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Creating new constants in differential field extensions via superfluous solutions to a D.E.
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I am reading the lecture series book Lectures on Differential Galois Theory by Andy Magid. I came across an example, and I am wondering if it generalizes.
The example
Let $F_1=mathbbC(z)$ be the field of rational functions in indeterminate $z$, and $F_2=mathbbC((z))$ the corresponding field of formal power series. Then with the derivation $D(z)=1$, and $D$ trivial on $mathbbC$, $F_2$ is a differential field, and $F_1$ a differential sub-field. Let $f$ be the usual exponential series, so that $D(f)=f$, and let $F=mathbbClangle f rangle$ be the field obtained by adjoining $f$ to $mathbbC$. Consider the differential equation $$Y'-Y=0$$ over $F$. The universal solution algebra is $R=F[y]$ with $D(y)=y$. This construction is superfluous since $F$ already contains the solutions to the differential equation. The existence of these two solutions gives the following equations on their ratios:
$$Dleft(fracyfright)=fracfD(y)-yD(f)f^2=fracfy-yff^2=0$$
In other words, the extra solution gives rise to a new constant.
My question
Suppose $L$ is a monic homogeneous linear differential operator of order $n$ over a differential field $F$ which contains $n$ solutions to $L=0$ linearly independent over constants. Does the adjunction of a new formal solution to $L=0$ result in a new constant? Is this the reason for the no new constant condition in the definition for a Picard Vessiot extension?
differential-equations field-theory extension-field differential-algebra
asked Jul 19 at 20:02
J.Kennedy
486
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up vote
1
down vote
favorite
I am reading the lecture series book Lectures on Differential Galois Theory by Andy Magid. I came across an example, and I am wondering if it generalizes.
The example
Let $F_1=mathbbC(z)$ be the field of rational functions in indeterminate $z$, and $F_2=mathbbC((z))$ the corresponding field of formal power series. Then with the derivation $D(z)=1$, and $D$ trivial on $mathbbC$, $F_2$ is a differential field, and $F_1$ a differential sub-field. Let $f$ be the usual exponential series, so that $D(f)=f$, and let $F=mathbbClangle f rangle$ be the field obtained by adjoining $f$ to $mathbbC$. Consider the differential equation $$Y'-Y=0$$ over $F$. The universal solution algebra is $R=F[y]$ with $D(y)=y$. This construction is superfluous since $F$ already contains the solutions to the differential equation. The existence of these two solutions gives the following equations on their ratios:
$$Dleft(fracyfright)=fracfD(y)-yD(f)f^2=fracfy-yff^2=0$$
In other words, the extra solution gives rise to a new constant.
My question
Suppose $L$ is a monic homogeneous linear differential operator of order $n$ over a differential field $F$ which contains $n$ solutions to $L=0$ linearly independent over constants. Does the adjunction of a new formal solution to $L=0$ result in a new constant? Is this the reason for the no new constant condition in the definition for a Picard Vessiot extension?
differential-equations field-theory extension-field differential-algebra
asked Jul 19 at 20:02
J.Kennedy
486
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading the lecture series book Lectures on Differential Galois Theory by Andy Magid. I came across an example, and I am wondering if it generalizes.
The example
Let $F_1=mathbbC(z)$ be the field of rational functions in indeterminate $z$, and $F_2=mathbbC((z))$ the corresponding field of formal power series. Then with the derivation $D(z)=1$, and $D$ trivial on $mathbbC$, $F_2$ is a differential field, and $F_1$ a differential sub-field. Let $f$ be the usual exponential series, so that $D(f)=f$, and let $F=mathbbClangle f rangle$ be the field obtained by adjoining $f$ to $mathbbC$. Consider the differential equation $$Y'-Y=0$$ over $F$. The universal solution algebra is $R=F[y]$ with $D(y)=y$. This construction is superfluous since $F$ already contains the solutions to the differential equation. The existence of these two solutions gives the following equations on their ratios:
$$Dleft(fracyfright)=fracfD(y)-yD(f)f^2=fracfy-yff^2=0$$
In other words, the extra solution gives rise to a new constant.
My question
Suppose $L$ is a monic homogeneous linear differential operator of order $n$ over a differential field $F$ which contains $n$ solutions to $L=0$ linearly independent over constants. Does the adjunction of a new formal solution to $L=0$ result in a new constant? Is this the reason for the no new constant condition in the definition for a Picard Vessiot extension?
differential-equations field-theory extension-field differential-algebra
asked Jul 19 at 20:02
J.Kennedy
486
I am reading the lecture series book Lectures on Differential Galois Theory by Andy Magid. I came across an example, and I am wondering if it generalizes.
The example
Let $F_1=mathbbC(z)$ be the field of rational functions in indeterminate $z$, and $F_2=mathbbC((z))$ the corresponding field of formal power series. Then with the derivation $D(z)=1$, and $D$ trivial on $mathbbC$, $F_2$ is a differential field, and $F_1$ a differential sub-field. Let $f$ be the usual exponential series, so that $D(f)=f$, and let $F=mathbbClangle f rangle$ be the field obtained by adjoining $f$ to $mathbbC$. Consider the differential equation $$Y'-Y=0$$ over $F$. The universal solution algebra is $R=F[y]$ with $D(y)=y$. This construction is superfluous since $F$ already contains the solutions to the differential equation. The existence of these two solutions gives the following equations on their ratios:
$$Dleft(fracyfright)=fracfD(y)-yD(f)f^2=fracfy-yff^2=0$$
In other words, the extra solution gives rise to a new constant.
My question
Suppose $L$ is a monic homogeneous linear differential operator of order $n$ over a differential field $F$ which contains $n$ solutions to $L=0$ linearly independent over constants. Does the adjunction of a new formal solution to $L=0$ result in a new constant? Is this the reason for the no new constant condition in the definition for a Picard Vessiot extension?
differential-equations field-theory extension-field differential-algebra
asked Jul 19 at 20:02
J.Kennedy
486
asked Jul 19 at 20:02
J.Kennedy
486
asked Jul 19 at 20:02
J.Kennedy
486
asked Jul 19 at 20:02
J.Kennedy
486
486
add a comment |Â
add a comment |Â
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