Mixing
![[qemu]Stay black screen!I think this phenomenon from kernel's config file](https://lh4.googleusercontent.com/-bCbHOPDhewQ/AAAAAAAAAAI/AAAAAAAAAAA/AGDgw-iAsAVfT7twCtvxou7QW_LjhuFpdw/s72-c-mo/photo.jpg?sz=32)
Algebraic structure, division ring
$begingroup$
In the set $mathbb{R}times mathbb{R}^3$ addition is defined by components. We define multiplication $*$ by $$(lambda,mathbf{x})*(mu,mathbf{y})=(lambdamu - mathbf{x}cdot mathbf{y}, lambda mathbf{y}+mumathbf{x}+mathbf{x}times mathbf{y}).$$
You need not prove that $(mathbb{R}times mathbb{R}^3,+,*)$ is a division ring. Also let $mathbf{a}in mathbb{R}^3$ with $lVertmathbf{a}rVert=1$.
- Find the multiplicative identity of the division ring $(mathbb{R}times mathbb{R}^3,+,*)$.
- Find the inverse of $(1,mathbf{a})$.
- Find all elements of the division ring $mathbb{R}times mathbb{R}^3$ that commute with $(0,mathbf{a})$, and prove that they also form a division ring, which is isomorphic to the division ring $mathbb{C}$.
I know that posting full problems isn't desirable here, but I have absolutely no clue on how to tackle this. I would be thankful if anyone could help me solve this.
A ring $K$ is a division ring $Leftrightarrow$ $Ksetminus{0}$ is a group for multiplication.
abstract-algebra vectors ring-isomorphism division-ring
asked Jan 22 at 0:59
lork251lork251
255
$endgroup$
add a comment |
$begingroup$
In the set $mathbb{R}times mathbb{R}^3$ addition is defined by components. We define multiplication $*$ by $$(lambda,mathbf{x})*(mu,mathbf{y})=(lambdamu - mathbf{x}cdot mathbf{y}, lambda mathbf{y}+mumathbf{x}+mathbf{x}times mathbf{y}).$$
You need not prove that $(mathbb{R}times mathbb{R}^3,+,*)$ is a division ring. Also let $mathbf{a}in mathbb{R}^3$ with $lVertmathbf{a}rVert=1$.
- Find the multiplicative identity of the division ring $(mathbb{R}times mathbb{R}^3,+,*)$.
- Find the inverse of $(1,mathbf{a})$.
- Find all elements of the division ring $mathbb{R}times mathbb{R}^3$ that commute with $(0,mathbf{a})$, and prove that they also form a division ring, which is isomorphic to the division ring $mathbb{C}$.
I know that posting full problems isn't desirable here, but I have absolutely no clue on how to tackle this. I would be thankful if anyone could help me solve this.
A ring $K$ is a division ring $Leftrightarrow$ $Ksetminus{0}$ is a group for multiplication.
abstract-algebra vectors ring-isomorphism division-ring
asked Jan 22 at 0:59
lork251lork251
255
$endgroup$
$begingroup$
@MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$.
$endgroup$
– lork251
Jan 22 at 1:15
$begingroup$
I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them.
$endgroup$
– lork251
Jan 22 at 1:23
2
$begingroup$
$(1,mathbf{0})$? Thanks!
$endgroup$
– lork251
Jan 22 at 1:27
$begingroup$
Converted to an answer.
$endgroup$
– Morgan Rodgers
Jan 22 at 1:41
add a comment |
$begingroup$
In the set $mathbb{R}times mathbb{R}^3$ addition is defined by components. We define multiplication $*$ by $$(lambda,mathbf{x})*(mu,mathbf{y})=(lambdamu - mathbf{x}cdot mathbf{y}, lambda mathbf{y}+mumathbf{x}+mathbf{x}times mathbf{y}).$$
You need not prove that $(mathbb{R}times mathbb{R}^3,+,*)$ is a division ring. Also let $mathbf{a}in mathbb{R}^3$ with $lVertmathbf{a}rVert=1$.
- Find the multiplicative identity of the division ring $(mathbb{R}times mathbb{R}^3,+,*)$.
- Find the inverse of $(1,mathbf{a})$.
- Find all elements of the division ring $mathbb{R}times mathbb{R}^3$ that commute with $(0,mathbf{a})$, and prove that they also form a division ring, which is isomorphic to the division ring $mathbb{C}$.
I know that posting full problems isn't desirable here, but I have absolutely no clue on how to tackle this. I would be thankful if anyone could help me solve this.
A ring $K$ is a division ring $Leftrightarrow$ $Ksetminus{0}$ is a group for multiplication.
abstract-algebra vectors ring-isomorphism division-ring
asked Jan 22 at 0:59
lork251lork251
255
$endgroup$
In the set $mathbb{R}times mathbb{R}^3$ addition is defined by components. We define multiplication $*$ by $$(lambda,mathbf{x})*(mu,mathbf{y})=(lambdamu - mathbf{x}cdot mathbf{y}, lambda mathbf{y}+mumathbf{x}+mathbf{x}times mathbf{y}).$$
You need not prove that $(mathbb{R}times mathbb{R}^3,+,*)$ is a division ring. Also let $mathbf{a}in mathbb{R}^3$ with $lVertmathbf{a}rVert=1$.
- Find the multiplicative identity of the division ring $(mathbb{R}times mathbb{R}^3,+,*)$.
- Find the inverse of $(1,mathbf{a})$.
- Find all elements of the division ring $mathbb{R}times mathbb{R}^3$ that commute with $(0,mathbf{a})$, and prove that they also form a division ring, which is isomorphic to the division ring $mathbb{C}$.
I know that posting full problems isn't desirable here, but I have absolutely no clue on how to tackle this. I would be thankful if anyone could help me solve this.
A ring $K$ is a division ring $Leftrightarrow$ $Ksetminus{0}$ is a group for multiplication.
abstract-algebra vectors ring-isomorphism division-ring
abstract-algebra vectors ring-isomorphism division-ring
asked Jan 22 at 0:59
lork251lork251
255
asked Jan 22 at 0:59
lork251lork251
255
edited Jan 22 at 1:19
lork251
asked Jan 22 at 0:59
lork251lork251
255
asked Jan 22 at 0:59
lork251lork251
255
asked Jan 22 at 0:59
lork251lork251
255
255
$begingroup$
@MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$.
$endgroup$
– lork251
Jan 22 at 1:15
$begingroup$
I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them.
$endgroup$
– lork251
Jan 22 at 1:23
2
$begingroup$
$(1,mathbf{0})$? Thanks!
$endgroup$
– lork251
Jan 22 at 1:27
$begingroup$
Converted to an answer.
$endgroup$
– Morgan Rodgers
Jan 22 at 1:41
add a comment |
$begingroup$
@MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$.
$endgroup$
– lork251
Jan 22 at 1:15
$begingroup$
I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them.
$endgroup$
– lork251
Jan 22 at 1:23
2
$begingroup$
$(1,mathbf{0})$? Thanks!
$endgroup$
– lork251
Jan 22 at 1:27
$begingroup$
Converted to an answer.
$endgroup$
– Morgan Rodgers
Jan 22 at 1:41
$begingroup$
@MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$.
$endgroup$
– lork251
Jan 22 at 1:15
$begingroup$
@MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$.
$endgroup$
– lork251
Jan 22 at 1:15
$begingroup$
I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them.
$endgroup$
– lork251
Jan 22 at 1:23
$begingroup$
I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them.
$endgroup$
– lork251
Jan 22 at 1:23
2
2
$begingroup$
$(1,mathbf{0})$? Thanks!
$endgroup$
– lork251
Jan 22 at 1:27
$begingroup$
$(1,mathbf{0})$? Thanks!
$endgroup$
– lork251
Jan 22 at 1:27
$begingroup$
Converted to an answer.
$endgroup$
– Morgan Rodgers
Jan 22 at 1:41
$begingroup$
Converted to an answer.
$endgroup$
– Morgan Rodgers
Jan 22 at 1:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you take an arbitrary element $(lambda, mathbf{x})$ and assume that $(mu,mathbf{y})$ is the multiplicative identity, then
$$(lambda, mathbf{x})*(mu,mathbf{y}) = (lambdamu-mathbf{x}mathbf{y},lambdamathbf{y}+mumathbf{x}+mathbf{x}timesmathbf{y}) = (lambda, mathbf{x}).$$
This gives two equations to solve, since you are assuming this is in fact a division ring the solution should be unique. You should check that the obvious solution to the first equation also works in the second.
This should help to solve 2., since now you know what $(1,mathbf{a})*(lambda,mathbf{x})$ needs to equal for $(lambda,mathbf{x})$ to be the inverse of $(1, mathbf{a})$.
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
$endgroup$
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
add a comment |
$begingroup$
You got $1$ (it's $(1,0)$).
For $2$, $(1,a)cdot (mu,y)=(1mu-acdot y,1y+mu a+a×y)=(1,0)$. So $(frac12, -frac12a)$ does the trick.
Finally, $(0,a)cdot (mu,y)=(mu,y)cdot (0,a)implies (-acdot y,mu a+a× y)=(-ycdot a,mu a+y×a)$.
This implies $(mu,y)=(mu,ta)$.
We get $Bbb R^2 cong Bbb C$.
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
1
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
add a 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%2f3082625%2falgebraic-structure-division-ring%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you take an arbitrary element $(lambda, mathbf{x})$ and assume that $(mu,mathbf{y})$ is the multiplicative identity, then
$$(lambda, mathbf{x})*(mu,mathbf{y}) = (lambdamu-mathbf{x}mathbf{y},lambdamathbf{y}+mumathbf{x}+mathbf{x}timesmathbf{y}) = (lambda, mathbf{x}).$$
This gives two equations to solve, since you are assuming this is in fact a division ring the solution should be unique. You should check that the obvious solution to the first equation also works in the second.
This should help to solve 2., since now you know what $(1,mathbf{a})*(lambda,mathbf{x})$ needs to equal for $(lambda,mathbf{x})$ to be the inverse of $(1, mathbf{a})$.
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
$endgroup$
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
add a comment |
$begingroup$
If you take an arbitrary element $(lambda, mathbf{x})$ and assume that $(mu,mathbf{y})$ is the multiplicative identity, then
$$(lambda, mathbf{x})*(mu,mathbf{y}) = (lambdamu-mathbf{x}mathbf{y},lambdamathbf{y}+mumathbf{x}+mathbf{x}timesmathbf{y}) = (lambda, mathbf{x}).$$
This gives two equations to solve, since you are assuming this is in fact a division ring the solution should be unique. You should check that the obvious solution to the first equation also works in the second.
This should help to solve 2., since now you know what $(1,mathbf{a})*(lambda,mathbf{x})$ needs to equal for $(lambda,mathbf{x})$ to be the inverse of $(1, mathbf{a})$.
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
$endgroup$
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
add a comment |
$begingroup$
If you take an arbitrary element $(lambda, mathbf{x})$ and assume that $(mu,mathbf{y})$ is the multiplicative identity, then
$$(lambda, mathbf{x})*(mu,mathbf{y}) = (lambdamu-mathbf{x}mathbf{y},lambdamathbf{y}+mumathbf{x}+mathbf{x}timesmathbf{y}) = (lambda, mathbf{x}).$$
This gives two equations to solve, since you are assuming this is in fact a division ring the solution should be unique. You should check that the obvious solution to the first equation also works in the second.
This should help to solve 2., since now you know what $(1,mathbf{a})*(lambda,mathbf{x})$ needs to equal for $(lambda,mathbf{x})$ to be the inverse of $(1, mathbf{a})$.
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
$endgroup$
If you take an arbitrary element $(lambda, mathbf{x})$ and assume that $(mu,mathbf{y})$ is the multiplicative identity, then
$$(lambda, mathbf{x})*(mu,mathbf{y}) = (lambdamu-mathbf{x}mathbf{y},lambdamathbf{y}+mumathbf{x}+mathbf{x}timesmathbf{y}) = (lambda, mathbf{x}).$$
This gives two equations to solve, since you are assuming this is in fact a division ring the solution should be unique. You should check that the obvious solution to the first equation also works in the second.
This should help to solve 2., since now you know what $(1,mathbf{a})*(lambda,mathbf{x})$ needs to equal for $(lambda,mathbf{x})$ to be the inverse of $(1, mathbf{a})$.
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
answered Jan 22 at 1:40
Morgan RodgersMorgan Rodgers
9,76621440
9,76621440
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
add a comment |
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
$begingroup$
Any clues for 3.?
$endgroup$
– lork251
Jan 22 at 1:47
add a comment |
$begingroup$
You got $1$ (it's $(1,0)$).
For $2$, $(1,a)cdot (mu,y)=(1mu-acdot y,1y+mu a+a×y)=(1,0)$. So $(frac12, -frac12a)$ does the trick.
Finally, $(0,a)cdot (mu,y)=(mu,y)cdot (0,a)implies (-acdot y,mu a+a× y)=(-ycdot a,mu a+y×a)$.
This implies $(mu,y)=(mu,ta)$.
We get $Bbb R^2 cong Bbb C$.
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
1
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
add a comment |
$begingroup$
You got $1$ (it's $(1,0)$).
For $2$, $(1,a)cdot (mu,y)=(1mu-acdot y,1y+mu a+a×y)=(1,0)$. So $(frac12, -frac12a)$ does the trick.
Finally, $(0,a)cdot (mu,y)=(mu,y)cdot (0,a)implies (-acdot y,mu a+a× y)=(-ycdot a,mu a+y×a)$.
This implies $(mu,y)=(mu,ta)$.
We get $Bbb R^2 cong Bbb C$.
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
1
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
add a comment |
$begingroup$
You got $1$ (it's $(1,0)$).
For $2$, $(1,a)cdot (mu,y)=(1mu-acdot y,1y+mu a+a×y)=(1,0)$. So $(frac12, -frac12a)$ does the trick.
Finally, $(0,a)cdot (mu,y)=(mu,y)cdot (0,a)implies (-acdot y,mu a+a× y)=(-ycdot a,mu a+y×a)$.
This implies $(mu,y)=(mu,ta)$.
We get $Bbb R^2 cong Bbb C$.
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
You got $1$ (it's $(1,0)$).
For $2$, $(1,a)cdot (mu,y)=(1mu-acdot y,1y+mu a+a×y)=(1,0)$. So $(frac12, -frac12a)$ does the trick.
Finally, $(0,a)cdot (mu,y)=(mu,y)cdot (0,a)implies (-acdot y,mu a+a× y)=(-ycdot a,mu a+y×a)$.
This implies $(mu,y)=(mu,ta)$.
We get $Bbb R^2 cong Bbb C$.
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
edited Jan 22 at 2:50
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
answered Jan 22 at 2:05
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
1
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
add a comment |
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
1
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
$begingroup$
$(1,mathbf{a})*(1,-mathbf{a})neq (1,mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$.
$endgroup$
– lork251
Jan 22 at 2:12
1
1
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
$begingroup$
Ok. I messed up. I'm getting $(frac12,-frac12a)$ with the correction.
$endgroup$
– Chris Custer
Jan 22 at 2:52
add a 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%2f3082625%2falgebraic-structure-division-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
$begingroup$
@MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$.
$endgroup$
– lork251
Jan 22 at 1:15
$begingroup$
I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them.
$endgroup$
– lork251
Jan 22 at 1:23
2
$begingroup$
$(1,mathbf{0})$? Thanks!
$endgroup$
– lork251
Jan 22 at 1:27
$begingroup$
Converted to an answer.
$endgroup$
– Morgan Rodgers
Jan 22 at 1:41