Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options answers only not deleted user 2083
4 votes
Accepted

flatness and reduction

The point is the so-called miracle flatness, (in the graded form for this example to ease notations). Let $S=k[l_1,...,l_d]$ where the $l$s form a system of parameters for $R/I$ (and also $R/J$). …
Hailong Dao's user avatar
  • 30.5k
10 votes

Is the support of a flat sheaf flat?

Here is an algebraic construction. The way I think about it is based on these two facts: 1) when $A$ is regular domain, a module-finite A-algebra is flat iff it is Cohen-Macaulay (CM) of same dimensi …
Hailong Dao's user avatar
  • 30.5k
2 votes

Quotient of flat module is flat - a property in Mumford's Red book

Flatness can be checked locally on maximal ideals, so we may as well let $R=A_m, k = R/mR$ and considering: If $Tor_i^R(N,k) =0$ for $i>0$, when is $N$ flat? … You can also check out the paper "A local flatness criterion ..." by Hans Schoutens (available on his website). …
Hailong Dao's user avatar
  • 30.5k
4 votes

Torsion-free tensor powers

It is hard for me not to mention the following splendid result by Auslander-Lichtenbaum: If $R$ is regular local with Krull dimension $d$, then for any finitely generated module $M$, $M^{\otimes n}$ …
Community's user avatar
  • 1
4 votes

The inverse limit of locally free module

The answer is yes. 1) $A$ is $I$-adic complete implies that $I \subset rad(A)$, the intersection of all maximal primes. Indeed, pick any $a \in I$. Look at $1-a + a^2 -a^3 ... \in A $ (this is where …
Hailong Dao's user avatar
  • 30.5k