Mixing
Image of polynomial in $F[x]$ under derivation
2
$begingroup$
Let $F$ be a field of characteristic zero and let $D$ be the formal polynomial differentiation map so that
$$D(a_0+a_1x+a_2x^2+....+a_nx^n)=a_1+2a_2x+3a_3x^2+....+na_nx^{n-1}$$
Find the image of $F[x]$ under $D$.
The answer is that the image of $F[x]$ under $D$ is $F[x]$, Im$(D)$
I understand as far as Im$(D)<F[x]$ (i.e., Im$(D)$ is a subring of $F[x]$) but I don't see why
Im$(D)=F[x]$. Can you guys please help?
abstract-algebra group-theory field-theory
edited Jan 22 at 21:17
darij grinberg
11.1k33167
asked Jan 22 at 13:51
MathQMathQ
112
$endgroup$
add a comment |
2
$begingroup$
Let $F$ be a field of characteristic zero and let $D$ be the formal polynomial differentiation map so that
$$D(a_0+a_1x+a_2x^2+....+a_nx^n)=a_1+2a_2x+3a_3x^2+....+na_nx^{n-1}$$
Find the image of $F[x]$ under $D$.
The answer is that the image of $F[x]$ under $D$ is $F[x]$, Im$(D)$
I understand as far as Im$(D)<F[x]$ (i.e., Im$(D)$ is a subring of $F[x]$) but I don't see why
Im$(D)=F[x]$. Can you guys please help?
abstract-algebra group-theory field-theory
edited Jan 22 at 21:17
darij grinberg
11.1k33167
asked Jan 22 at 13:51
MathQMathQ
112
$endgroup$
4
$begingroup$
What is your question? Of course, every given polynomial is a derivation of some polynomial. Try with the monomials first.
$endgroup$
– Dietrich Burde
Jan 22 at 13:52
1
$begingroup$
Let $p = p_0 + p_1 x + p_2 x^2 + dots + p_k x^k$. Assume that $D(a_0 + a_1x + a_2 x^2 + dots + a_n x^n) = p$, substitute the definition of $D$, and compare corresponding powers of $x$.
$endgroup$
– lisyarus
Jan 22 at 15:08
1
$begingroup$
I think the main thing to observe and use is that $F$ contains (an isomorphic copy of) the rational numbers as its prime subfield, so you can divide by non-zero integers in $F$.
$endgroup$
– Lukas Geyer
Jan 22 at 21:06
1
$begingroup$
Also, should $F$ have a positive characteristic, then $Im(D)$ will fail to be a subring. Therefore that claim may be less clear than you think in characteristic zero as well.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:27
$begingroup$
Can you also formally integrate a polynomial in this case?
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:28
add a comment |
2
2
2
1
$begingroup$
Let $F$ be a field of characteristic zero and let $D$ be the formal polynomial differentiation map so that
$$D(a_0+a_1x+a_2x^2+....+a_nx^n)=a_1+2a_2x+3a_3x^2+....+na_nx^{n-1}$$
Find the image of $F[x]$ under $D$.
The answer is that the image of $F[x]$ under $D$ is $F[x]$, Im$(D)$
I understand as far as Im$(D)<F[x]$ (i.e., Im$(D)$ is a subring of $F[x]$) but I don't see why
Im$(D)=F[x]$. Can you guys please help?
abstract-algebra group-theory field-theory
edited Jan 22 at 21:17
darij grinberg
11.1k33167
asked Jan 22 at 13:51
MathQMathQ
112
$endgroup$
Let $F$ be a field of characteristic zero and let $D$ be the formal polynomial differentiation map so that
$$D(a_0+a_1x+a_2x^2+....+a_nx^n)=a_1+2a_2x+3a_3x^2+....+na_nx^{n-1}$$
Find the image of $F[x]$ under $D$.
The answer is that the image of $F[x]$ under $D$ is $F[x]$, Im$(D)$
I understand as far as Im$(D)<F[x]$ (i.e., Im$(D)$ is a subring of $F[x]$) but I don't see why
Im$(D)=F[x]$. Can you guys please help?
abstract-algebra group-theory field-theory
abstract-algebra group-theory field-theory
edited Jan 22 at 21:17
darij grinberg
11.1k33167
asked Jan 22 at 13:51
MathQMathQ
112
edited Jan 22 at 21:17
darij grinberg
11.1k33167
asked Jan 22 at 13:51
MathQMathQ
112
edited Jan 22 at 21:17
darij grinberg
11.1k33167
edited Jan 22 at 21:17
darij grinberg
11.1k33167
edited Jan 22 at 21:17
darij grinberg
11.1k33167
11.1k33167
asked Jan 22 at 13:51
MathQMathQ
112
asked Jan 22 at 13:51
MathQMathQ
112
asked Jan 22 at 13:51
MathQMathQ
112
112
4
$begingroup$
What is your question? Of course, every given polynomial is a derivation of some polynomial. Try with the monomials first.
$endgroup$
– Dietrich Burde
Jan 22 at 13:52
1
$begingroup$
Let $p = p_0 + p_1 x + p_2 x^2 + dots + p_k x^k$. Assume that $D(a_0 + a_1x + a_2 x^2 + dots + a_n x^n) = p$, substitute the definition of $D$, and compare corresponding powers of $x$.
$endgroup$
– lisyarus
Jan 22 at 15:08
1
$begingroup$
I think the main thing to observe and use is that $F$ contains (an isomorphic copy of) the rational numbers as its prime subfield, so you can divide by non-zero integers in $F$.
$endgroup$
– Lukas Geyer
Jan 22 at 21:06
1
$begingroup$
Also, should $F$ have a positive characteristic, then $Im(D)$ will fail to be a subring. Therefore that claim may be less clear than you think in characteristic zero as well.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:27
$begingroup$
Can you also formally integrate a polynomial in this case?
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:28
add a comment |
4
$begingroup$
What is your question? Of course, every given polynomial is a derivation of some polynomial. Try with the monomials first.
$endgroup$
– Dietrich Burde
Jan 22 at 13:52
1
$begingroup$
Let $p = p_0 + p_1 x + p_2 x^2 + dots + p_k x^k$. Assume that $D(a_0 + a_1x + a_2 x^2 + dots + a_n x^n) = p$, substitute the definition of $D$, and compare corresponding powers of $x$.
$endgroup$
– lisyarus
Jan 22 at 15:08
1
$begingroup$
I think the main thing to observe and use is that $F$ contains (an isomorphic copy of) the rational numbers as its prime subfield, so you can divide by non-zero integers in $F$.
$endgroup$
– Lukas Geyer
Jan 22 at 21:06
1
$begingroup$
Also, should $F$ have a positive characteristic, then $Im(D)$ will fail to be a subring. Therefore that claim may be less clear than you think in characteristic zero as well.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:27
$begingroup$
Can you also formally integrate a polynomial in this case?
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:28
4
4
$begingroup$
What is your question? Of course, every given polynomial is a derivation of some polynomial. Try with the monomials first.
$endgroup$
– Dietrich Burde
Jan 22 at 13:52
$begingroup$
What is your question? Of course, every given polynomial is a derivation of some polynomial. Try with the monomials first.
$endgroup$
– Dietrich Burde
Jan 22 at 13:52
1
1
$begingroup$
Let $p = p_0 + p_1 x + p_2 x^2 + dots + p_k x^k$. Assume that $D(a_0 + a_1x + a_2 x^2 + dots + a_n x^n) = p$, substitute the definition of $D$, and compare corresponding powers of $x$.
$endgroup$
– lisyarus
Jan 22 at 15:08
$begingroup$
Let $p = p_0 + p_1 x + p_2 x^2 + dots + p_k x^k$. Assume that $D(a_0 + a_1x + a_2 x^2 + dots + a_n x^n) = p$, substitute the definition of $D$, and compare corresponding powers of $x$.
$endgroup$
– lisyarus
Jan 22 at 15:08
1
1
$begingroup$
I think the main thing to observe and use is that $F$ contains (an isomorphic copy of) the rational numbers as its prime subfield, so you can divide by non-zero integers in $F$.
$endgroup$
– Lukas Geyer
Jan 22 at 21:06
$begingroup$
I think the main thing to observe and use is that $F$ contains (an isomorphic copy of) the rational numbers as its prime subfield, so you can divide by non-zero integers in $F$.
$endgroup$
– Lukas Geyer
Jan 22 at 21:06
1
1
$begingroup$
Also, should $F$ have a positive characteristic, then $Im(D)$ will fail to be a subring. Therefore that claim may be less clear than you think in characteristic zero as well.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:27
$begingroup$
Also, should $F$ have a positive characteristic, then $Im(D)$ will fail to be a subring. Therefore that claim may be less clear than you think in characteristic zero as well.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:27
$begingroup$
Can you also formally integrate a polynomial in this case?
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:28
$begingroup$
Can you also formally integrate a polynomial in this case?
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:28
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Hint:
$$
DBigl(frac{x^{n+1}}{n+1}Bigr)=
$$
Further hint: as soon as you have proved that $x^ninoperatorname{Im}(D)$, for every $nge0$, you can apply the fact that $D$ is linear.
answered Jan 22 at 21:54
egregegreg
184k1486205
$endgroup$
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
0
$begingroup$
Hint:
$$
DBigl(frac{x^{n+1}}{n+1}Bigr)=
$$
Further hint: as soon as you have proved that $x^ninoperatorname{Im}(D)$, for every $nge0$, you can apply the fact that $D$ is linear.
answered Jan 22 at 21:54
egregegreg
184k1486205
$endgroup$
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
add a comment |
0
$begingroup$
Hint:
$$
DBigl(frac{x^{n+1}}{n+1}Bigr)=
$$
Further hint: as soon as you have proved that $x^ninoperatorname{Im}(D)$, for every $nge0$, you can apply the fact that $D$ is linear.
answered Jan 22 at 21:54
egregegreg
184k1486205
$endgroup$
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
add a comment |
0
0
0
$begingroup$
Hint:
$$
DBigl(frac{x^{n+1}}{n+1}Bigr)=
$$
Further hint: as soon as you have proved that $x^ninoperatorname{Im}(D)$, for every $nge0$, you can apply the fact that $D$ is linear.
answered Jan 22 at 21:54
egregegreg
184k1486205
$endgroup$
Hint:
$$
DBigl(frac{x^{n+1}}{n+1}Bigr)=
$$
Further hint: as soon as you have proved that $x^ninoperatorname{Im}(D)$, for every $nge0$, you can apply the fact that $D$ is linear.
answered Jan 22 at 21:54
egregegreg
184k1486205
edited Jan 23 at 10:05
answered Jan 22 at 21:54
egregegreg
184k1486205
answered Jan 22 at 21:54
egregegreg
184k1486205
answered Jan 22 at 21:54
egregegreg
184k1486205
184k1486205
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
add a comment |
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
...equals x^n. And D(1)=0. But why would that imply that Im(D)=F[x]?
$endgroup$
– MathQ
Jan 23 at 6:16
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
@MathQ The map $D$ is linear and the polynomials $x^n$ (for $nge0$) span $F[x]$ as an $F$-vector space.
$endgroup$
– egreg
Jan 23 at 7:46
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
(1,x,x^2,...,x^n) spans F[x] but (1,x,x^2,...,x^(n-1) span Im(D). For me Im(D) is a subring of lower dimension and not equal to F[x]. i still dont get it :(
$endgroup$
– MathQ
Jan 23 at 10:02
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
$begingroup$
@MathQ There's no limitation on the degree in either case.
$endgroup$
– egreg
Jan 23 at 10:03
add a comment |
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4
$begingroup$
What is your question? Of course, every given polynomial is a derivation of some polynomial. Try with the monomials first.
$endgroup$
– Dietrich Burde
Jan 22 at 13:52
1
$begingroup$
Let $p = p_0 + p_1 x + p_2 x^2 + dots + p_k x^k$. Assume that $D(a_0 + a_1x + a_2 x^2 + dots + a_n x^n) = p$, substitute the definition of $D$, and compare corresponding powers of $x$.
$endgroup$
– lisyarus
Jan 22 at 15:08
1
$begingroup$
I think the main thing to observe and use is that $F$ contains (an isomorphic copy of) the rational numbers as its prime subfield, so you can divide by non-zero integers in $F$.
$endgroup$
– Lukas Geyer
Jan 22 at 21:06
1
$begingroup$
Also, should $F$ have a positive characteristic, then $Im(D)$ will fail to be a subring. Therefore that claim may be less clear than you think in characteristic zero as well.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:27
$begingroup$
Can you also formally integrate a polynomial in this case?
$endgroup$
– Jyrki Lahtonen
Jan 22 at 21:28