Timeline for How to show that the intersection of two certain affine varieties is reduced?
Current License: CC BY-SA 4.0
10 events
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| Feb 13, 2023 at 23:41 | comment | added | Sándor Kovács | My rationale for suggesting that was that the $Y_i$ and $Z_j$ are linear spaces,so are their pairwise intersections. If that is not contained in $X$ (which is also linear), then they intersect transversally, so one does not expect "excess" intersection. In other words, I would expect the intersection to be reduced. Of course,this would only tell you that individually the intersections are reduced,but that same assumption also suggests that the union of these intersections are still reduced. I am not claiming this is a proof,just that this was the heuristic thinking that led to that suggestion | |
| Feb 12, 2023 at 22:50 | vote | accept | Utf | ||
| Feb 12, 2023 at 22:49 | comment | added | Utf | Hi @SándorKovács: I'd say this new example is actually again a counterexample to my optimistic guess (indeed it is pretty similar to the very first counterexample that you gave me). Thanks for realizing me that my conditions were not enough, I am now trying to understand if my setting actually fulfills also the condition that you hinted as possibly sufficient, namely $Y_i \cap Z_j \nsubseteq X$. Could you just explain me a bit more why this condition should be enough to ensure that $X\cap (\bigcup_i Y_i\cup\bigcup_j Z_j)$ is reduced? What kind of argument should I use to prove that? | |
| Feb 10, 2023 at 8:19 | comment | added | Sándor Kovács | @Utf, what do you think of this new example? | |
| Feb 9, 2023 at 20:39 | history | edited | Sándor Kovács | CC BY-SA 4.0 |
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| Feb 8, 2023 at 21:33 | comment | added | Utf | actually I am confused about your new example...I think now both components of $Y$ intersect $X$ in a single point, which is the same for both. So this would be no more a counterexample to my question, since it does not satisfy the condition $(i)$ that I put (the codimensions are not $1$, and more importantly the two intersections are not distinct). According to my intuition (which may be easily wrong, as you had just proven), it should not be possible to find a counterexample having just $Y$'s components, i.e. components satisfying condition $(i)$. | |
| Feb 8, 2023 at 20:56 | history | edited | Sándor Kovács | CC BY-SA 4.0 |
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| Feb 8, 2023 at 20:55 | comment | added | Sándor Kovács | Hi @Utf, you're right, the example I gave didn't work, but I think it does now. In fact, this was the example I first came up with (when I still remembered to check condition (i)), but then I tried to simplify it.... As for the condition that might work: yes, I think that has a good chance. | |
| Feb 8, 2023 at 20:48 | comment | added | Utf | Dear @SándorKovács thanks a lot for your answer (the very last tip included :)), which actually gives a counterexample to what I was (way too optimistically) hoping to be true. Nevertheless, if I correctly understood what you said, I'd say that this is not a counterexample having only $Y$'s components: in fact $Y=Z(z,t)$ certainly intersects $X=Z(x-z,t)$ in a line, but $Z=Z(x,y)$ intersects it just in a point. So, how should I modify the condition you suggested accordingly? Should it become something like $Y_i\cap Z_j \nsubseteq X$ $\forall i,j$? | |
| Feb 8, 2023 at 6:47 | history | answered | Sándor Kovács | CC BY-SA 4.0 |