Let $X$ be a geometrically integral surface over a field $k$ and let $C$ be an integral curve on $X$. I want to know whether $C$ is geometrically irreducible over $k$ or not.
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$\begingroup$ Welcome new contributor. There are examples where $C$ is not a geometrically irreducible curve. Consider the case where$X$ equals $\mathbb{A}^2_k$ with a coordinate projection to $\mathbb{A}^1_k$. By the Primitive Element Theorem, every finite, separable extension field $L/k$ is realized as an irreducible divisor in $\mathbb{A}^1_k$. The inverse image of that irreducible divisor under the coordinate projection is an integral curve in $\mathbb{A}^2_k$ that is not geometrically irreducible. $\endgroup$– Jason StarrCommented Jun 23, 2021 at 14:55
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1$\begingroup$ More explicitly: $x^2 + y^2 = 0$ is integral but not geometrically irreducible over $\mathbb{R}$. $\endgroup$– Daniel LoughranCommented Jun 23, 2021 at 15:10
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$\begingroup$ Thanks @ Lughran and @starr. I want to know can we put some condition on the surface or the ground field(not algebraically closed) so that all integral curves are geometrically irreducible over the ground field? $\endgroup$– MaddyCommented Jun 23, 2021 at 15:19
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3$\begingroup$ Separably closed would suffice, i guess, because totally inseparable extensions can make integral curves non-reduced but not reducible. $\endgroup$– Will SawinCommented Jun 23, 2021 at 15:22
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$\begingroup$ Yes separable closure of the ground field is enoughThanks@ Will Sawin. $\endgroup$– MaddyCommented Jun 25, 2021 at 2:50
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