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Duchamp Gérard H. E.
Duchamp Gérard H. E.
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"Frac" and "Const" commute: A differential ring is simply a pair $(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

See there for a discussion and counterexamples.

"Frac" and "Const" commute A differential ring is simply a pair $(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

See there for a discussion and counterexamples.

"Frac" and "Const" commute: A differential ring is simply a pair $(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

See there for a discussion and counterexamples.

I elaborated the false belief
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Duchamp Gérard H. E.
Duchamp Gérard H. E.
  • 3.7k
  • 34
  • 40

"Frac" and "Const" commute A differential ring is simply a pair See there for a counterexample$(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see abovebelow).

See there for a discussion and counterexamples.

"Frac" and "Const" commute See there for a counterexample.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see above).

"Frac" and "Const" commute A differential ring is simply a pair $(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

See there for a discussion and counterexamples.

Source Link
Duchamp Gérard H. E.
Duchamp Gérard H. E.
  • 3.7k
  • 34
  • 40

"Frac" and "Const" commute See there for a counterexample.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see above).

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