Mixing
Algebras finite-dimensional over their base field are Artinian.
$begingroup$
The standard proof that a $k$-algebra $A$ that is finite-dimensional as a $k$-vector space is Artinian goes as follows:
"Suppose we have an infinite descending chain of ideals $I_1 supseteq I_2 supseteq I_3 supseteq ...$ in $A$. Then each $I_j$ is also a finite-dimensional $k$--vector space (since $A$ is), so $dim_k I_1 ge dim_k I_2 ge dim_k I_3 ge ...$. Since we cannot have an infinite strictly descending sequence of non-negative integers, the sequence of dimensions must stabilize, so the chain of ideals must stabilize."
What I'm wondering is why it isn't possible to eventually have $I_n supseteq I_{n+1} supseteq I_{n+2} supseteq ...$ where $dim_k I_n = dim_k I_{n+1} = dim_k I_{n+2} = ...$ but $I_n$, $I_{n+1}$, $I_{n+2}$, etc. are all distinct ideals in $A$. In other words, why should the fact that the dimensions stabilize imply that the ideals also stabilize?
Thanks in advance for any help.
ring-theory ideals artinian
asked Jan 23 at 0:18
user514014user514014
1336
$endgroup$
add a comment |
$begingroup$
The standard proof that a $k$-algebra $A$ that is finite-dimensional as a $k$-vector space is Artinian goes as follows:
"Suppose we have an infinite descending chain of ideals $I_1 supseteq I_2 supseteq I_3 supseteq ...$ in $A$. Then each $I_j$ is also a finite-dimensional $k$--vector space (since $A$ is), so $dim_k I_1 ge dim_k I_2 ge dim_k I_3 ge ...$. Since we cannot have an infinite strictly descending sequence of non-negative integers, the sequence of dimensions must stabilize, so the chain of ideals must stabilize."
What I'm wondering is why it isn't possible to eventually have $I_n supseteq I_{n+1} supseteq I_{n+2} supseteq ...$ where $dim_k I_n = dim_k I_{n+1} = dim_k I_{n+2} = ...$ but $I_n$, $I_{n+1}$, $I_{n+2}$, etc. are all distinct ideals in $A$. In other words, why should the fact that the dimensions stabilize imply that the ideals also stabilize?
Thanks in advance for any help.
ring-theory ideals artinian
asked Jan 23 at 0:18
user514014user514014
1336
$endgroup$
1
$begingroup$
How many $n$-dimensional subspaces are there of an $n$-dimensional vector space? Adapt the answer to your chain of ideals. Can the $I_{n+i}$, for non-negative integers $i,$ be distinct?
$endgroup$
– Chris Leary
Jan 23 at 0:25
1
$begingroup$
note that if an algebral is unital then any ideal must be a subspace.
$endgroup$
– David Holden
Jan 23 at 0:30
add a comment |
$begingroup$
The standard proof that a $k$-algebra $A$ that is finite-dimensional as a $k$-vector space is Artinian goes as follows:
"Suppose we have an infinite descending chain of ideals $I_1 supseteq I_2 supseteq I_3 supseteq ...$ in $A$. Then each $I_j$ is also a finite-dimensional $k$--vector space (since $A$ is), so $dim_k I_1 ge dim_k I_2 ge dim_k I_3 ge ...$. Since we cannot have an infinite strictly descending sequence of non-negative integers, the sequence of dimensions must stabilize, so the chain of ideals must stabilize."
What I'm wondering is why it isn't possible to eventually have $I_n supseteq I_{n+1} supseteq I_{n+2} supseteq ...$ where $dim_k I_n = dim_k I_{n+1} = dim_k I_{n+2} = ...$ but $I_n$, $I_{n+1}$, $I_{n+2}$, etc. are all distinct ideals in $A$. In other words, why should the fact that the dimensions stabilize imply that the ideals also stabilize?
Thanks in advance for any help.
ring-theory ideals artinian
asked Jan 23 at 0:18
user514014user514014
1336
$endgroup$
The standard proof that a $k$-algebra $A$ that is finite-dimensional as a $k$-vector space is Artinian goes as follows:
"Suppose we have an infinite descending chain of ideals $I_1 supseteq I_2 supseteq I_3 supseteq ...$ in $A$. Then each $I_j$ is also a finite-dimensional $k$--vector space (since $A$ is), so $dim_k I_1 ge dim_k I_2 ge dim_k I_3 ge ...$. Since we cannot have an infinite strictly descending sequence of non-negative integers, the sequence of dimensions must stabilize, so the chain of ideals must stabilize."
What I'm wondering is why it isn't possible to eventually have $I_n supseteq I_{n+1} supseteq I_{n+2} supseteq ...$ where $dim_k I_n = dim_k I_{n+1} = dim_k I_{n+2} = ...$ but $I_n$, $I_{n+1}$, $I_{n+2}$, etc. are all distinct ideals in $A$. In other words, why should the fact that the dimensions stabilize imply that the ideals also stabilize?
Thanks in advance for any help.
ring-theory ideals artinian
ring-theory ideals artinian
asked Jan 23 at 0:18
user514014user514014
1336
asked Jan 23 at 0:18
user514014user514014
1336
asked Jan 23 at 0:18
user514014user514014
1336
asked Jan 23 at 0:18
user514014user514014
1336
asked Jan 23 at 0:18
user514014user514014
1336
1336
1
$begingroup$
How many $n$-dimensional subspaces are there of an $n$-dimensional vector space? Adapt the answer to your chain of ideals. Can the $I_{n+i}$, for non-negative integers $i,$ be distinct?
$endgroup$
– Chris Leary
Jan 23 at 0:25
1
$begingroup$
note that if an algebral is unital then any ideal must be a subspace.
$endgroup$
– David Holden
Jan 23 at 0:30
add a comment |
1
$begingroup$
How many $n$-dimensional subspaces are there of an $n$-dimensional vector space? Adapt the answer to your chain of ideals. Can the $I_{n+i}$, for non-negative integers $i,$ be distinct?
$endgroup$
– Chris Leary
Jan 23 at 0:25
1
$begingroup$
note that if an algebral is unital then any ideal must be a subspace.
$endgroup$
– David Holden
Jan 23 at 0:30
1
1
$begingroup$
How many $n$-dimensional subspaces are there of an $n$-dimensional vector space? Adapt the answer to your chain of ideals. Can the $I_{n+i}$, for non-negative integers $i,$ be distinct?
$endgroup$
– Chris Leary
Jan 23 at 0:25
$begingroup$
How many $n$-dimensional subspaces are there of an $n$-dimensional vector space? Adapt the answer to your chain of ideals. Can the $I_{n+i}$, for non-negative integers $i,$ be distinct?
$endgroup$
– Chris Leary
Jan 23 at 0:25
1
1
$begingroup$
note that if an algebral is unital then any ideal must be a subspace.
$endgroup$
– David Holden
Jan 23 at 0:30
$begingroup$
note that if an algebral is unital then any ideal must be a subspace.
$endgroup$
– David Holden
Jan 23 at 0:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $V$ is a finite-dimensional vector space and $Wsubseteq V$ is a subspace with $dim W=dim V$, then $W=V$. Indeed, $dim V=dim W+dim V/W$ so $dim V/W=0$ which means $V/W$ is trivial so $W=V$.
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
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%2f3083920%2falgebras-finite-dimensional-over-their-base-field-are-artinian%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
$begingroup$
If $V$ is a finite-dimensional vector space and $Wsubseteq V$ is a subspace with $dim W=dim V$, then $W=V$. Indeed, $dim V=dim W+dim V/W$ so $dim V/W=0$ which means $V/W$ is trivial so $W=V$.
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
add a comment |
$begingroup$
If $V$ is a finite-dimensional vector space and $Wsubseteq V$ is a subspace with $dim W=dim V$, then $W=V$. Indeed, $dim V=dim W+dim V/W$ so $dim V/W=0$ which means $V/W$ is trivial so $W=V$.
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
add a comment |
$begingroup$
If $V$ is a finite-dimensional vector space and $Wsubseteq V$ is a subspace with $dim W=dim V$, then $W=V$. Indeed, $dim V=dim W+dim V/W$ so $dim V/W=0$ which means $V/W$ is trivial so $W=V$.
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
$endgroup$
If $V$ is a finite-dimensional vector space and $Wsubseteq V$ is a subspace with $dim W=dim V$, then $W=V$. Indeed, $dim V=dim W+dim V/W$ so $dim V/W=0$ which means $V/W$ is trivial so $W=V$.
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
answered Jan 23 at 0:25
Eric WofseyEric Wofsey
189k14216347
189k14216347
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
add a comment |
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
$begingroup$
Thanks, that clears it up!
$endgroup$
– user514014
Jan 23 at 1:03
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%2f3083920%2falgebras-finite-dimensional-over-their-base-field-are-artinian%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
$begingroup$
How many $n$-dimensional subspaces are there of an $n$-dimensional vector space? Adapt the answer to your chain of ideals. Can the $I_{n+i}$, for non-negative integers $i,$ be distinct?
$endgroup$
– Chris Leary
Jan 23 at 0:25
1
$begingroup$
note that if an algebral is unital then any ideal must be a subspace.
$endgroup$
– David Holden
Jan 23 at 0:30