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Let $k$ be a perfect field.

We say that $k$ has property $(*)$ if for every geometrically irreducible morphism locally of finite type $X\rightarrow \mathrm{Spec}\, k$ admitting a section, the set of points $x\in |X|$ such that the canonical injection $k\rightarrow k(x)$ is a surjection is dense.

Do there exist non-algebraically closed fields satisfying $(*)$?

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    $\begingroup$ Welcome new contributor. Every nonempty geometrically irreducible $k$-scheme $Y$ that is locally of finite type contains at least one closed point $y:\text{Spec}\ \kappa \to Y$ contained in an open affine. You can form the "cofiber coproduct" of $y:\text{Spec}\ \kappa \to Y$ and of $\text{Spec}\ \kappa \to \text{Spec}\ k$ to obtain a new geometrically irreducible $k$-scheme $X$ that is locally of finite type and has a $k$-point. Assuming (*), then $X$ has a dense set of $k$-points. Thus also $Y$ has a dense set of $k$-points. So your notion is the same as "PAC". $\endgroup$ Mar 13, 2019 at 8:57
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    $\begingroup$ @AknazarKazhymurat. I do not understand your question. I see no morphism $\text{Spec}\ k \to X$ in my comment nor in the post except for those that are explicitly $k$-morphisms. $\endgroup$ Mar 13, 2019 at 9:17
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    $\begingroup$ @AknazarKazhymurat. "so for a non-perfect field, a closed point $y:\text{Spec}\ k \to Y$ is a $k$-point?" No. However, there is no place in the post nor in the comment where that is implied or used. If the composite morphism $\text{Spec}\ \kappa \to Y \to \text{Spec}\ k$ happens to be an isomorphism, then $X$ equals $Y$. But in every case, the cofiber coproduct $X$ of $\text{Spec}\ \kappa \xrightarrow{y} Y$ and the morphism $\text{Spec}\ \kappa \to \text{Spec}\ k$ is a geometrically irreducible $k$-scheme that is locally of finite type. $\endgroup$ Mar 13, 2019 at 9:35

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