Why $mathbb{C}(f(t),g(t))=mathbb{C}(t)$ implies that $gcd(f(t)-a,g(t)-b)=t-c$, for some $a,b,c in mathbb{C}$?
$begingroup$
Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:
(1) $deg(f) geq 2$ and $deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
In this question it was mentioned that in that case, there exist
$a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.
Unfortunately, I do not see why this is true.
Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).
Edit: Just to make sure:
Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?
Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.
Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
W.l.o.g., there exist finitely many $a in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.
This is impossible from the following reason:
Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.
So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
Any hints are welcome!
polynomials commutative-algebra field-theory greatest-common-divisor
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
$endgroup$
add a comment |
$begingroup$
Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:
(1) $deg(f) geq 2$ and $deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
In this question it was mentioned that in that case, there exist
$a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.
Unfortunately, I do not see why this is true.
Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).
Edit: Just to make sure:
Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?
Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.
Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
W.l.o.g., there exist finitely many $a in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.
This is impossible from the following reason:
Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.
So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
Any hints are welcome!
polynomials commutative-algebra field-theory greatest-common-divisor
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
$endgroup$
add a comment |
$begingroup$
Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:
(1) $deg(f) geq 2$ and $deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
In this question it was mentioned that in that case, there exist
$a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.
Unfortunately, I do not see why this is true.
Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).
Edit: Just to make sure:
Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?
Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.
Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
W.l.o.g., there exist finitely many $a in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.
This is impossible from the following reason:
Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.
So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
Any hints are welcome!
polynomials commutative-algebra field-theory greatest-common-divisor
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
$endgroup$
Assume that $f(t),g(t) in mathbb{C}[t]$ satisfy the following two conditions:
(1) $deg(f) geq 2$ and $deg(g) geq 2$.
(2) $mathbb{C}(f(t),g(t))=mathbb{C}(t)$.
In this question it was mentioned that in that case, there exist
$a,b,c in mathbb{C}$ such that $gcd(f(t)-a,g(t)-b)=t-c$.
Unfortunately, I do not see why this is true.
Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).
Edit: Just to make sure:
Is it true that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c in mathbb{C}$?
Choose $c in mathbb{C}$ such that $f'(c) neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.
Asumme that there exist finitely many $a in mathbb{C}$ or finitely many $b in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
W.l.o.g., there exist finitely many $a in mathbb{C}$
such that $gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.
This is impossible from the following reason:
Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.
So after all, I think that I have proved that there exist infinitely many $a in mathbb{C}$ and infinitely many $b in mathbb{C}$ such that
$gcd(f(t)-a,g(t)-b)=t-c$, $c in mathbb{C}$.
Any hints are welcome!
polynomials commutative-algebra field-theory greatest-common-divisor
polynomials commutative-algebra field-theory greatest-common-divisor
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
edited Dec 24 '18 at 10:26
user237522
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
asked Dec 15 '18 at 22:08
user237522user237522
2,1631617
2,1631617
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We may assume that $f$ and $g$ are monic.
There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).
Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.
Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042016%2fwhy-mathbbcft-gt-mathbbct-implies-that-gcdft-a-gt-b-t-c%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We may assume that $f$ and $g$ are monic.
There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).
Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.
Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
|
show 1 more comment
$begingroup$
We may assume that $f$ and $g$ are monic.
There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).
Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.
Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
|
show 1 more comment
$begingroup$
We may assume that $f$ and $g$ are monic.
There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).
Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.
Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
$endgroup$
We may assume that $f$ and $g$ are monic.
There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).
Let $c$ be such that $f’(c) neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.
Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
edited Dec 15 '18 at 22:59
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
answered Dec 15 '18 at 22:47
MindlackMindlack
4,885211
4,885211
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
|
show 1 more comment
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q in mathbb{C}^{times}$?
$endgroup$
– user237522
Dec 15 '18 at 22:51
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation.
$endgroup$
– Mindlack
Dec 15 '18 at 22:54
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
Thanks; please, could you slightly elaborate your answer?
$endgroup$
– user237522
Dec 15 '18 at 22:55
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
I edited. Is there enough detail?
$endgroup$
– Mindlack
Dec 15 '18 at 22:59
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
$begingroup$
Thank you. I think I now understand your answer, very nice!
$endgroup$
– user237522
Dec 15 '18 at 23:07
|
show 1 more comment
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042016%2fwhy-mathbbcft-gt-mathbbct-implies-that-gcdft-a-gt-b-t-c%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
