Exercise 1.20

Proof of $i)$. Let $U,V\subseteq \bar{Y}$ open. Since $U$ is the neighborhood of some $x\in \bar{Y}$, we see that $U\cap Y\ne \varnothing$; similarly for $V\cap Y$. But since $U\cap Y,V\cap Y$ are open, $U\cap V\supseteq (U\cap Y)\cap (V\cap Y)\ne \varnothing$ by the irreducibility of $Y$. □

Proof of $\mathit{ii})$. Let $Z$ be irreducible; we order the set $\mathrm{\Sigma }$ of irreducible subspaces of $X$ containing $Z$ by inclusion. $\mathrm{\Sigma }\ne \varnothing$ since $Z\in \mathrm{\Sigma }$, and so by Zorn’s lemma, it suffices to show any ascending chain ${\{Y_{\alpha }\}}_{\alpha \in A}$ has an upper bound $Y={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in A}{Y}_{\alpha }$ in $\mathrm{\Sigma }$. So let $U\cap Y,V\cap Y$ be non-empty opens in $Y$. Then, there exist $\alpha ,\beta$ such that $U\cap Y_{\alpha },V\cap Y_{\beta }$ are nonempty. Assuming without loss of generality that $\alpha \le \beta$, $U\cap Y_{\beta }\supseteq U\cap Y_{\alpha }$ is also nonempty, and $U\cap V\cap Y\supseteq U\cap V\cap Y_{\beta }\ne \varnothing$ by the irreducibility of $Y_{\beta }$. □

Proof of $\mathit{iii})$. Maximal irreducible subspaces $Y$ are closed since $\bar{Y}$ is also irreducible by $i)$, hence $Y=\bar{Y}$. They cover $X$ since $\{x\}$ is irreducible for any $x\in X$ (every nonempty open subset of $\{x\}$ contains $x$), and so $x$ is contained in some maximal irreducible subspace by $\mathit{ii})$.

Suppose $X$is Hausdorff, and $Y\subseteq X$ is irreducible. If $Y$ is a one-point set, it is irreducible by above. If $Y\ni x,y$ for $x\ne y$, the Hausdorff axiom implies there exists open $U\ni x,V\ni y$ such that $U\cap V=\varnothing$, and so $Y$ is reducible. Thus the irreducible components are one-point sets. □

Proof of $\mathit{iv})$. First, $X_f\cap V(𝔭)\ne \varnothing$ if and only if $f\notin 𝔮$ for some prime ideal $𝔮\supseteq 𝔭$, which occurs if and only if $f\notin 𝔭$.

Now to show $V(𝔭)$ is irreducible, it suffices to show any non-empty basis elements $X_f\cap V(𝔭)$ and $X_g\cap V(𝔭)$ intersect. By the above, $f,g\notin 𝔭$, and by primeness, $\mathit{fg}\notin 𝔭$, hence $𝔭\in X_{\mathit{fg}}\cap V(𝔭)=X_f\cap X_g\cap V(𝔭)$.

Conversely, suppose $Y\subseteq \mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(A)$ is irreducible. $Y=V(r(𝔞))$ for some ideal $𝔞\subseteq A$ by Exercise $15i)$. Suppose $r(𝔞)$ is not prime, and $f,g\notin r(𝔞)$ such that $\mathit{fg}\in r(𝔞)$. There then exist $𝔭\not\ni f$ and $𝔮\not\ni g$ in $V(𝔞)$ by Prop.  $1.14$, and so by the first paragraph, $X_f\cap V(𝔭)$ and $X_g\cap V(𝔮)$ are non-empty. Thus $X_f\cap V(r(𝔞))$ and $X_g\cap V(r(𝔞))$ are non-empty. But by Exercise $17i)$, their intersection $X_{\mathit{fg}}\cap V(r(𝔞))$ is empty since $\mathit{fg}\in r(𝔞)$ implies $\mathit{fg}\in 𝔭$ for any $𝔭\in V(r(𝔞))$, a contradiction.

So, the irreducible subspaces of $X$ are of the form $V(𝔭)$ for $𝔭$ prime. Moreover, $V(𝔭)$ is maximal among these subspaces if and only if $𝔭$ is a minimal prime ideal, and so the irreducible components of $X$ are of the form $V(𝔭)$ for $𝔭$ a minimal prime. □