# Prove that an affine variety is irreducible if and only its projective closure is irreducible.

Prove that an affine variety $X \subseteq \mathbb{A^n}$ is irreducible if and only its projective closure $\bar X \subset \mathbb{P}^n$ is irreducible.

attempt: suppose by contradiction that $X$ is reducible. And that $\bar X$ is reducible. Then $\bar X = A \cup B$, where $A,B$ are close sets. Then $X = (X \cap A) \cup (X \cap B)$. But this tell tell us that because $X$ is irreducible, we have $X = X \cup A$ or $X = X \cup B$. Then we have $X = X \cup A$ implies $X \subseteq A$ or $X = X \cup B$ implies $X \subseteq B$, then since $A,B$ are close, then $A = \bar A$ and $B = \bar B$, so we have $\bar X \subseteq A$ or $\bar X \subseteq B$, which we have a contradiction, so we must have that $X$ is irreducible.

Conversely, let $\bar X$ be reducible, so that $\bar X = A \cup B$, then $X$ is contained in $A \cup B$.

could someone please help me? I don't know if this makes sense. I need help on the converse part. Thank you.

Assume $\overline{X}$ is irreducible. Let $A, B$ be two closed sets of $X$ whose union is $X$. We want to show that either $A = X$ or $B = X$. Since $X = A \cup B$, we have $\overline{X} =\overline{A} \cup \overline{B}$. Since $\overline{X}$ is irreducible, we have say, $\overline{X} = \overline{A}$. The closure of $A$ as a subset of $X$ is $\overline{A} \cap X = \overline{X} \cap X = X$. But $A$ is already closed in $X$, so $A = X$.
• this shows that $X$ is reducible? because $X$ can be expressed as a union of tow closed proper sets. and it's a contradiction so we have that $X$ is irreducible? – Mahidevran Mar 3 '17 at 6:04
• but we have a contradiction because we must have closed subsets in $X$ such that either $\bar X$ is not contained either on $A$ or $B$? – Mahidevran Mar 3 '17 at 6:20