About module decomposition
$begingroup$
I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. There there is a module decomposition
$$
A=mathfrak{a}_1opluscdotsoplusmathfrak{a}_n
$$
of a commutative ring $A$. Then the last sentence on the page says "The identity element $e_i$ of $mathfrak{a}_i$ is an idempotent in $A$, and $mathfrak{a}_i=(e_i)$." But do ideals have identity elements?
commutative-algebra
edited Dec 31 '18 at 15:17
user26857
39.3k124183
asked Dec 31 '18 at 10:15
saisai
1376
$endgroup$
add a comment |
$begingroup$
I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. There there is a module decomposition
$$
A=mathfrak{a}_1opluscdotsoplusmathfrak{a}_n
$$
of a commutative ring $A$. Then the last sentence on the page says "The identity element $e_i$ of $mathfrak{a}_i$ is an idempotent in $A$, and $mathfrak{a}_i=(e_i)$." But do ideals have identity elements?
commutative-algebra
edited Dec 31 '18 at 15:17
user26857
39.3k124183
asked Dec 31 '18 at 10:15
saisai
1376
$endgroup$
1
$begingroup$
Yes, sometimes they have identity elements. Not always. The ideals which are direct summands of a commutative ring all have identities.
$endgroup$
– rschwieb
Dec 31 '18 at 11:16
add a comment |
$begingroup$
I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. There there is a module decomposition
$$
A=mathfrak{a}_1opluscdotsoplusmathfrak{a}_n
$$
of a commutative ring $A$. Then the last sentence on the page says "The identity element $e_i$ of $mathfrak{a}_i$ is an idempotent in $A$, and $mathfrak{a}_i=(e_i)$." But do ideals have identity elements?
commutative-algebra
edited Dec 31 '18 at 15:17
user26857
39.3k124183
asked Dec 31 '18 at 10:15
saisai
1376
$endgroup$
I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. There there is a module decomposition
$$
A=mathfrak{a}_1opluscdotsoplusmathfrak{a}_n
$$
of a commutative ring $A$. Then the last sentence on the page says "The identity element $e_i$ of $mathfrak{a}_i$ is an idempotent in $A$, and $mathfrak{a}_i=(e_i)$." But do ideals have identity elements?
commutative-algebra
commutative-algebra
edited Dec 31 '18 at 15:17
user26857
39.3k124183
asked Dec 31 '18 at 10:15
saisai
1376
edited Dec 31 '18 at 15:17
user26857
39.3k124183
asked Dec 31 '18 at 10:15
saisai
1376
edited Dec 31 '18 at 15:17
user26857
39.3k124183
edited Dec 31 '18 at 15:17
user26857
39.3k124183
edited Dec 31 '18 at 15:17
user26857
39.3k124183
39.3k124183
asked Dec 31 '18 at 10:15
saisai
1376
asked Dec 31 '18 at 10:15
saisai
1376
asked Dec 31 '18 at 10:15
saisai
1376
1376
1
$begingroup$
Yes, sometimes they have identity elements. Not always. The ideals which are direct summands of a commutative ring all have identities.
$endgroup$
– rschwieb
Dec 31 '18 at 11:16
add a comment |
1
$begingroup$
Yes, sometimes they have identity elements. Not always. The ideals which are direct summands of a commutative ring all have identities.
$endgroup$
– rschwieb
Dec 31 '18 at 11:16
1
1
$begingroup$
Yes, sometimes they have identity elements. Not always. The ideals which are direct summands of a commutative ring all have identities.
$endgroup$
– rschwieb
Dec 31 '18 at 11:16
$begingroup$
Yes, sometimes they have identity elements. Not always. The ideals which are direct summands of a commutative ring all have identities.
$endgroup$
– rschwieb
Dec 31 '18 at 11:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
More context is important here. From the beginning of the paragraph, $A = prod A_i$ is a product of rings, and the ideals $mathfrak a_i$ are the image of the non-unit preserving maps $A_i to A$ sending $a_i mapsto (0,0,ldots, 0,a_i,0,ldots,0,0)$. This map identifies $mathfrak a_i simeq A_i$ as sets, so $e_i$ just means the element corresponding to the identity of $A_i$.
No ideals do not have identity elements, generally speaking they are non-unital rings. Even if they do have a unit, they are not considered subrings of the ring they are an ideal for (because they have a different unit).
answered Dec 31 '18 at 10:37
BenBen
4,178617
$endgroup$
add a comment |
$begingroup$
You forgot the sentence before that quote:
Each $mathfrak a_i$ is a ring (isomorphic to $A/mathfrak b_i$).
(where $mathfrak b_i = bigoplus_{j ne i} mathfrak a_j$)
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
$endgroup$
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%2f3057571%2fabout-module-decomposition%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
$begingroup$
More context is important here. From the beginning of the paragraph, $A = prod A_i$ is a product of rings, and the ideals $mathfrak a_i$ are the image of the non-unit preserving maps $A_i to A$ sending $a_i mapsto (0,0,ldots, 0,a_i,0,ldots,0,0)$. This map identifies $mathfrak a_i simeq A_i$ as sets, so $e_i$ just means the element corresponding to the identity of $A_i$.
No ideals do not have identity elements, generally speaking they are non-unital rings. Even if they do have a unit, they are not considered subrings of the ring they are an ideal for (because they have a different unit).
answered Dec 31 '18 at 10:37
BenBen
4,178617
$endgroup$
add a comment |
$begingroup$
More context is important here. From the beginning of the paragraph, $A = prod A_i$ is a product of rings, and the ideals $mathfrak a_i$ are the image of the non-unit preserving maps $A_i to A$ sending $a_i mapsto (0,0,ldots, 0,a_i,0,ldots,0,0)$. This map identifies $mathfrak a_i simeq A_i$ as sets, so $e_i$ just means the element corresponding to the identity of $A_i$.
No ideals do not have identity elements, generally speaking they are non-unital rings. Even if they do have a unit, they are not considered subrings of the ring they are an ideal for (because they have a different unit).
answered Dec 31 '18 at 10:37
BenBen
4,178617
$endgroup$
add a comment |
$begingroup$
More context is important here. From the beginning of the paragraph, $A = prod A_i$ is a product of rings, and the ideals $mathfrak a_i$ are the image of the non-unit preserving maps $A_i to A$ sending $a_i mapsto (0,0,ldots, 0,a_i,0,ldots,0,0)$. This map identifies $mathfrak a_i simeq A_i$ as sets, so $e_i$ just means the element corresponding to the identity of $A_i$.
No ideals do not have identity elements, generally speaking they are non-unital rings. Even if they do have a unit, they are not considered subrings of the ring they are an ideal for (because they have a different unit).
answered Dec 31 '18 at 10:37
BenBen
4,178617
$endgroup$
More context is important here. From the beginning of the paragraph, $A = prod A_i$ is a product of rings, and the ideals $mathfrak a_i$ are the image of the non-unit preserving maps $A_i to A$ sending $a_i mapsto (0,0,ldots, 0,a_i,0,ldots,0,0)$. This map identifies $mathfrak a_i simeq A_i$ as sets, so $e_i$ just means the element corresponding to the identity of $A_i$.
No ideals do not have identity elements, generally speaking they are non-unital rings. Even if they do have a unit, they are not considered subrings of the ring they are an ideal for (because they have a different unit).
answered Dec 31 '18 at 10:37
BenBen
4,178617
answered Dec 31 '18 at 10:37
BenBen
4,178617
answered Dec 31 '18 at 10:37
BenBen
4,178617
answered Dec 31 '18 at 10:37
BenBen
4,178617
4,178617
add a comment |
add a comment |
$begingroup$
You forgot the sentence before that quote:
Each $mathfrak a_i$ is a ring (isomorphic to $A/mathfrak b_i$).
(where $mathfrak b_i = bigoplus_{j ne i} mathfrak a_j$)
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
$endgroup$
add a comment |
$begingroup$
You forgot the sentence before that quote:
Each $mathfrak a_i$ is a ring (isomorphic to $A/mathfrak b_i$).
(where $mathfrak b_i = bigoplus_{j ne i} mathfrak a_j$)
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
$endgroup$
add a comment |
$begingroup$
You forgot the sentence before that quote:
Each $mathfrak a_i$ is a ring (isomorphic to $A/mathfrak b_i$).
(where $mathfrak b_i = bigoplus_{j ne i} mathfrak a_j$)
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
$endgroup$
You forgot the sentence before that quote:
Each $mathfrak a_i$ is a ring (isomorphic to $A/mathfrak b_i$).
(where $mathfrak b_i = bigoplus_{j ne i} mathfrak a_j$)
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
answered Dec 31 '18 at 10:28
Kenny LauKenny Lau
19.9k2160
19.9k2160
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%2f3057571%2fabout-module-decomposition%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
$begingroup$
Yes, sometimes they have identity elements. Not always. The ideals which are direct summands of a commutative ring all have identities.
$endgroup$
– rschwieb
Dec 31 '18 at 11:16