When $mathbb{C}(f(t),g(t))=mathbb{C}(t)$ implies that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$?
$begingroup$
Let $f=f(t), g=g(t) in mathbb{C}[t]$.
Assume that:
(1) $deg(f) geq 2,deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
(3) The set of zeros of $f$, $S_f$, equals the set of zeros of $g$,
$S_g$; denote $S_f=S_g={c_1,ldots,c_m}$.
(4) $langle f'(t),g'(t) rangle = mathbb{C}[t]$ (namely, the ideal generated by $f'$ and $g'$ is the unit ideal).
Now, (4) implies that $f'$ and $g'$ do not have a common zero (there is no need to apply Hilbert's weak Nullstellensatz), so from (3) we obtain that:
$f(t)=lambda(t-c_1)^{d_1}cdots(t-c_l)^{d_l}(t-c_{l+1})^{d_{l+1}}cdots(t-c_m)^{d_m}$,
$g(t)=mu(t-c_1)^{e_1}cdots(t-c_l)^{e_l}(t-c_{l+1})^{e_{l+1}}cdots(t-c_m)^{e_m}$,
where $lambda, mu in mathbb{C}^{times}$,
and $d_i,e_i geq 1$ are such that if $d_i geq 2$ then $e_i=1$ and vice versa (= if $e_i geq 2$ then $d_i=1$).
The reason is as follows: If, for example, $d_1 geq 2$ then $c_1$ is also a zero of $f'(t)$, and if $e_1 geq 2$ then $c_1$ would also be a zero of $g'(t)$, but then $c_1$ would be a common zero of $f'(t)$ and $g'(t)$, contrary to assumption (4).
Is it true that necessarily $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? Or is there a counterexample?
Notice that $f(t)=t^2$, $g(t)=t^3$ is not a counterexample, since conditions (1), (2), (3) are satisfied, but condition (4) is not satisfied.
See also this question.
Edit: After receivig a counterexample, do you think that there exists an additional (not too strong) condition that will guarantee that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? There is a nice condition which says that $f'(t),g'(t) in mathbb{C}[f(t),g(t)]$, so the fifth condition, together with the former four conditions, should imply that $f'(t),g'(t) in mathbb{C}[f,g]$.
A 'plausible' fifth condition: (5) $f(t)$ and $g(t)$ are separable polynomials.
In that case we must have: $d_i=e_i=1$ for every $1 leq i leq m$. Therefore, $f(t)=g(t)$, so from (2) we get that $mathbb{C}(f(t))= mathbb{C}(f(t),g(t))=mathbb{C}(t)$, hence $deg(f)=1$ which contradicts condition (1)... Therefore, if we omit condition (1) and add condition (5), we obtain that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$.
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
$endgroup$
add a comment |
$begingroup$
Let $f=f(t), g=g(t) in mathbb{C}[t]$.
Assume that:
(1) $deg(f) geq 2,deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
(3) The set of zeros of $f$, $S_f$, equals the set of zeros of $g$,
$S_g$; denote $S_f=S_g={c_1,ldots,c_m}$.
(4) $langle f'(t),g'(t) rangle = mathbb{C}[t]$ (namely, the ideal generated by $f'$ and $g'$ is the unit ideal).
Now, (4) implies that $f'$ and $g'$ do not have a common zero (there is no need to apply Hilbert's weak Nullstellensatz), so from (3) we obtain that:
$f(t)=lambda(t-c_1)^{d_1}cdots(t-c_l)^{d_l}(t-c_{l+1})^{d_{l+1}}cdots(t-c_m)^{d_m}$,
$g(t)=mu(t-c_1)^{e_1}cdots(t-c_l)^{e_l}(t-c_{l+1})^{e_{l+1}}cdots(t-c_m)^{e_m}$,
where $lambda, mu in mathbb{C}^{times}$,
and $d_i,e_i geq 1$ are such that if $d_i geq 2$ then $e_i=1$ and vice versa (= if $e_i geq 2$ then $d_i=1$).
The reason is as follows: If, for example, $d_1 geq 2$ then $c_1$ is also a zero of $f'(t)$, and if $e_1 geq 2$ then $c_1$ would also be a zero of $g'(t)$, but then $c_1$ would be a common zero of $f'(t)$ and $g'(t)$, contrary to assumption (4).
Is it true that necessarily $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? Or is there a counterexample?
Notice that $f(t)=t^2$, $g(t)=t^3$ is not a counterexample, since conditions (1), (2), (3) are satisfied, but condition (4) is not satisfied.
See also this question.
Edit: After receivig a counterexample, do you think that there exists an additional (not too strong) condition that will guarantee that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? There is a nice condition which says that $f'(t),g'(t) in mathbb{C}[f(t),g(t)]$, so the fifth condition, together with the former four conditions, should imply that $f'(t),g'(t) in mathbb{C}[f,g]$.
A 'plausible' fifth condition: (5) $f(t)$ and $g(t)$ are separable polynomials.
In that case we must have: $d_i=e_i=1$ for every $1 leq i leq m$. Therefore, $f(t)=g(t)$, so from (2) we get that $mathbb{C}(f(t))= mathbb{C}(f(t),g(t))=mathbb{C}(t)$, hence $deg(f)=1$ which contradicts condition (1)... Therefore, if we omit condition (1) and add condition (5), we obtain that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$.
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
$endgroup$
3
$begingroup$
$f(t) = (t+1)^2t,g(t) = t^2(t+1)$ ?
$endgroup$
– reuns
Dec 23 '18 at 1:29
$begingroup$
Thank you. How to show that $mathbb{C}(f,g)=mathbb{C}(t)$? Should I apply math.stackexchange.com/questions/2803632/…?
$endgroup$
– user237522
Dec 23 '18 at 1:39
2
$begingroup$
$f(t)/g(t) = ?$
$endgroup$
– reuns
Dec 23 '18 at 1:45
$begingroup$
Nice; easier than I thought. (The argument that I have suggested also works, but your argument is easier). Please, do you think that there is an additional condition that necessarily implies that $mathbb{C}[f,g]=mathbb{C}[t]$?
$endgroup$
– user237522
Dec 23 '18 at 1:47
add a comment |
$begingroup$
Let $f=f(t), g=g(t) in mathbb{C}[t]$.
Assume that:
(1) $deg(f) geq 2,deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
(3) The set of zeros of $f$, $S_f$, equals the set of zeros of $g$,
$S_g$; denote $S_f=S_g={c_1,ldots,c_m}$.
(4) $langle f'(t),g'(t) rangle = mathbb{C}[t]$ (namely, the ideal generated by $f'$ and $g'$ is the unit ideal).
Now, (4) implies that $f'$ and $g'$ do not have a common zero (there is no need to apply Hilbert's weak Nullstellensatz), so from (3) we obtain that:
$f(t)=lambda(t-c_1)^{d_1}cdots(t-c_l)^{d_l}(t-c_{l+1})^{d_{l+1}}cdots(t-c_m)^{d_m}$,
$g(t)=mu(t-c_1)^{e_1}cdots(t-c_l)^{e_l}(t-c_{l+1})^{e_{l+1}}cdots(t-c_m)^{e_m}$,
where $lambda, mu in mathbb{C}^{times}$,
and $d_i,e_i geq 1$ are such that if $d_i geq 2$ then $e_i=1$ and vice versa (= if $e_i geq 2$ then $d_i=1$).
The reason is as follows: If, for example, $d_1 geq 2$ then $c_1$ is also a zero of $f'(t)$, and if $e_1 geq 2$ then $c_1$ would also be a zero of $g'(t)$, but then $c_1$ would be a common zero of $f'(t)$ and $g'(t)$, contrary to assumption (4).
Is it true that necessarily $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? Or is there a counterexample?
Notice that $f(t)=t^2$, $g(t)=t^3$ is not a counterexample, since conditions (1), (2), (3) are satisfied, but condition (4) is not satisfied.
See also this question.
Edit: After receivig a counterexample, do you think that there exists an additional (not too strong) condition that will guarantee that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? There is a nice condition which says that $f'(t),g'(t) in mathbb{C}[f(t),g(t)]$, so the fifth condition, together with the former four conditions, should imply that $f'(t),g'(t) in mathbb{C}[f,g]$.
A 'plausible' fifth condition: (5) $f(t)$ and $g(t)$ are separable polynomials.
In that case we must have: $d_i=e_i=1$ for every $1 leq i leq m$. Therefore, $f(t)=g(t)$, so from (2) we get that $mathbb{C}(f(t))= mathbb{C}(f(t),g(t))=mathbb{C}(t)$, hence $deg(f)=1$ which contradicts condition (1)... Therefore, if we omit condition (1) and add condition (5), we obtain that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$.
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
$endgroup$
Let $f=f(t), g=g(t) in mathbb{C}[t]$.
Assume that:
(1) $deg(f) geq 2,deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
(3) The set of zeros of $f$, $S_f$, equals the set of zeros of $g$,
$S_g$; denote $S_f=S_g={c_1,ldots,c_m}$.
(4) $langle f'(t),g'(t) rangle = mathbb{C}[t]$ (namely, the ideal generated by $f'$ and $g'$ is the unit ideal).
Now, (4) implies that $f'$ and $g'$ do not have a common zero (there is no need to apply Hilbert's weak Nullstellensatz), so from (3) we obtain that:
$f(t)=lambda(t-c_1)^{d_1}cdots(t-c_l)^{d_l}(t-c_{l+1})^{d_{l+1}}cdots(t-c_m)^{d_m}$,
$g(t)=mu(t-c_1)^{e_1}cdots(t-c_l)^{e_l}(t-c_{l+1})^{e_{l+1}}cdots(t-c_m)^{e_m}$,
where $lambda, mu in mathbb{C}^{times}$,
and $d_i,e_i geq 1$ are such that if $d_i geq 2$ then $e_i=1$ and vice versa (= if $e_i geq 2$ then $d_i=1$).
The reason is as follows: If, for example, $d_1 geq 2$ then $c_1$ is also a zero of $f'(t)$, and if $e_1 geq 2$ then $c_1$ would also be a zero of $g'(t)$, but then $c_1$ would be a common zero of $f'(t)$ and $g'(t)$, contrary to assumption (4).
Is it true that necessarily $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? Or is there a counterexample?
Notice that $f(t)=t^2$, $g(t)=t^3$ is not a counterexample, since conditions (1), (2), (3) are satisfied, but condition (4) is not satisfied.
See also this question.
Edit: After receivig a counterexample, do you think that there exists an additional (not too strong) condition that will guarantee that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$? There is a nice condition which says that $f'(t),g'(t) in mathbb{C}[f(t),g(t)]$, so the fifth condition, together with the former four conditions, should imply that $f'(t),g'(t) in mathbb{C}[f,g]$.
A 'plausible' fifth condition: (5) $f(t)$ and $g(t)$ are separable polynomials.
In that case we must have: $d_i=e_i=1$ for every $1 leq i leq m$. Therefore, $f(t)=g(t)$, so from (2) we get that $mathbb{C}(f(t))= mathbb{C}(f(t),g(t))=mathbb{C}(t)$, hence $deg(f)=1$ which contradicts condition (1)... Therefore, if we omit condition (1) and add condition (5), we obtain that $mathbb{C}[f(t),g(t)]=mathbb{C}[t]$.
Thank you very much!
algebraic-geometry polynomials commutative-algebra
algebraic-geometry polynomials commutative-algebra
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
edited Dec 23 '18 at 2:26
user237522
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
asked Dec 23 '18 at 1:07
user237522user237522
2,1651617
2,1651617
3
$begingroup$
$f(t) = (t+1)^2t,g(t) = t^2(t+1)$ ?
$endgroup$
– reuns
Dec 23 '18 at 1:29
$begingroup$
Thank you. How to show that $mathbb{C}(f,g)=mathbb{C}(t)$? Should I apply math.stackexchange.com/questions/2803632/…?
$endgroup$
– user237522
Dec 23 '18 at 1:39
2
$begingroup$
$f(t)/g(t) = ?$
$endgroup$
– reuns
Dec 23 '18 at 1:45
$begingroup$
Nice; easier than I thought. (The argument that I have suggested also works, but your argument is easier). Please, do you think that there is an additional condition that necessarily implies that $mathbb{C}[f,g]=mathbb{C}[t]$?
$endgroup$
– user237522
Dec 23 '18 at 1:47
add a comment |
3
$begingroup$
$f(t) = (t+1)^2t,g(t) = t^2(t+1)$ ?
$endgroup$
– reuns
Dec 23 '18 at 1:29
$begingroup$
Thank you. How to show that $mathbb{C}(f,g)=mathbb{C}(t)$? Should I apply math.stackexchange.com/questions/2803632/…?
$endgroup$
– user237522
Dec 23 '18 at 1:39
2
$begingroup$
$f(t)/g(t) = ?$
$endgroup$
– reuns
Dec 23 '18 at 1:45
$begingroup$
Nice; easier than I thought. (The argument that I have suggested also works, but your argument is easier). Please, do you think that there is an additional condition that necessarily implies that $mathbb{C}[f,g]=mathbb{C}[t]$?
$endgroup$
– user237522
Dec 23 '18 at 1:47
3
3
$begingroup$
$f(t) = (t+1)^2t,g(t) = t^2(t+1)$ ?
$endgroup$
– reuns
Dec 23 '18 at 1:29
$begingroup$
$f(t) = (t+1)^2t,g(t) = t^2(t+1)$ ?
$endgroup$
– reuns
Dec 23 '18 at 1:29
$begingroup$
Thank you. How to show that $mathbb{C}(f,g)=mathbb{C}(t)$? Should I apply math.stackexchange.com/questions/2803632/…?
$endgroup$
– user237522
Dec 23 '18 at 1:39
$begingroup$
Thank you. How to show that $mathbb{C}(f,g)=mathbb{C}(t)$? Should I apply math.stackexchange.com/questions/2803632/…?
$endgroup$
– user237522
Dec 23 '18 at 1:39
2
2
$begingroup$
$f(t)/g(t) = ?$
$endgroup$
– reuns
Dec 23 '18 at 1:45
$begingroup$
$f(t)/g(t) = ?$
$endgroup$
– reuns
Dec 23 '18 at 1:45
$begingroup$
Nice; easier than I thought. (The argument that I have suggested also works, but your argument is easier). Please, do you think that there is an additional condition that necessarily implies that $mathbb{C}[f,g]=mathbb{C}[t]$?
$endgroup$
– user237522
Dec 23 '18 at 1:47
$begingroup$
Nice; easier than I thought. (The argument that I have suggested also works, but your argument is easier). Please, do you think that there is an additional condition that necessarily implies that $mathbb{C}[f,g]=mathbb{C}[t]$?
$endgroup$
– user237522
Dec 23 '18 at 1:47
add a comment |
0
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3
$begingroup$
$f(t) = (t+1)^2t,g(t) = t^2(t+1)$ ?
$endgroup$
– reuns
Dec 23 '18 at 1:29
$begingroup$
Thank you. How to show that $mathbb{C}(f,g)=mathbb{C}(t)$? Should I apply math.stackexchange.com/questions/2803632/…?
$endgroup$
– user237522
Dec 23 '18 at 1:39
2
$begingroup$
$f(t)/g(t) = ?$
$endgroup$
– reuns
Dec 23 '18 at 1:45
$begingroup$
Nice; easier than I thought. (The argument that I have suggested also works, but your argument is easier). Please, do you think that there is an additional condition that necessarily implies that $mathbb{C}[f,g]=mathbb{C}[t]$?
$endgroup$
– user237522
Dec 23 '18 at 1:47