$, we deduce that, if $(F_i)_{1 \leq i \leq n}$ is a finite family of closed sets of $X$ such that $E \subset \bigcup_{i=1}^n F_i$, there exists an index $i$ such that $E \subset F_i$.

For an open set of $X$ to meet $E$, it is necessary and sufficient that it meet $\overline{E}$; then the proposition follows from the above remarks.

(ii) Let $(U_\alpha)_{\alpha \in A}$ be a non-empty covering of a topological space $X$ consisting of open sets such that $U_\alpha \cap U_\beta = \emptyset$ for every pair of indices $(\alpha, \beta)$. If the sets $U_\alpha$ are irreducible, the space $X$ is irreducible.

(i) If $X$ is irreducible, $U \subset X$ non-empty and open in $X$ and $V \subset U$ non-empty and open in $U$, $V$ is also open in $X$, therefore dense in $X$ and a fortiori dense in $U$. Then $U$ is irreducible (Proposition 1).

(ii) Let us show that, for every non-empty open set $V$ in $X$, $V \cap U_\alpha \neq \emptyset$ for all $\alpha \in A$: it follows that $V \cap U_\alpha$ is dense in $U_\alpha$, by hypothesis, hence that $V$ is dense in $X$ and this proves that $X$ is irreducible (Proposition 1). Now there exists at least one index $y$ such that $V \cap U_y \neq \emptyset$; as $U_\alpha \cap U_y \neq \emptyset$ for all $\alpha$ and $V \cap U_y$ is dense in $U_y$, $U_y \cap V \neq \emptyset$ and a fortiori $U_y \cap V \neq \emptyset$, which completes the proof of (ii).

If $U, V$ are two open sets of $Y$ which meet $f(E)$, $f^{-1}(U)$ and $f^{-1}(V)$ are open sets of $X$ which meet $E$. Consequently, $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V)$ meets $E$, which implies that $U \cap V$ meets $f(E)$ and proves the proposition.

To prove the first assertion, it is sufficient, by virtue of Zorn's Lemma, to prove that the set 3 of irreducible subsets of $X$ is inductive. Let $\mathscr{G}$ be a subset of $\mathfrak{S}$ totally ordered by inclusion; we show that the union $E$ of the sets $F \in \mathscr{G}$ is irreducible. Let $U, V$ be two open sets of $X$ which meet $E$; as $\mathscr{G}$ is totally ordered, there exists a set $F \in \mathscr{G}$ meeting $U$ and $V$; as $F$ is irreducible, $U \cap V$ meets $F$ and hence also $E$, which proves that $E$ is irreducible and hence that 3 is inductive. The second assertion follows from the first, for every subset of $X$ consisting of a single point is irreducible.

Every irreducible subspace of $X$ is connected by Proposition 1 and hence is contained in a connected component of $X$.

Note that two distinct irreducible components of $X$ may have points in common (Exercise 11).

We may restrict ourselves to the case where no two $P_i$ are comparable. If $E$ is an irreducible subset of $X$, then $E \subset \bigcup_{i=1}^n P_i$, and hence $E$ is contained in one of the closed sets $P_i$; this proves that the $P_i$ are the only maximal irreducible subsets of $X$.

$\overline{E}$ is the union of the $\overline{Q}_i$, which are irreducible (Proposition 2); as $Q_i$ is closed in $E$, $\overline{Q}_i \cap E = Q_i$; as $Q_i \notin Q_j$ for $i \neq j$, $\overline{Q}_i \notin \overline{Q}_j$, whence the corollary by virtue of Proposition 6.

If $V$ is closed in $U$ and irreducible, $\overline{V}$ is irreducible (Proposition 2) and $V = \overline{V} \cap U$. Conversely, if $Z$ is irreducible and closed in $X$ and meets $U$, $Z \cap U$ is a non-empty open subset of $Z$, hence irreducible (Proposition 3) and dense in $Z$ and, as $Z$ is closed, $Z = \overline{Z} \cap U$. This proves the proposition.

(i) Let $X$ be a Noetherian space, $A$ a subspace of $X$ and $(F_i)$ a decreasing sequence of subsets of $A$ which are closed in $A$; then $F_i = \overline{F_n} \cap A$ and the closures $\overline{F_n}$ of the $F_i$ in $X$ form a decreasing sequence of closed subsets of $X$. As this sequence is stationary, so is the sequence $(F_i)$.

(ii) Let $(G_n)_{n \geq 0}$ be a decreasing sequence of closed subsets of $X$; by hypothesis, each of the sequences $(G_n \cap A_i)_{n \geq 0}$ is stationary. As $I$ is finite, there is an integer $n_0$ such that, for $n \geq n_0$, $G_n \cap A_i = G_{n_0} \cap A_i$, for all $i \in I$. But

and hence the sequence $(G_n)$ is stationary and $X$ is Noetherian.

To show that the condition is necessary, it is sufficient, by virtue of Proposition 8, to prove that every Noetherian space $X$ is quasi-compact. Let $(U_i)_{i \in I}$ be an open covering of $X$; the set of finite unions of sets $U_i$ is non-empty and hence admits a maximal element $V = \bigcup_{i \in H} U_i$, where $H$ is a finite subset of $I$. By definition, $V \cup U_i = V$ for all $i \in I$ and hence $V = X$.

Conversely, suppose that every open set of $X$ is quasi-compact and let $(U_i)$ be an increasing sequence of open subsets of $X$. The union $V$ of the $U_i$ is open and hence quasi-compact; as $(U_i)$ is an open covering of $V$, there is a finite subfamily of $(U_i)$ which is a covering of $V$ and hence $V = U_n$ for some index $n$, which proves that the sequence $(U_i)$ is stationary.

Lemma 1 ("Principle of Noetherian induction"). Let E be an ordered set every non-empty subset of which admits a minimal element. Let F be a subset of E with the following property : if $a \in E$ is such that the relation $x < a$ implies $x \in F$, then $a \in F$. Then $F = E$.

Suppose $F \neq E$; then $CF$ would have a minimal element $b$. By definition, $x \in F$ for all $x < b$, which implies that $b \in F$, which is a contradiction.

It is sufficient to prove that X is a finite union of irreducible closed subsets (no. 1, Proposition 6). Let us show that the principle of Noetherian induction can be applied taking E to be the set of closed subsets of X, ordered by inclusion, and F to be the set of finite unions of irreducible closed subsets. Let Y be a closed subset of X such that every closed subset $\neq Y$ of Y belongs to F. If Y is irreducible, then $Y \in F$ by definition; otherwise, Y is the union of two closed subsets $Y_1,\ Y_2$ which are distinct from Y. Then $Y_1 \in F$ and $Y_2 \in F$ by hypothesis, whence $Y \in F$ by definition of F.

In particular it follows that a Hausdorff Noetherian space is necessarily finite.

(iii) The mappings 3 and V define decreasing bijections, one of which is the inverse of the other, between the set of closed subsets of X and the set of ideals of A which are equal to their radicals.

Assertion (i) and the first assertion of (ii) follow from the definitions and § 2, no. 6, Corollary 1 to Proposition 13. If a closed set V(M) (for some M ⊂ A) contains Y, then M ⊂ p for every prime ideal p ∈ Y, whence M ⊂ J(Y) and consequently V(M) ⊃ V(J(Y)); as Y ⊂ V(J(Y)), V(J(Y)) is the smallest closed set of X containing Y, which completes the proof of (ii). Finally, it follows from (ii) that, if a is a prime ideal equal to its radical, then J(V(a)) = a and that, if Y is closed in X, then V(J(Y)) = Y; this proves (iii).

It follows immediately from Proposition 11 that, if M is any subset of A and Y is any subset of X, then V(M) = V(J(V(M))) and J(Y) = J(V(J(Y))).

It follows from Proposition 11 (iii) that J(∩_{λ ∈ L} Y_λ) is the smallest ideal which is equal to its radical and contains all the J(Y_λ); this ideal then contains ∑_{λ ∈ L} J(Y_λ) and therefore also the radical of ∑_{λ ∈ L} J(Y_λ) (§ 2, no. 6, Corollary 2 to Proposition 13), whence the corollary.

It is immediate that the relations b ⊂ r(a) and r(b) ⊂ r(a) are equivalent and, as V(a) = V(r(a)), the corollary follows immediately from Proposition 11, (iii).

The relation X_g ⊂ ∪_{λ ∈ L} X_{f_λ} is equivalent to V(g) ⊃ ∩_{λ ∈ L} V(f_λ) and it is sufficient to apply Corollary 2.

COROLLARY 4, For two elements f, g of A to satisfy X_f = X_g, it is necessary and sufficient that there exist two integers m > 0, n > 0 such that f^m ∈ Ag and g^n ∈ Af.

This follows immediately from Corollary 4.

As the $X_g$ form a base of the topology, it is sufficient to prove that, if $(g_\lambda)_{\lambda \in L}$ is a family of elements of $A$ such that $X_f \subset \bigcup_{\lambda \in L} X_{g_\lambda}$, then there exists a finite subfamily $(g_{\lambda_i})_{i \in I}$ such that $X_f = \bigcup_{i \in I} X_{g_{\lambda_i}}$. But the relation $X_f = \bigcup_{\lambda \in L} X_{g_\lambda}$ means that there exists an integer $n > 0$ and a finite subfamily $(g_{\lambda_i})_{i \in I}$ such that "f" belongs to the ideal generated by that subfamily (Corollary 3 to Proposition 11); whence the proposition.

For $M \subset A$, the set $(^a h)^{-1}(V(M))$ is the set of prime ideals $p'$ of $A'$ such that $M \subset \overline{h}(p')$, which is equivalent to $h(M) \subset p'$; this set is then equal to $V(h(M))$ and is therefore closed.

We call $^a h$ the mapping associated with the homomorphism $h$.

Let $f' = f/s$, where $f \in A, s \in S$; then $X'_{f'} = X'_{f/1}$ since $s/1$ is invertible in $A'$. We know already that $^a h$ is injective and that, for all $p' \in X'$, the relations $f/1 \in p'$ and $f \in h^{-1}(p') = ^a h(p')$ are equivalent and hence the conditions $p' \in X'_{f/1}$ and $^a h(p') \in X_f$ are equivalent; this shows that $^a h(X'_{f'})$ is equal to $X_f \cap ^a h(X')$, whence the first assertion, since the $X, (resp.\ X'_{f'})$ form a base of the topology of $X$ (resp. $X$). The second assertion follows from , no. 5, Proposition 11 (ii)

Writing $p = \mathfrak{J}(Y)$, we note that, for an element $f \in A$, the relation $f \in p$ is equivalent to $Y \subset V(f)$. Suppose that $Y$ is irreducible and let $f, g$ be elements of $A$ such that $fg \in p$. Then

as $Y$ is irreducible and $V(f)$ and $V(g)$ are closed, $Y \subset V(f)$ or $Y \subset V(g)$, hence $f \in p$ or $g \in p$, which proves that $p$ is prime.

Suppose now that $p$ is prime; then $\overline{Y} = V(p)$ (Proposition 11 (ii)) and, as $p$ is prime, $p = \mathfrak{J}(\{p\})$, whence $\overline{Y} = V(\mathfrak{J}(\{p\})) = \{p\}$ (Proposition 11 (ii)). As a set consisting of a single point is irreducible, $Y$ is irreducible (no. 1, Proposition 2).

By Proposition 11 (i), $\mathfrak{J}(X)$ is the radical of the ideal $(0)$, that is $\mathfrak{N}$.

As $\mathfrak{J}(V(p)) = p$ for every prime ideal $p$ of $A$ and $Y = V(\mathfrak{J}(Y))$ for every closed subset $Y$ of $X$, the first assertion follows from Proposition 14; on the other hand, for $Y \supset V(p)$, it is necessary and sufficient that

(Proposition 11), whence the second assertion.

$X = \mathrm{Spec}(A)$ has only a finite number of irreducible components (Corollary 7 to Proposition 11 and no. 2, Proposition 10) and the corollary follows from Corollary 2 above.

Let $(e_i)_{i \in I}$ be an element of $E$ and set $Y_i = \mathrm{V}(A(1 - e_i))$; if $i \neq j$, then $1 = 1 - e_i + e_i(1 - e_j) \in A(1 - e_i) + A(1 - e_j)$, whence $Y_i \cap Y_j = \varnothing$ (formulae (1) and (2)). On the other hand,

(formula (3));

by hypothesis $\prod_{i \in I} (1 - e_i) = 1 - \sum_{i \in I} e_i = 0$, whence $\bigcup_{i \in I} Y_i = X$ (formula (I)). As the $Y_i$ are closed, they are also open, whence $\overline{\omega}(E) \subset P$. Also, obviously $A = \sum_{i \in I} Ae_i$; if $0 = \sum_{i \in I} a_i e_i$ where $a_i \in A$, we obtain, by multiplying by $e_i$, $O = a_i e_i^2 = a_i e_i$ for all $i$; this proves that $\sigma(E) \subset S$.

There exists by hypothesis integers $m \geq 0, n \geq 0$ such that $e = e^m \in Af$ and $f = f^n \in Ae$; let $x, y$ be elements of $A$ such that $e = xf, f = ye$; then $ef = x^2 = x = e$ and similarly $ef = ye^2 = ye = f$, whence $e = f$.

Lemma 2 and Corollary 2 to Proposition 11 show that the mappings $\overline{\omega}$ and $\sigma$ are injective.

Let us show that $\sigma$ is surjective. If $(a_i)_{i \in I}$ is an element of $S$, there are elements $e_i \in a_i$ such that $1 = \sum_{i \in I} e_i$; if $i \neq j$, then $e_i e_j \in a_i \cap a_j = \{0\}$, whence

finally, $Ae_i \subset a_i$, for all $i \in I$ and $\sum_{i \in I} Ae_i = A$, whence $Ae_i = a_i$.

It remains to prove that $\overline{\omega}$ is surjective. Let $(U_i)_{i \in I}$ be an element of $P$ and set $Z_i = \bigcup_{j \neq i} U_j$; as $U_i$ and $Z_i$ are closed, there exist ideals $a_i, b_i$ of $A$ such that $U_i = \mathrm{V}(a_i), Z_i = \mathrm{V}(b_i)$. We now show that it is possible to suppose further that $a_i \cap b_i = 0$. Now $U_i \cap Z_i = \varnothing$, whence $a_i + b_i = A$; let $a_i, b_i \in a_i, b_i$ be such that $a_i + b_i = 1$. Then $X = U_i \cup Z_i = \mathrm{V}(a_i b_i)$ (formula (3)); every element of $a_i b_i$ is therefore nilpotent (Corollary 2 to Proposition 11); let $p$ be an integer such that $a_i^p b_i^p = 0$. Then $U_i \subset \mathrm{V}(Aa_i) = \mathrm{V}(Aa_i^p)$,

and $V(Aa_i) \cap V(Ab_i) = V(Aa_i + Ab_i) = \varnothing$, hence $U_i = V(Aa_i^p)$ and $Z_i = V(Ab_i^p)$, which establishes our assertion by replacing $a_i$ by $Aa_i^p$ and $b_i$ by $Ab_i^p$. The ideals $a_i$ and $b_i$ thus chosen, it follows from the fact that $\sigma$ is bijective that there exist two idempotents $f_i \in a_i, e_i \in b_i$ such that $1 = e_i + f_i, e_i f_i = 0, a_i = Af_i, b_i = Ae_i$. If $i \neq j$, then $X = Z, \cup Z_j = V(Ae_i e_j)$, and as $e_i e_j$ is idempotent, Lemma 2 shows that $e_i e_j = 0$. Finally $e = \sum_{i \in I} e_i$ is idempotent and $e_i \in Ae$ for all $i \in I$, whence $V(Ae) \subset Z$, for all $i$; it follows that

and Lemma 2 shows also that $e = 1$.

We write $A' = A/r$. We know (Remark following Proposition 13) that

is a homeomorphism, every prime ideal of $A$ containing $r$ by hypothesis. Proposition 15 shows that there exists in $A$ a finite orthogonal family $(e_i)_{i \in I}$, of idempotents such that $\sum_{i \in I} e_i = 1$ and that the image under $ah$ of $V(A'(1 - e'_i))$ is $V(A(1 - e_i))$. But clearly $V(A(1 - e_i))$ is also the image under $ah$ of

as $1 - e'_i$ and $1 - h(e_i)$ are idempotent, Lemma 2 shows that $e'_i = h(e_i)$, whence the corollary.

To say that $X$ is not connected means that there exists in $X$ a set which is open and closed and distinct from $\varnothing$ and $X$.