Mixing
Equality of derivation on prime ring
Suppose that $R$ is a prime ring and charR $neq2$.
$d_{1},d_{2}$ are derivations of $R$ such that for some non-zero ideal of $I$ of $R$ and some $cin C$ ($C$: extended centroid of $R$).
$d_{1}d_{2}(x)=cx$ for all $x in I$.
By computing in different ways $d_{1}d_{2}(xty)$ for $x,t,yin I$, making use of the facts that $d_{i}$'s are derivations and $d_{1}d_{2}(x)=cx$ for $xin I$ and some $cin C$, one can easily obtain $$0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$$ My question is how can we show that $hspace{0.1cm}$ $0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$?
$textbf{My attempt:}$
begin{align}
begin{aligned}
d_1d_2(xty) =& d_1[d_2(x)ty+xd_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(ty)]+d_1(x)[d_2(ty)]+xd_1[d_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(t)y+td_1(y)]+d_1(x)[d_2(t)y+td_2(y)]+xd_1[d_2(t)y+td_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+d_1d_2(x)ty+d_2(x)d_1(t)y +
x[d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+3cxyt+[d_2(x)d_1(t)+d_1(x)d_2(t)]y+x[d_2(t)d_1(y)+d_1(t)d_2(y)]
end{aligned}
end{align}
Now,$cty=d_1d_2(ty)=d_1[d_2(t)y+td_2(y)]=d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)$.
By hypothesis $cty=cty+d_2(t)d_1(y)+d_1(t)d_2(y)+cty$
Hence, we obtain $d_2(t)d_1(y)=-cty-d_1(t)d_2(y)$ and $d_2(x)d_1(t)=-ctx-d_1(x)d_2(t)$.
ring-theory noncommutative-algebra
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
add a comment |
Suppose that $R$ is a prime ring and charR $neq2$.
$d_{1},d_{2}$ are derivations of $R$ such that for some non-zero ideal of $I$ of $R$ and some $cin C$ ($C$: extended centroid of $R$).
$d_{1}d_{2}(x)=cx$ for all $x in I$.
By computing in different ways $d_{1}d_{2}(xty)$ for $x,t,yin I$, making use of the facts that $d_{i}$'s are derivations and $d_{1}d_{2}(x)=cx$ for $xin I$ and some $cin C$, one can easily obtain $$0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$$ My question is how can we show that $hspace{0.1cm}$ $0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$?
$textbf{My attempt:}$
begin{align}
begin{aligned}
d_1d_2(xty) =& d_1[d_2(x)ty+xd_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(ty)]+d_1(x)[d_2(ty)]+xd_1[d_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(t)y+td_1(y)]+d_1(x)[d_2(t)y+td_2(y)]+xd_1[d_2(t)y+td_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+d_1d_2(x)ty+d_2(x)d_1(t)y +
x[d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+3cxyt+[d_2(x)d_1(t)+d_1(x)d_2(t)]y+x[d_2(t)d_1(y)+d_1(t)d_2(y)]
end{aligned}
end{align}
Now,$cty=d_1d_2(ty)=d_1[d_2(t)y+td_2(y)]=d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)$.
By hypothesis $cty=cty+d_2(t)d_1(y)+d_1(t)d_2(y)+cty$
Hence, we obtain $d_2(t)d_1(y)=-cty-d_1(t)d_2(y)$ and $d_2(x)d_1(t)=-ctx-d_1(x)d_2(t)$.
ring-theory noncommutative-algebra
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
add a comment |
Suppose that $R$ is a prime ring and charR $neq2$.
$d_{1},d_{2}$ are derivations of $R$ such that for some non-zero ideal of $I$ of $R$ and some $cin C$ ($C$: extended centroid of $R$).
$d_{1}d_{2}(x)=cx$ for all $x in I$.
By computing in different ways $d_{1}d_{2}(xty)$ for $x,t,yin I$, making use of the facts that $d_{i}$'s are derivations and $d_{1}d_{2}(x)=cx$ for $xin I$ and some $cin C$, one can easily obtain $$0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$$ My question is how can we show that $hspace{0.1cm}$ $0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$?
$textbf{My attempt:}$
begin{align}
begin{aligned}
d_1d_2(xty) =& d_1[d_2(x)ty+xd_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(ty)]+d_1(x)[d_2(ty)]+xd_1[d_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(t)y+td_1(y)]+d_1(x)[d_2(t)y+td_2(y)]+xd_1[d_2(t)y+td_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+d_1d_2(x)ty+d_2(x)d_1(t)y +
x[d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+3cxyt+[d_2(x)d_1(t)+d_1(x)d_2(t)]y+x[d_2(t)d_1(y)+d_1(t)d_2(y)]
end{aligned}
end{align}
Now,$cty=d_1d_2(ty)=d_1[d_2(t)y+td_2(y)]=d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)$.
By hypothesis $cty=cty+d_2(t)d_1(y)+d_1(t)d_2(y)+cty$
Hence, we obtain $d_2(t)d_1(y)=-cty-d_1(t)d_2(y)$ and $d_2(x)d_1(t)=-ctx-d_1(x)d_2(t)$.
ring-theory noncommutative-algebra
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
Suppose that $R$ is a prime ring and charR $neq2$.
$d_{1},d_{2}$ are derivations of $R$ such that for some non-zero ideal of $I$ of $R$ and some $cin C$ ($C$: extended centroid of $R$).
$d_{1}d_{2}(x)=cx$ for all $x in I$.
By computing in different ways $d_{1}d_{2}(xty)$ for $x,t,yin I$, making use of the facts that $d_{i}$'s are derivations and $d_{1}d_{2}(x)=cx$ for $xin I$ and some $cin C$, one can easily obtain $$0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$$ My question is how can we show that $hspace{0.1cm}$ $0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$?
$textbf{My attempt:}$
begin{align}
begin{aligned}
d_1d_2(xty) =& d_1[d_2(x)ty+xd_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(ty)]+d_1(x)[d_2(ty)]+xd_1[d_2(ty)]\
=& d_1d_2(x)ty+d_2(x)[d_1(t)y+td_1(y)]+d_1(x)[d_2(t)y+td_2(y)]+xd_1[d_2(t)y+td_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+d_1d_2(x)ty+d_2(x)d_1(t)y +
x[d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)] \
=& d_1(x)td_2(y)+d_2(x)td_1(y)+3cxyt+[d_2(x)d_1(t)+d_1(x)d_2(t)]y+x[d_2(t)d_1(y)+d_1(t)d_2(y)]
end{aligned}
end{align}
Now,$cty=d_1d_2(ty)=d_1[d_2(t)y+td_2(y)]=d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)$.
By hypothesis $cty=cty+d_2(t)d_1(y)+d_1(t)d_2(y)+cty$
Hence, we obtain $d_2(t)d_1(y)=-cty-d_1(t)d_2(y)$ and $d_2(x)d_1(t)=-ctx-d_1(x)d_2(t)$.
ring-theory noncommutative-algebra
ring-theory noncommutative-algebra
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
edited Nov 21 '18 at 8:53
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
asked Dec 2 '15 at 15:31
1ENİGMA1
958416
958416
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%2f1556618%2fequality-of-derivation-on-prime-ring%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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1556618%2fequality-of-derivation-on-prime-ring%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
