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I have asked the following question in MSE:

Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$. Assume that $Q(A) \cap B=A$.

Define property $P$ as follows:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.

By the assumption that $Q(A) \cap B=A$, property $P$ becomes:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.

Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that $A \subseteq B$ satisfies property $P$?

Non-answer: The following four conditions do not guarantee that $A \subseteq B$ satisfies property $P$:

(i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$.

(ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist $a_{n-1},\ldots,a_1,a_0 \in A$ such that $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$.

(iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only.

(iv) $A \subseteq B$ is flat.

(v) Both $A$ and $B$ are UFD's.

Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.

Plausible answers:

(1.) $A \subseteq B$ is a separable ring extension?

(2.) $Q(A)=Q(B)$. Indeed, in this case we get $B=Q(B) \cap B=Q(A) \cap B=A$. Then $b \in B=A$.

Remarks:

(1) Notice that if $n=0$ then property $P$ is satisfied; indeed, if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$, so $\tilde{a_0}= w \in Q(A) \cap B =A$.

(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.

Thank you very much for any hints and comments!

I have asked the following question in MSE:

Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$. Assume that $Q(A) \cap B=A$.

Define property $P$ as follows:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.

By the assumption that $Q(A) \cap B=A$, property $P$ becomes:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.

Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that $A \subseteq B$ satisfies property $P$?

Non-answer: The following four conditions do not guarantee that $A \subseteq B$ satisfies property $P$:

(i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$.

(ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist $a_{n-1},\ldots,a_1,a_0 \in A$ such that $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$.

(iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only.

(iv) $A \subseteq B$ is flat.

(v) Both $A$ and $B$ are UFD's.

Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.

Plausible answers:

(1.) $A \subseteq B$ is a separable ring extension? (or separable and flat? Flatness alone is not enough).

(2.) $Q(A)=Q(B)$. Indeed, in this case we get $B=Q(B) \cap B=Q(A) \cap B=A$. Then $b \in B=A$.

Remarks:

(1) Notice that if $n=0$ then property $P$ is satisfied; indeed, if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$, so $\tilde{a_0}= w \in Q(A) \cap B =A$.

(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.

Thank you very much for any hints and comments!

I have asked the following question in MSE:

Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$. Assume that $Q(A) \cap B=A$.

Define property $P$ as follows:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.

By the assumption that $Q(A) \cap B=A$, property $P$ becomes:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.

Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that $A \subseteq B$ satisfies property $P$?

Non-answer: The following three conditions do not guarantee that $A \subseteq B$ satisfies property $P$:  (i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$.  (ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist $a_{n-1},\ldots,a_1,a_0 \in A$ such that $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$.  (iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only. Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.

Plausible answers:

(1) $A \subseteq B$ is a separable ring extension?

(2) $A \subseteq B$ is faithfully flat?

(3) $Q(A)=Q(B)$. Indeed, in this case we get $B=Q(B) \cap B=Q(A) \cap B=A$. Then $b \in B=A$.

Remarks:

(1) Notice that if $n=0$ then property $P$ is satisfied; indeed, if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$, so $\tilde{a_0}= w \in Q(A) \cap B =A$.

(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.

Thank you very much for any hints and comments!

I have asked the following question in MSE:

Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$. Assume that $Q(A) \cap B=A$.

Define property $P$ as follows:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.

By the assumption that $Q(A) \cap B=A$, property $P$ becomes:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.

Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that $A \subseteq B$ satisfies property $P$?

Non-answer: The following four conditions do not guarantee that $A \subseteq B$ satisfies property $P$: 

(i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$. 

(ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist $a_{n-1},\ldots,a_1,a_0 \in A$ such that $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$. 

(iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only.

(iv) $A \subseteq B$ is flat.

(v) Both $A$ and $B$ are UFD's.

Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.

Plausible answers:

(1.) $A \subseteq B$ is a separable ring extension?

(2.) $Q(A)=Q(B)$. Indeed, in this case we get $B=Q(B) \cap B=Q(A) \cap B=A$. Then $b \in B=A$.

Remarks:

(1) Notice that if $n=0$ then property $P$ is satisfied; indeed, if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$, so $\tilde{a_0}= w \in Q(A) \cap B =A$.

(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.

Thank you very much for any hints and comments!

I have asked the following question in MSE:

Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$. Assume that $Q(A) \cap B=A$.

Define property $P$ as follows:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.

By the assumption that $Q(A) \cap B=A$, property $P$ becomes:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.

Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that $A \subseteq B$ satisfies property $P$?

Non-answer: The following three conditions do not guarantee that $A \subseteq B$ satisfies property $P$: (i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$. (ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist $a_{n-1},\ldots,a_1,a_0 \in A$ such that $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$. (iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only. Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.

Plausible answers: $A \subseteq B$ is a separable ring extension?  $A \subseteq B$ is faithfully flat?

Remarks:

(1) Notice that if $n=0$ then property $P$ is satisfied; indeed, if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$, so $\tilde{a_0}= w \in Q(A) \cap B =A$.

(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.

Thank you very much for any hints and comments!

I have asked the following question in MSE:

Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$. Assume that $Q(A) \cap B=A$.

Define property $P$ as follows:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.

By the assumption that $Q(A) \cap B=A$, property $P$ becomes:

If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$, then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.

Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that $A \subseteq B$ satisfies property $P$?

Non-answer: The following three conditions do not guarantee that $A \subseteq B$ satisfies property $P$: (i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$. (ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist $a_{n-1},\ldots,a_1,a_0 \in A$ such that $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$. (iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only. Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.

Plausible answers:

(1) $A \subseteq B$ is a separable ring extension?

(2) $A \subseteq B$ is faithfully flat?

(3) $Q(A)=Q(B)$. Indeed, in this case we get $B=Q(B) \cap B=Q(A) \cap B=A$. Then $b \in B=A$.

Remarks:

(1) Notice that if $n=0$ then property $P$ is satisfied; indeed, if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$, so $\tilde{a_0}= w \in Q(A) \cap B =A$.

(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.

Thank you very much for any hints and comments!