What is the non trivial example of monosemiring?
$begingroup$
I considered the definition of monosemiring as: a semiring $(R, +, .)$ is said to be a monosemiring if $x+y=xy$ $forall~x, yin R,$ where $(R,+)$ and $(R, .)$ are semi groups. I also know that distributive laws will be satisfied in a monosemiring but it seems that every semi group can also be considered to be monosemiring where distributive laws will be trivially satisfied. What is the example of semi group which cannot be extended to a monosemiring? Or What is the non trivial example of monosemiring? I am bit scared as many questions went unanswered. Please someone answer this. Thank you in advance
examples-counterexamples semigroups semiring
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
asked Dec 24 '18 at 11:18
getegete
847
$endgroup$
add a comment |
$begingroup$
I considered the definition of monosemiring as: a semiring $(R, +, .)$ is said to be a monosemiring if $x+y=xy$ $forall~x, yin R,$ where $(R,+)$ and $(R, .)$ are semi groups. I also know that distributive laws will be satisfied in a monosemiring but it seems that every semi group can also be considered to be monosemiring where distributive laws will be trivially satisfied. What is the example of semi group which cannot be extended to a monosemiring? Or What is the non trivial example of monosemiring? I am bit scared as many questions went unanswered. Please someone answer this. Thank you in advance
examples-counterexamples semigroups semiring
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
asked Dec 24 '18 at 11:18
getegete
847
$endgroup$
$begingroup$
No, distributive rule isn't trivially satisfied: it translates to the identities $abc=abac$ and $abc=acbc$. Basically, monosemigroups are exactly the semigroups satisfying these identities.
$endgroup$
– Berci
Dec 24 '18 at 12:08
add a comment |
$begingroup$
I considered the definition of monosemiring as: a semiring $(R, +, .)$ is said to be a monosemiring if $x+y=xy$ $forall~x, yin R,$ where $(R,+)$ and $(R, .)$ are semi groups. I also know that distributive laws will be satisfied in a monosemiring but it seems that every semi group can also be considered to be monosemiring where distributive laws will be trivially satisfied. What is the example of semi group which cannot be extended to a monosemiring? Or What is the non trivial example of monosemiring? I am bit scared as many questions went unanswered. Please someone answer this. Thank you in advance
examples-counterexamples semigroups semiring
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
asked Dec 24 '18 at 11:18
getegete
847
$endgroup$
I considered the definition of monosemiring as: a semiring $(R, +, .)$ is said to be a monosemiring if $x+y=xy$ $forall~x, yin R,$ where $(R,+)$ and $(R, .)$ are semi groups. I also know that distributive laws will be satisfied in a monosemiring but it seems that every semi group can also be considered to be monosemiring where distributive laws will be trivially satisfied. What is the example of semi group which cannot be extended to a monosemiring? Or What is the non trivial example of monosemiring? I am bit scared as many questions went unanswered. Please someone answer this. Thank you in advance
examples-counterexamples semigroups semiring
examples-counterexamples semigroups semiring
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
asked Dec 24 '18 at 11:18
getegete
847
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
asked Dec 24 '18 at 11:18
getegete
847
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
edited Dec 24 '18 at 11:20
Shaun
11.1k113688
11.1k113688
asked Dec 24 '18 at 11:18
getegete
847
asked Dec 24 '18 at 11:18
getegete
847
asked Dec 24 '18 at 11:18
getegete
847
847
$begingroup$
No, distributive rule isn't trivially satisfied: it translates to the identities $abc=abac$ and $abc=acbc$. Basically, monosemigroups are exactly the semigroups satisfying these identities.
$endgroup$
– Berci
Dec 24 '18 at 12:08
add a comment |
$begingroup$
No, distributive rule isn't trivially satisfied: it translates to the identities $abc=abac$ and $abc=acbc$. Basically, monosemigroups are exactly the semigroups satisfying these identities.
$endgroup$
– Berci
Dec 24 '18 at 12:08
$begingroup$
No, distributive rule isn't trivially satisfied: it translates to the identities $abc=abac$ and $abc=acbc$. Basically, monosemigroups are exactly the semigroups satisfying these identities.
$endgroup$
– Berci
Dec 24 '18 at 12:08
$begingroup$
No, distributive rule isn't trivially satisfied: it translates to the identities $abc=abac$ and $abc=acbc$. Basically, monosemigroups are exactly the semigroups satisfying these identities.
$endgroup$
– Berci
Dec 24 '18 at 12:08
add a comment |
0
active
oldest
votes
Your Answer
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%2f3051158%2fwhat-is-the-non-trivial-example-of-monosemiring%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%2f3051158%2fwhat-is-the-non-trivial-example-of-monosemiring%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
$begingroup$
No, distributive rule isn't trivially satisfied: it translates to the identities $abc=abac$ and $abc=acbc$. Basically, monosemigroups are exactly the semigroups satisfying these identities.
$endgroup$
– Berci
Dec 24 '18 at 12:08