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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. …
user237522's user avatar
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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 …
user237522's user avatar
  • 2,837
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) \ …
user237522's user avatar
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-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 $\langle p,q \rangle= \langle r \rangle$: …
user237522's user avatar
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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! …
user237522's user avatar
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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 …
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0 votes
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. …
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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_ …
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1 vote
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 …
user237522's user avatar
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0 votes
1 answer
429 views

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. …
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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. …
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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 …
user237522's user avatar
  • 2,837
3 votes
1 answer
236 views

Flatness of certain subrings

But what about flatness? I guess (?) that there exists a counterexample in dimension two. …
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