# Is $\varnothing$ an affine variety with these definitions?

A topological space $X$ is called reducible if $X=X_1 \cup X_2$ where $X_1, X_2$ are non-empty, proper subsets of $X$ and closed.

An irreducible algebraic set in $\mathbb{A}^n$ is called an affine variety.

$\varnothing = Z(1)$, so $\varnothing$ is an algebraic set, and obviously $\varnothing \subseteq \mathbb{A}^n$.

If we suppose that $\varnothing$ is reducible, then the only options for $X_1$ and $X_2$ are $\varnothing$, but this can't happen, because $X_1$ and $X_2$ must be non-empty.

So, am I right saying that

$\varnothing$ is an affine variety?

• possible duplicate of is the empty set an (irreducible) variety? – Alex Kruckman Mar 9 '15 at 21:04
• I saw that question but the definition of ''reducible'' is different, I don't suppose that $X \neq \varnothing$ – Leafar Mar 9 '15 at 21:08
• @Leafar. You are not entitled to change the meaning of definitions. This is like asking "if I call $3$ the number $4$, is it true that $2+2=3$" ? So, yes, $\emptyset=Z(1)$ is algebraic, not irreducible and that's the end of the (uninteresting) story. – Georges Elencwajg Mar 9 '15 at 21:23
• Why uninteresting? There are many things associated, like the dimension, we have to consider all pathological cases... – Leafar Mar 9 '15 at 21:27

However, there are good reasons for considering this to be the wrong definition of "irreducible", as explained in this question. Here's another: the defining property of sober spaces (examples include algebraic varieties with the Zariski topology and Hausdorff spaces) is "every irreducible closed set has a unique generic point". This is nonsense if we allow $\emptyset$ to be irreducible.