Mixing
What is meant by $langle X_1,…,X_nrangle$ in algebra?
$begingroup$
I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.
abstract-algebra
edited Jan 22 at 15:25
Clayton
19.3k33287
asked Jan 22 at 15:14
DoleDole
931614
$endgroup$
add a comment |
$begingroup$
I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.
abstract-algebra
edited Jan 22 at 15:25
Clayton
19.3k33287
asked Jan 22 at 15:14
DoleDole
931614
$endgroup$
2
$begingroup$
As a tip, you can uselangleandrangle, respectively, to produce $langle,rangle$.
$endgroup$
– Clayton
Jan 22 at 15:26
add a comment |
$begingroup$
I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.
abstract-algebra
edited Jan 22 at 15:25
Clayton
19.3k33287
asked Jan 22 at 15:14
DoleDole
931614
$endgroup$
I tried searching what the definition of the set $langle X_1,...,X_nrangle$ in Algebra but came up with nothing. As an example of use, the task is to prove that this set is an ideal and equal to $ker(F)$, $F$ being the zero argument function from $R[X_1,...,X_n]rightarrow R$.
abstract-algebra
abstract-algebra
edited Jan 22 at 15:25
Clayton
19.3k33287
asked Jan 22 at 15:14
DoleDole
931614
edited Jan 22 at 15:25
Clayton
19.3k33287
asked Jan 22 at 15:14
DoleDole
931614
edited Jan 22 at 15:25
Clayton
19.3k33287
edited Jan 22 at 15:25
Clayton
19.3k33287
edited Jan 22 at 15:25
Clayton
19.3k33287
19.3k33287
asked Jan 22 at 15:14
DoleDole
931614
asked Jan 22 at 15:14
DoleDole
931614
asked Jan 22 at 15:14
DoleDole
931614
931614
2
$begingroup$
As a tip, you can uselangleandrangle, respectively, to produce $langle,rangle$.
$endgroup$
– Clayton
Jan 22 at 15:26
add a comment |
2
$begingroup$
As a tip, you can uselangleandrangle, respectively, to produce $langle,rangle$.
$endgroup$
– Clayton
Jan 22 at 15:26
2
2
$begingroup$
As a tip, you can use
langle and rangle, respectively, to produce $langle,rangle$.$endgroup$
– Clayton
Jan 22 at 15:26
$begingroup$
As a tip, you can use
langle and rangle, respectively, to produce $langle,rangle$.$endgroup$
– Clayton
Jan 22 at 15:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.
It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.
1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.
2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.
3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.
4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
$endgroup$
1
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
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%2f3083286%2fwhat-is-meant-by-langle-x-1-x-n-rangle-in-algebra%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
$begingroup$
In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.
It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.
It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.
It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
$endgroup$
In ring theory, angle brackets are usually used to denote the ideal generated by $X_1, X_2,ldots$ inside the ring that $X_1, X_2,ldots$ comes from. In that case it would be silly to prove it is an ideal: it's an ideal by definition.
It looks more like the task is to show that the kernel of the evaluation homomorphism that evaluates polynomials at $0$ is equal to the ideal generated by these elements. Surely you can prove they mutually contain each other, and then you're already done.
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
answered Jan 22 at 15:21
rschwiebrschwieb
107k12102251
107k12102251
add a comment |
add a comment |
$begingroup$
There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.
1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.
2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.
3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.
4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
$endgroup$
1
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
add a comment |
$begingroup$
There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.
1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.
2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.
3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.
4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
$endgroup$
1
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
add a comment |
$begingroup$
There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.
1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.
2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.
3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.
4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
$endgroup$
There are different but related meanings of $langle X_1,ldots,X_nrangle$ in algebra.
1) The first one is the smallest ideal containing $X_1,ldots,X_n$, which, when inside a ring, is generated by all the combinations of sums and products of the elements $X_1,ldots,X_n$.
2) But you can be also inside another structure different from a ring, for example inside a monoid, in which case the ideal $langle X_1,ldots,X_nrangle$ is formed just by all the products of the elements $X_1,ldots,X_n$.
3) In relation to this, you also encounter this notation meaning that the $X_1,ldots,X_n$ are free variables and $langle X_1,ldots,X_nrangle$ is the free monoid on those variables, meaning the monoid formed by products of the variables (i.e., words on them and their compositions), with no other relations imposed whatsoever.
4) In addition, you can also see the notation in $Clangle X_1,ldots,X_nrangle$, for $C$ a commutative ring, meaning the free associative algebra over $C$, which is formed by first taking the free monoid on $X_1,ldots,X_n$ (i.e., forming monomials via products) and then taking all the $C$-linear combinations of elements of the monoid (i.e., forming polynomials via sums). There are no relations imposed on the algebra, other than those of being associative. In particular, the free algebra is not commutative: the commutative free associative algebra on $X_1,ldots,X_n$ is denoted as $C[X_1,ldots,X_n]$ and is the ring of polynomials (here $[X_1,ldots,X_n]$ stands for the free commutative monoid on $X_1,ldots,X_n$).
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
answered Jan 22 at 15:31
Jose BroxJose Brox
3,15711128
3,15711128
1
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
add a comment |
1
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
1
1
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
$begingroup$
+1 very thorough
$endgroup$
– rschwieb
Jan 22 at 15:46
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%2f3083286%2fwhat-is-meant-by-langle-x-1-x-n-rangle-in-algebra%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
2
$begingroup$
As a tip, you can use
langleandrangle, respectively, to produce $langle,rangle$.$endgroup$
– Clayton
Jan 22 at 15:26