Timeline for Subschemes of the affine line over a domain
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 13, 2020 at 19:05 | comment | added | David Lampert | @lefuneste I think I don't have much in general to add to what's already formulated in your question. I merely gave an example to show that special fibers over the base can have various different qualities from the generic fiber. Think of more examples with special fibers where the leading coefficient of $f$ vanishes but not all coefficients. | |
| May 13, 2020 at 18:22 | history | edited | lefuneste | CC BY-SA 4.0 |
I have added an Edit
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| May 13, 2020 at 18:16 | comment | added | lefuneste | Perfect, @David and thanks a lot. I would be very happy if you wrote down the answer in general, guided by your edifying example. (I tried to upvote your comment, but unfortunately couldn't, probably because I am a new user on this site) | |
| May 13, 2020 at 14:53 | comment | added | David Lampert | @lefuneste Sorry I forgot about $P \cap R = 0$. Instead $P=(aT^3-bT, cT^3-dT)$ and $f=T, Y$ as above $\cup \{T=0\}$. | |
| May 13, 2020 at 14:42 | comment | added | lefuneste | @David Your example is very interesting but it doesn't quite satisfy my hypotheses. Indeed you have $P\cap R\neq 0$, contrary to what I suppose, since $c(aT^2-b)-a(cT^2-d)= -bc+ad\neq 0$ is in $P\cap R$. Nevertheless I'm sure one could extract a nice example from your situation and I thank you for this comment. | |
| May 13, 2020 at 13:39 | comment | added | David Lampert | Here maybe is a helpful example: $R=k[a,b,c,d], P=(aT^2-b, cT^2-d)$. Then $f=1$ (so the generic fiber of $Y \rightarrow S$ is empty), $Y$ is finite degree $2$ over $\{ad-bc=0\} - \{a=c=0\}$ $\cup$ line over $\{a=b=c=d=0\}$. | |
| May 13, 2020 at 9:02 | comment | added | lefuneste | "The hypothesis says, (f)=P∩I where I∩R≠0" No hypothesis says that for the good reason that $I$ is never mentioned in my question and you don't say what it represents. In the most charitable interpretation I guess $I$ means the ideal defining $Y$ and then your inequality is not always true. Anyway I'm getting a bit tired of trying to understand unjustified statements that I find quite enigmatic. this is not the way this site works: please write a complete answer in the answer box if you are capable and willing. Else, no hard feelings and I wish you all the best . | |
| May 13, 2020 at 1:37 | comment | added | Mohan | The hypothesis says, $(f)=P\cap I$ where $I\cap R\neq 0$ and then $\sqrt{I\cap R} R[T]=\sqrt{I}$, since $I$ has height one. | |
| May 12, 2020 at 21:56 | comment | added | lefuneste | @mohan I can't make head or tail of what you write. Of course the minimal primes containing $f$ have height one but what has this got to do with the morphism to $S$ , the pull back of closed subsets of $S$ and the hypothesis that $P\cap R=0$? | |
| May 12, 2020 at 20:47 | comment | added | Mohan | Then, use Krull's principal ideal theroem and look at the prime decomposition of $(f)$. Set theoretically this is the union of $Y$ and the pull back of a closed subset of $S$, since all minimal primes containing $f$ have height one. | |
| May 12, 2020 at 20:01 | comment | added | lefuneste | No, I was not assuming $R$ noetherian. But I would be happy to read a proof of your claim assuming noetherianness. | |
| May 12, 2020 at 17:52 | comment | added | Mohan | Are you assuming $R$ is Noetherian? I was. | |
| May 12, 2020 at 17:50 | comment | added | lefuneste | @Mohan: Well of course, that's the obvious guess, as I hinted in my question. I'm asking for a detailed, rigorous proof. | |
| May 12, 2020 at 17:34 | comment | added | Mohan | $V(f)$ is the union of $Y$ and a subscheme of $X$ which is the pull back of some subscheme from $S$. | |
| May 12, 2020 at 16:30 | review | First posts | |||
| May 12, 2020 at 16:42 | |||||
| May 12, 2020 at 16:29 | history | asked | lefuneste | CC BY-SA 4.0 |