Determining if a set is a ring
$begingroup$
My Abstract Algebra professor was going through an example in class today that I really don't understand. The problem is:
Let $B(X)$ be the set of all subsets of $X$. Consider two binary operations on $B(X):$ $Ycap Z$ and $Ycup Z$, for $Y$, $Zin B(X)$. Can these two operations be the binary operations of some ring on $B(X)$?
Can someone explain how I would start to begin this example or how to determine if this is, in fact, a ring? I think it's not a ring because I cannot figure out what the inverse would be.
abstract-algebra ring-theory
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
$endgroup$
add a comment |
$begingroup$
My Abstract Algebra professor was going through an example in class today that I really don't understand. The problem is:
Let $B(X)$ be the set of all subsets of $X$. Consider two binary operations on $B(X):$ $Ycap Z$ and $Ycup Z$, for $Y$, $Zin B(X)$. Can these two operations be the binary operations of some ring on $B(X)$?
Can someone explain how I would start to begin this example or how to determine if this is, in fact, a ring? I think it's not a ring because I cannot figure out what the inverse would be.
abstract-algebra ring-theory
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
$endgroup$
$begingroup$
Every element in the ring need not have an multiplicative inverse. Take the ring of integers for e.g.
$endgroup$
– thedilated
Dec 5 '18 at 0:11
$begingroup$
take a look at hasse diagrams
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 0:31
add a comment |
$begingroup$
My Abstract Algebra professor was going through an example in class today that I really don't understand. The problem is:
Let $B(X)$ be the set of all subsets of $X$. Consider two binary operations on $B(X):$ $Ycap Z$ and $Ycup Z$, for $Y$, $Zin B(X)$. Can these two operations be the binary operations of some ring on $B(X)$?
Can someone explain how I would start to begin this example or how to determine if this is, in fact, a ring? I think it's not a ring because I cannot figure out what the inverse would be.
abstract-algebra ring-theory
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
$endgroup$
My Abstract Algebra professor was going through an example in class today that I really don't understand. The problem is:
Let $B(X)$ be the set of all subsets of $X$. Consider two binary operations on $B(X):$ $Ycap Z$ and $Ycup Z$, for $Y$, $Zin B(X)$. Can these two operations be the binary operations of some ring on $B(X)$?
Can someone explain how I would start to begin this example or how to determine if this is, in fact, a ring? I think it's not a ring because I cannot figure out what the inverse would be.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
edited Dec 5 '18 at 0:26
lmckal45
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
asked Dec 4 '18 at 23:59
lmckal45lmckal45
244
244
$begingroup$
Every element in the ring need not have an multiplicative inverse. Take the ring of integers for e.g.
$endgroup$
– thedilated
Dec 5 '18 at 0:11
$begingroup$
take a look at hasse diagrams
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 0:31
add a comment |
$begingroup$
Every element in the ring need not have an multiplicative inverse. Take the ring of integers for e.g.
$endgroup$
– thedilated
Dec 5 '18 at 0:11
$begingroup$
take a look at hasse diagrams
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 0:31
$begingroup$
Every element in the ring need not have an multiplicative inverse. Take the ring of integers for e.g.
$endgroup$
– thedilated
Dec 5 '18 at 0:11
$begingroup$
Every element in the ring need not have an multiplicative inverse. Take the ring of integers for e.g.
$endgroup$
– thedilated
Dec 5 '18 at 0:11
$begingroup$
take a look at hasse diagrams
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 0:31
$begingroup$
take a look at hasse diagrams
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 0:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a simple diagram of the Power set of ${a,b,c}$
To find the union of two sets, then we follow the latice upward until we find an node above our sets with edges that connect to the set.
To find the intersection, move downward.
Hopefully this will help to visualize the problem.
To prove that it is a ring, intersection and union are both associative and commutative.
The distributive property holds $Acap (Bcup C) = (Acap B)cup (Acap C)$
That there is an identity element for each operation. What are your identity elements for each operation?
And that there is an inverse associated with one operation.
Rings do not require multiplicative inverses.
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
$endgroup$
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%2f3026375%2fdetermining-if-a-set-is-a-ring%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
$begingroup$
Here is a simple diagram of the Power set of ${a,b,c}$
To find the union of two sets, then we follow the latice upward until we find an node above our sets with edges that connect to the set.
To find the intersection, move downward.
Hopefully this will help to visualize the problem.
To prove that it is a ring, intersection and union are both associative and commutative.
The distributive property holds $Acap (Bcup C) = (Acap B)cup (Acap C)$
That there is an identity element for each operation. What are your identity elements for each operation?
And that there is an inverse associated with one operation.
Rings do not require multiplicative inverses.
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
$begingroup$
Here is a simple diagram of the Power set of ${a,b,c}$
To find the union of two sets, then we follow the latice upward until we find an node above our sets with edges that connect to the set.
To find the intersection, move downward.
Hopefully this will help to visualize the problem.
To prove that it is a ring, intersection and union are both associative and commutative.
The distributive property holds $Acap (Bcup C) = (Acap B)cup (Acap C)$
That there is an identity element for each operation. What are your identity elements for each operation?
And that there is an inverse associated with one operation.
Rings do not require multiplicative inverses.
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
$begingroup$
Here is a simple diagram of the Power set of ${a,b,c}$
To find the union of two sets, then we follow the latice upward until we find an node above our sets with edges that connect to the set.
To find the intersection, move downward.
Hopefully this will help to visualize the problem.
To prove that it is a ring, intersection and union are both associative and commutative.
The distributive property holds $Acap (Bcup C) = (Acap B)cup (Acap C)$
That there is an identity element for each operation. What are your identity elements for each operation?
And that there is an inverse associated with one operation.
Rings do not require multiplicative inverses.
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
$endgroup$
Here is a simple diagram of the Power set of ${a,b,c}$
To find the union of two sets, then we follow the latice upward until we find an node above our sets with edges that connect to the set.
To find the intersection, move downward.
Hopefully this will help to visualize the problem.
To prove that it is a ring, intersection and union are both associative and commutative.
The distributive property holds $Acap (Bcup C) = (Acap B)cup (Acap C)$
That there is an identity element for each operation. What are your identity elements for each operation?
And that there is an inverse associated with one operation.
Rings do not require multiplicative inverses.
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
answered Dec 5 '18 at 0:14
Doug MDoug M
44.7k31854
44.7k31854
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.
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%2f3026375%2fdetermining-if-a-set-is-a-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
$begingroup$
Every element in the ring need not have an multiplicative inverse. Take the ring of integers for e.g.
$endgroup$
– thedilated
Dec 5 '18 at 0:11
$begingroup$
take a look at hasse diagrams
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 0:31