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0
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0
answers
175
views
Flatness of a simple ring extension
I have tried to show flatness using the follwong ideas:
(1) Corollary 9 (and Theorem 5) of Wang. …
2
votes
1
answer
279
views
Deciding whether a non-f.g. non-divisible flat module is projective or not
Assume $S= R[T]/(f)= R[w]$ is a flat non-divisible $R$-module, where $R$ is a noetherian UFD, $T$ is an indeterminate over $R$, and $f\in R[T]$ is a non-monic polynomial of positive degree.
Can we sa …
2
votes
1
answer
343
views
Projective dimension of a quotient ring
Assume $A$ and $B$ are commutative algebras with $1$, $B = A[z] = A[Z]/(h(Z))$,
$Z$ an indeterminate.
The first comment in this question says that, if $A$ is noetherian, then
$pd_{B\otimes_A B}(B) \ …
-1
votes
1
answer
199
views
Flatness of certain quotient rings
"Answer": Perhaps the followig additional condition (5) would guarantee
flatness of $\frac{\mathbb{C}[x,y]}{\langle p,q \rangle}$ over $\mathbb{C}[x]$
or $\langle p,q \rangle= \langle r \rangle$: …
3
votes
0
answers
59
views
When flatness of $S$ over $R_i$ implies flatness of $S$ over the ring generated by $R_1,R_2$
Are there conditions that will guarantee flatness of $S$ over $R$? … See also this relevant question about flatness over tensor products.
Thank you very much! …
2
votes
2
answers
737
views
When $mB \neq B$? $m$ is a maximal ideal of $A$, $A \subseteq B$
The following is a question I have asked here without receiving any comments, therefore I post it here:
Let $A \subseteq B$ be commutative rings, $m$ a maximal ideal of $A$.
When $mB \neq B$?
This i …
0
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1
answer
205
views
Separable non-flat simple ring extension
Is there a fifth condition that would guarantee flatness of $R \subseteq S$? … Perhaps adding a fifth condition (5) $R$ is a UFD (or at least integrally closed) would guarantee
flatness of $R \subseteq S$? (I am not sure).
The above is (almost) question 3 of this question. …
1
vote
1
answer
254
views
Flat and algebraic (non-integral) local rings extension $R \subseteq S$ with $m_RS=m_S$
Let $R \subseteq S$ be two Noetherian local rings, not necessarily regular, which are integral domains,
with $m_RS=m_S$, namely, the ideal in $S$ generated by $m_R$ (= the maximal ideal of $R$) is $m_ …
1
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0
answers
192
views
Is the following local map unramified?
Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.
In comments to this question it was claimed that in such situation …
0
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1
answer
429
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Separability of $\mathbb{C}[x]$ over its $\mathbb{C}$-subalgebras
Here $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$ is not flat; I am not sure if there is a connection between flatness or non-flatness of $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$ and $\mathbb{C}[x] \ … {C}[x^2,x^3]} \mathbb{C}[x] \subset \mathbb{C}[x]$ would imply flatness of $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$, which is false. …
3
votes
0
answers
150
views
Flatness of certain $R \subseteq \mathbb{C}[x,y]$
I am not sure if one of the known proofs for flatness of $\mathbb{C}[p,q] \subseteq \mathbb{C}[x,y]$ can be adjusted here.
I have also asked the above question in MSE. …
1
vote
0
answers
114
views
$A \to B$ with $A$ regular imply that $B$ is CM
The answer to this question says the following:
"The general statement is if $A \to B$ is finite and injective, and $A$ is noetherian and regular, then $B$ is CM if and only if $A \to B$ is flat.
The …
3
votes
1
answer
236
views
Flatness of certain subrings
But what about flatness?
I guess (?) that there exists a counterexample in dimension two. …