What algebraic structure does the set of endomorphisms of a ring have?
$begingroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
11 hours ago
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
10 hours ago
add a comment |
$begingroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
$endgroup$
Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:
$+$, defined by $(f+g)(x)=f(x)+g(x)$
$cdot$, defined by $(fcdot g)(x)=f(g(x))$
$*$, defiend by $(f*g)(x)=f(x)g(x)$
Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?
Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?
abstract-algebra ring-theory modules terminology ring-homomorphism
abstract-algebra ring-theory modules terminology ring-homomorphism
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
asked 11 hours ago
Keshav SrinivasanKeshav Srinivasan
2,04811441
2,04811441
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
11 hours ago
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
10 hours ago
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
11 hours ago
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
10 hours ago
1
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
11 hours ago
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
11 hours ago
1
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
10 hours ago
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
10 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
$endgroup$
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%2f3074950%2fwhat-algebraic-structure-does-the-set-of-endomorphisms-of-a-ring-have%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
$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
$endgroup$
add a comment |
$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
$endgroup$
add a comment |
$begingroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
$endgroup$
Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).
The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
answered 11 hours ago
Eric WofseyEric Wofsey
181k12209337
181k12209337
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%2f3074950%2fwhat-algebraic-structure-does-the-set-of-endomorphisms-of-a-ring-have%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
1
$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
11 hours ago
1
$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
10 hours ago