A quasi-affine variety $X $ is an open subset of some affine variety $X'$. Thus, a closed subset of the quasi-affine variety $X$ is of the form $X \cap Z'$, using a closed subset $Z'$ of $X'$. If the closed subset $X \cap Z'$ of this quasi-affine variety is irreducible, is $Z'$ irreducible? I hope it is irreducible. If so, then an irreducible closed subset of the quasi-affine variety $X$ can be given as $X\cap Z''$ using an irreducible closed subset $Z''$ of $X'$.
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Take $X' = \mathbb{A}_{\mathbb{C}}^2$, with $L_x$, $L_y$ being the axes. Then, let $X = \mathbb{A}^2 - L_x$ and $Z' = L_x \cup L_y$. Then, $Z' \cap X = L_y - \{(0,0)\} \cong \mathbb{C}^*$ is irreducible but $Z'$ is clearly not.
However, you may always take $Z'$ to be irreducible. In the above example, we may have taken $Z'$ to be $L_y$ and the intersection would have been the same.
This is due the the fact that if $S \subset X'$ is an irreducible subset (in subspace topology) of $X'$, its closure $\overline{S}$ is irreducible. Hence, if $Z \subset X$ is irreducible, then $Z = \overline{Z} \cap X$ so we may take $Z' = \overline{Z}$ which is irreducible.
Daniel
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Thank you. I newly learned that when X is a projective (affine) variety and X' is its open set, a nonempty irreducible closed subset of X' corresponds one-to-one to a irreducible closed subset of X whose intersection with X' is nonempty. – ねここねこ Nov 30 '23 at 16:09