Hausdorff-ness of irreducible affine algebraic sets

Question

In what case is $X$, an irreducible affine algebraic set not a Hausdorff topological space with respect to the Zariski topology?

My understanding is that this would only occur if $X$ is infinite, however I am stuck on how one might prove this.

Answer 1

Recall affine $n$-space is a set of tuples: $$A^n_K := \bigg\{(a_1,...,a_n)\text{ }\bigg|\text{ }a_i\in K_{alg.closedfield}\bigg\}$$ and $A:= K[x_1,...,x_n]$ is a polynomial ring in $n$ variables over $K$.

An algebraic set is the zero set of a collection of such polys. That is: $$\bigg(Y_{algebraic}\subseteq A^n \bigg)\leftrightarrow \bigg(\exists T\subseteq A: Y = Z(T):= \bigg\{p\in A^n\text{ }\bigg|\text{ }\forall f\in T:f(p) = 0\bigg\}\bigg).$$ The Zariski Topology on $A^n$ is defined by: $$\mathcal{O}_{open}\in \tau_{A^n} \leftrightarrow \mathcal{O}^c_{algebraic}.$$ Next, for any topological space, $X$, we have $Y\subseteq X$ is irreducible if $Y\neq Y_1\cup Y_2$ for any proper subsets of $Y$ that are closed with respect to the subspace topology on $Y$.

Lastly, a top space is Hausdorff if $\forall p,q\in X: \exists \mathcal{O}_p,\mathcal{O}_q\in\tau_X$ such that $\mathcal{O}_p\cap\mathcal{O}_q = \emptyset$.

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With this bag of definitions, if $\color{blue}{X\text{ is an irreducible algebraic set}}$, then we have: $$\exists T \subseteq A: Z(T) = X \neq X_1\cup X_2$$ for any $X_{i,proper.closed}\subseteq X$.

Notice that closed sets in $A^n$ (w.r.t. Zariski) are algebraic sets. Closed sets in the subspace topology would be then just algebraic sets restricted to $X$. Proper closed subsets of $X$ then are not empty and not equal to $Z(T)$. Rewriting: $$\color{darkblue}{\forall T_i\subseteq A: (\emptyset\neq T_i \neq T)\wedge\bigg[X := Z(T)\neq (Z(T_1)\cup Z(T_2))\cap X\bigg]}$$

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WTS now that $X$ is not Hausdorff w.r.t. the subspace topology. Accordingly, let $p,q\in X$ and suppose $\mathcal{O}_p\cap \mathcal{O}_q = \emptyset$. Since $\color{darkred}{\mathcal{O}_p := Z(T_1)^c\cap X}$ and $\color{darkred}{\mathcal{O}_q := Z(T_2)^c\cap X}$ for subsets $T_i\subseteq A$, by DeMorgan, we have: $$\mathcal{O}_p\cap \mathcal{O}_q = (Z(T_1)^c\cap X) \cap (Z(T_2)^c\cap X) = (Z(T_1)\cup Z(T_2))^c\cap X = \emptyset$$ $$\implies (Z(T_1)\cup Z(T_2))^c = X^c$$ $$\implies (Z(T_1)\cup Z(T_2))\cap X = X$$ Thus if we can show our $\emptyset \neq T_i\neq T$, we've reached a contradiction.

WLOG we prove the result for $T_1$:
Assume for the moment that $T_1\neq \emptyset$. Since $p\in \mathcal{O}_p = Z(T_1)^c\cap X$, we know $p\in Z(T_1)^c$. So $\exists f\in T_1: f(p)\neq 0$. But since $p\in X=Z(T)$, we have $\forall f\in T:f(p)=0$. This implies $\exists f\in T_1$ that is not in $T$. Hence $T_1\neq T$.

Finally, suppose $T_1=\emptyset$, then $Z(T_1) = A^n$ (vacuously). So $\mathcal{O}_p = Z(T_1)^c\cap X = \emptyset \cap X = \emptyset$. Since $p\in \mathcal{O}_p$, this is a contradiction and we're done. $\square$

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So to answer the question, according to the above proof it seems all cases of irreducible algebraic sets $X$ are not Hausdorff in the Zariski topology.