Timeline for integral subschems of geometrically integeral schemes
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2021 at 2:50 | comment | added | Maddy | Yes separable closure of the ground field is enoughThanks@ Will Sawin. | |
| Jun 23, 2021 at 15:22 | comment | added | Will Sawin | Separably closed would suffice, i guess, because totally inseparable extensions can make integral curves non-reduced but not reducible. | |
| Jun 23, 2021 at 15:19 | comment | added | Maddy | 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? | |
| Jun 23, 2021 at 15:18 | review | Close votes | |||
| Jul 1, 2021 at 10:04 | |||||
| Jun 23, 2021 at 15:10 | comment | added | Daniel Loughran | More explicitly: $x^2 + y^2 = 0$ is integral but not geometrically irreducible over $\mathbb{R}$. | |
| Jun 23, 2021 at 14:57 | review | First posts | |||
| Jun 23, 2021 at 15:00 | |||||
| Jun 23, 2021 at 14:55 | comment | added | Jason Starr | 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. | |
| Jun 23, 2021 at 14:51 | history | asked | Maddy | CC BY-SA 4.0 |