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#4954 on tag 00HL by procvaustralia

procvaustralia — Mon, 24 Feb 2020 07:10:19 GMT

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#4953 on tag 02GW by awllower

awllower — Mon, 24 Feb 2020 12:01:17 GMT

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#4952 on tag 0B9J by SDIGR

SDIGR — Sun, 23 Feb 2020 03:04:44 GMT

Typo!?:
Last sentence:
The divisor $D$ corresponding to a $k$ -rational point of $\underline{\mathrm{Hilb}}_{X/k}^{d+1}$ must be of degree $d+1$ , not of degree $d$ .

#4951 on tag 0B9H by SDIGR

SDIGR — Sun, 23 Feb 2020 02:32:38 GMT

Sorry, I mean the other way around.

#4950 on tag 0B9H by SDIGR

SDIGR — Sun, 23 Feb 2020 02:31:53 GMT

First sentence of the Proof: $\underline{\mathrm{Hilb}_{X/S}^0}=S\mapsto\underline{\mathrm{Hilb}_{X/S}^d}=S$ .

#4949 on tag 0E6N by Min

Min — Sun, 23 Feb 2020 09:22:20 GMT

What is T in the functor the dualizing sheaf represents? Should it be U?

#4948 on tag 0E6N by Min

Min — Sun, 23 Feb 2020 09:21:07 GMT

What is $T$ in the functor the dualizing sheaf represents? Should it be $U$ ?

#4947 on tag 00LB by yogesh

yogesh — Sat, 22 Feb 2020 08:58:47 GMT

Oops, sorry disregard my previous comment. I got forgot we're talkng about the annihilator as an element of $M$ , not $M/M_{n-1}$ . I was thinking of using the previous result of associated primes of short exact sequences applied to $M_{n-1} \to M \to R/\mathfrak{p}_n$ so $Ass(M) \subseteq Ass(M_{n-1} \cup \{ \mathfrak{p}_n\}$ But the proof you give is basically the proof of that result, whose proof is omitted.

#4946 on tag 00LB by yogesh

yogesh — Sat, 22 Feb 2020 08:53:02 GMT

Oops, sorry disregard my previous comment. I got forgot we're talkng about the annihilator as an element of $M$ , not $M/M_{n-1}$ .

#4945 on tag 00LB by yogesh

yogesh — Sat, 22 Feb 2020 08:34:02 GMT

In the proof, I think the inclusion $\mathfrak{p} \subset \mathfrak{p}_n$ is going the wrong way and the rest of the proof is unnecessary. I think any nonzero element of $R/\mathfrak{p}_n$ is annihilated by exactly by $\mathfrak{p}_n$ .