Commutative ring with unity
I'm given the following question :
$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.
I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?
abstract-algebra
edited Dec 7 at 8:55
egreg
177k1484200
asked Dec 7 at 8:35
Jasmine
283
|
show 2 more comments
I'm given the following question :
$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.
I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?
abstract-algebra
edited Dec 7 at 8:55
egreg
177k1484200
asked Dec 7 at 8:35
Jasmine
283
First you should think about how multiplication and addition can be defined.
– Paul K
Dec 7 at 8:37
answers.yahoo.com/question/…
– 1ENİGMA1
Dec 7 at 8:45
Its component-wise.
– Wuestenfux
Dec 7 at 8:47
You do have an appropriate software to write it accurately, though; it is just done with pure text:$Bbb ZtimesBbb Z$gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
– Arthur
Dec 7 at 8:49
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
– coffeemath
Dec 7 at 10:10
|
show 2 more comments
I'm given the following question :
$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.
I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?
abstract-algebra
edited Dec 7 at 8:55
egreg
177k1484200
asked Dec 7 at 8:35
Jasmine
283
I'm given the following question :
$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.
I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?
abstract-algebra
abstract-algebra
edited Dec 7 at 8:55
egreg
177k1484200
asked Dec 7 at 8:35
Jasmine
283
edited Dec 7 at 8:55
egreg
177k1484200
asked Dec 7 at 8:35
Jasmine
283
edited Dec 7 at 8:55
egreg
177k1484200
edited Dec 7 at 8:55
egreg
177k1484200
edited Dec 7 at 8:55
egreg
177k1484200
177k1484200
asked Dec 7 at 8:35
Jasmine
283
asked Dec 7 at 8:35
Jasmine
283
asked Dec 7 at 8:35
Jasmine
283
283
First you should think about how multiplication and addition can be defined.
– Paul K
Dec 7 at 8:37
answers.yahoo.com/question/…
– 1ENİGMA1
Dec 7 at 8:45
Its component-wise.
– Wuestenfux
Dec 7 at 8:47
You do have an appropriate software to write it accurately, though; it is just done with pure text:$Bbb ZtimesBbb Z$gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
– Arthur
Dec 7 at 8:49
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
– coffeemath
Dec 7 at 10:10
|
show 2 more comments
First you should think about how multiplication and addition can be defined.
– Paul K
Dec 7 at 8:37
answers.yahoo.com/question/…
– 1ENİGMA1
Dec 7 at 8:45
Its component-wise.
– Wuestenfux
Dec 7 at 8:47
You do have an appropriate software to write it accurately, though; it is just done with pure text:$Bbb ZtimesBbb Z$gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
– Arthur
Dec 7 at 8:49
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
– coffeemath
Dec 7 at 10:10
First you should think about how multiplication and addition can be defined.
– Paul K
Dec 7 at 8:37
First you should think about how multiplication and addition can be defined.
– Paul K
Dec 7 at 8:37
answers.yahoo.com/question/…
– 1ENİGMA1
Dec 7 at 8:45
answers.yahoo.com/question/…
– 1ENİGMA1
Dec 7 at 8:45
Its component-wise.
– Wuestenfux
Dec 7 at 8:47
Its component-wise.
– Wuestenfux
Dec 7 at 8:47
You do have an appropriate software to write it accurately, though; it is just done with pure text:
$Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.– Arthur
Dec 7 at 8:49
You do have an appropriate software to write it accurately, though; it is just done with pure text:
$Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.– Arthur
Dec 7 at 8:49
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
– coffeemath
Dec 7 at 10:10
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
– coffeemath
Dec 7 at 10:10
|
show 2 more comments
1 Answer
1
active
oldest
votes
Before saying a set is a ring, you have to specify the operations.
There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.
Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
answered Dec 7 at 8:59
egreg
177k1484200
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%2f3029657%2fcommutative-ring-with-unity%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
Before saying a set is a ring, you have to specify the operations.
There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.
Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
answered Dec 7 at 8:59
egreg
177k1484200
add a comment |
Before saying a set is a ring, you have to specify the operations.
There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.
Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
answered Dec 7 at 8:59
egreg
177k1484200
add a comment |
Before saying a set is a ring, you have to specify the operations.
There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.
Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
answered Dec 7 at 8:59
egreg
177k1484200
Before saying a set is a ring, you have to specify the operations.
There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.
Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
answered Dec 7 at 8:59
egreg
177k1484200
answered Dec 7 at 8:59
egreg
177k1484200
answered Dec 7 at 8:59
egreg
177k1484200
answered Dec 7 at 8:59
egreg
177k1484200
177k1484200
add a comment |
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.
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%2f3029657%2fcommutative-ring-with-unity%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
First you should think about how multiplication and addition can be defined.
– Paul K
Dec 7 at 8:37
answers.yahoo.com/question/…
– 1ENİGMA1
Dec 7 at 8:45
Its component-wise.
– Wuestenfux
Dec 7 at 8:47
You do have an appropriate software to write it accurately, though; it is just done with pure text:
$Bbb ZtimesBbb Z$gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.– Arthur
Dec 7 at 8:49
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
– coffeemath
Dec 7 at 10:10