Mixing
Divisibility of $A(x,y)u(x)+B(x,y)v(x)$ by $y$
$begingroup$
Let $k$ be a field of characteristic zero, $A=A(x,y),B=B(x,y) in k[x,y]$, $u=u(x), v=v(x) in k[x]$.
Assume that:
- At least one of ${u,v}$ has degree $geq 1$.
$u$ and $v$ do not have common roots.
$y$ divides $Au+Bv$.
What can be said about such $A,B,u,v$? Should $y$ divide both $A$ and $B$?
My (partial) solution:
Write $A=a_ny^n+cdots+a_1y+a_0$, $B=b_my^m+cdots+b_1y+b_0$, where $a_i,b_j in k[x]$. By the third assumption, $y$ divides $Au+Bv=(a_ny^n+cdots+a_1y+a_0)u+(b_my^m+cdots+b_1y+b_0)v=(a_ny^n+cdots+a_1y)u+(b_my^m+cdots+b_1y)v+(a_0u+b_0v)
$
This shows that $y$ divides $a_0u+b_0v in k[x]$, so $a_0u+b_0v=0$.
The second assumption says that $u$ and $v$ are relatively prime, so $b_0=u tilde{u}$ and $a_0=v tilde{v}$, for some $tilde{u},tilde{v} in k[x]$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y+ u tilde{u}$.
Moreover, from $a_0u+b_0v=0$, we get that $v tilde{v}u+u tilde{u}v=0$,
so $vu(tilde{v}+tilde{u})=0$, which implies that $tilde{u}=-tilde{v}$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y- u tilde{v}$.
For example, $A=B=y$, $u=x+1$, $v=x$, so $Au+Bv=y(x+1)+yx=y(2x+1)$.
Non-examples,
$A=y+1$, $B=y-1$, $u=x+1$, $v=x$, so $Au+Bv=(y+1)(x+1)+(y-1)x=xy+y+x+1+xy-x=
2xy+y+1$.
$A=y+1$, $B=y+1$, $u=x-1$, $v=x+1$, so $Au+Bv=(y+1)(x+1)+(y+1)(-x+1)=
xy+y+x+1-xy+y-x+1=2y+2$.
(See also this similar question).
Any hints and comments are welcome!
polynomials divisibility
asked Jan 6 at 11:40
user237522user237522
2,1541617
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field of characteristic zero, $A=A(x,y),B=B(x,y) in k[x,y]$, $u=u(x), v=v(x) in k[x]$.
Assume that:
- At least one of ${u,v}$ has degree $geq 1$.
$u$ and $v$ do not have common roots.
$y$ divides $Au+Bv$.
What can be said about such $A,B,u,v$? Should $y$ divide both $A$ and $B$?
My (partial) solution:
Write $A=a_ny^n+cdots+a_1y+a_0$, $B=b_my^m+cdots+b_1y+b_0$, where $a_i,b_j in k[x]$. By the third assumption, $y$ divides $Au+Bv=(a_ny^n+cdots+a_1y+a_0)u+(b_my^m+cdots+b_1y+b_0)v=(a_ny^n+cdots+a_1y)u+(b_my^m+cdots+b_1y)v+(a_0u+b_0v)
$
This shows that $y$ divides $a_0u+b_0v in k[x]$, so $a_0u+b_0v=0$.
The second assumption says that $u$ and $v$ are relatively prime, so $b_0=u tilde{u}$ and $a_0=v tilde{v}$, for some $tilde{u},tilde{v} in k[x]$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y+ u tilde{u}$.
Moreover, from $a_0u+b_0v=0$, we get that $v tilde{v}u+u tilde{u}v=0$,
so $vu(tilde{v}+tilde{u})=0$, which implies that $tilde{u}=-tilde{v}$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y- u tilde{v}$.
For example, $A=B=y$, $u=x+1$, $v=x$, so $Au+Bv=y(x+1)+yx=y(2x+1)$.
Non-examples,
$A=y+1$, $B=y-1$, $u=x+1$, $v=x$, so $Au+Bv=(y+1)(x+1)+(y-1)x=xy+y+x+1+xy-x=
2xy+y+1$.
$A=y+1$, $B=y+1$, $u=x-1$, $v=x+1$, so $Au+Bv=(y+1)(x+1)+(y+1)(-x+1)=
xy+y+x+1-xy+y-x+1=2y+2$.
(See also this similar question).
Any hints and comments are welcome!
polynomials divisibility
asked Jan 6 at 11:40
user237522user237522
2,1541617
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field of characteristic zero, $A=A(x,y),B=B(x,y) in k[x,y]$, $u=u(x), v=v(x) in k[x]$.
Assume that:
- At least one of ${u,v}$ has degree $geq 1$.
$u$ and $v$ do not have common roots.
$y$ divides $Au+Bv$.
What can be said about such $A,B,u,v$? Should $y$ divide both $A$ and $B$?
My (partial) solution:
Write $A=a_ny^n+cdots+a_1y+a_0$, $B=b_my^m+cdots+b_1y+b_0$, where $a_i,b_j in k[x]$. By the third assumption, $y$ divides $Au+Bv=(a_ny^n+cdots+a_1y+a_0)u+(b_my^m+cdots+b_1y+b_0)v=(a_ny^n+cdots+a_1y)u+(b_my^m+cdots+b_1y)v+(a_0u+b_0v)
$
This shows that $y$ divides $a_0u+b_0v in k[x]$, so $a_0u+b_0v=0$.
The second assumption says that $u$ and $v$ are relatively prime, so $b_0=u tilde{u}$ and $a_0=v tilde{v}$, for some $tilde{u},tilde{v} in k[x]$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y+ u tilde{u}$.
Moreover, from $a_0u+b_0v=0$, we get that $v tilde{v}u+u tilde{u}v=0$,
so $vu(tilde{v}+tilde{u})=0$, which implies that $tilde{u}=-tilde{v}$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y- u tilde{v}$.
For example, $A=B=y$, $u=x+1$, $v=x$, so $Au+Bv=y(x+1)+yx=y(2x+1)$.
Non-examples,
$A=y+1$, $B=y-1$, $u=x+1$, $v=x$, so $Au+Bv=(y+1)(x+1)+(y-1)x=xy+y+x+1+xy-x=
2xy+y+1$.
$A=y+1$, $B=y+1$, $u=x-1$, $v=x+1$, so $Au+Bv=(y+1)(x+1)+(y+1)(-x+1)=
xy+y+x+1-xy+y-x+1=2y+2$.
(See also this similar question).
Any hints and comments are welcome!
polynomials divisibility
asked Jan 6 at 11:40
user237522user237522
2,1541617
$endgroup$
Let $k$ be a field of characteristic zero, $A=A(x,y),B=B(x,y) in k[x,y]$, $u=u(x), v=v(x) in k[x]$.
Assume that:
- At least one of ${u,v}$ has degree $geq 1$.
$u$ and $v$ do not have common roots.
$y$ divides $Au+Bv$.
What can be said about such $A,B,u,v$? Should $y$ divide both $A$ and $B$?
My (partial) solution:
Write $A=a_ny^n+cdots+a_1y+a_0$, $B=b_my^m+cdots+b_1y+b_0$, where $a_i,b_j in k[x]$. By the third assumption, $y$ divides $Au+Bv=(a_ny^n+cdots+a_1y+a_0)u+(b_my^m+cdots+b_1y+b_0)v=(a_ny^n+cdots+a_1y)u+(b_my^m+cdots+b_1y)v+(a_0u+b_0v)
$
This shows that $y$ divides $a_0u+b_0v in k[x]$, so $a_0u+b_0v=0$.
The second assumption says that $u$ and $v$ are relatively prime, so $b_0=u tilde{u}$ and $a_0=v tilde{v}$, for some $tilde{u},tilde{v} in k[x]$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y+ u tilde{u}$.
Moreover, from $a_0u+b_0v=0$, we get that $v tilde{v}u+u tilde{u}v=0$,
so $vu(tilde{v}+tilde{u})=0$, which implies that $tilde{u}=-tilde{v}$.
Therefore, $A=a_ny^n+cdots+a_1y+ v tilde{v}$ and $B=b_my^m+cdots+b_1y- u tilde{v}$.
For example, $A=B=y$, $u=x+1$, $v=x$, so $Au+Bv=y(x+1)+yx=y(2x+1)$.
Non-examples,
$A=y+1$, $B=y-1$, $u=x+1$, $v=x$, so $Au+Bv=(y+1)(x+1)+(y-1)x=xy+y+x+1+xy-x=
2xy+y+1$.
$A=y+1$, $B=y+1$, $u=x-1$, $v=x+1$, so $Au+Bv=(y+1)(x+1)+(y+1)(-x+1)=
xy+y+x+1-xy+y-x+1=2y+2$.
(See also this similar question).
Any hints and comments are welcome!
polynomials divisibility
polynomials divisibility
asked Jan 6 at 11:40
user237522user237522
2,1541617
asked Jan 6 at 11:40
user237522user237522
2,1541617
edited Jan 6 at 21:30
user237522
asked Jan 6 at 11:40
user237522user237522
2,1541617
asked Jan 6 at 11:40
user237522user237522
2,1541617
asked Jan 6 at 11:40
user237522user237522
2,1541617
2,1541617
add a comment |
add a comment |
0
active
oldest
votes
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%2f3063751%2fdivisibility-of-ax-yuxbx-yvx-by-y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3063751%2fdivisibility-of-ax-yuxbx-yvx-by-y%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
