Surjective ring homomorphism from $M_n(R)$ to $M_n(R/I)$ where $R$ is a ring and $I$ is an ideal for R?
I'm looking for such a surjective homomorphism. I was thinking of starting from the canonical surjection from $R$ to $R/I$ and induce one but somehow I get stuck... Can you help me?
Thank you very much!
linear-algebra ring-theory abstract-algebra
asked Oct 15 '14 at 16:27
Noome
15719
add a comment |
I'm looking for such a surjective homomorphism. I was thinking of starting from the canonical surjection from $R$ to $R/I$ and induce one but somehow I get stuck... Can you help me?
Thank you very much!
linear-algebra ring-theory abstract-algebra
asked Oct 15 '14 at 16:27
Noome
15719
2
Apply $Rto R/I$ to each entry.
– Hagen von Eitzen
Oct 15 '14 at 16:36
@HagenvonEitzen That'd make a good hint-answer. If you really hate rep you can always CW it.
– rschwieb
Oct 15 '14 at 16:52
Don't I get cosets as entries then? I don't understand that...
– Noome
Oct 15 '14 at 18:01
@Noome Don't I get cosets as entries then yes, and $R/I$ is a set of cosets. What's unexpected about this? :)
– rschwieb
Oct 15 '14 at 18:28
Example: let $n = 2$, $R=mathbb{Z}$, and $I= 2mathbb{Z}$, then for each 2x2 matrix, take the map that sends each entry to $mathbb{Z} /2mathbb{Z}$. That is, sends entries divisible by $2$ to $0$ and all other entries to $1$.
– Kevin Johnson
Oct 27 '14 at 6:23
add a comment |
I'm looking for such a surjective homomorphism. I was thinking of starting from the canonical surjection from $R$ to $R/I$ and induce one but somehow I get stuck... Can you help me?
Thank you very much!
linear-algebra ring-theory abstract-algebra
asked Oct 15 '14 at 16:27
Noome
15719
I'm looking for such a surjective homomorphism. I was thinking of starting from the canonical surjection from $R$ to $R/I$ and induce one but somehow I get stuck... Can you help me?
Thank you very much!
linear-algebra ring-theory abstract-algebra
linear-algebra ring-theory abstract-algebra
asked Oct 15 '14 at 16:27
Noome
15719
asked Oct 15 '14 at 16:27
Noome
15719
asked Oct 15 '14 at 16:27
Noome
15719
asked Oct 15 '14 at 16:27
Noome
15719
asked Oct 15 '14 at 16:27
Noome
15719
15719
2
Apply $Rto R/I$ to each entry.
– Hagen von Eitzen
Oct 15 '14 at 16:36
@HagenvonEitzen That'd make a good hint-answer. If you really hate rep you can always CW it.
– rschwieb
Oct 15 '14 at 16:52
Don't I get cosets as entries then? I don't understand that...
– Noome
Oct 15 '14 at 18:01
@Noome Don't I get cosets as entries then yes, and $R/I$ is a set of cosets. What's unexpected about this? :)
– rschwieb
Oct 15 '14 at 18:28
Example: let $n = 2$, $R=mathbb{Z}$, and $I= 2mathbb{Z}$, then for each 2x2 matrix, take the map that sends each entry to $mathbb{Z} /2mathbb{Z}$. That is, sends entries divisible by $2$ to $0$ and all other entries to $1$.
– Kevin Johnson
Oct 27 '14 at 6:23
add a comment |
2
Apply $Rto R/I$ to each entry.
– Hagen von Eitzen
Oct 15 '14 at 16:36
@HagenvonEitzen That'd make a good hint-answer. If you really hate rep you can always CW it.
– rschwieb
Oct 15 '14 at 16:52
Don't I get cosets as entries then? I don't understand that...
– Noome
Oct 15 '14 at 18:01
@Noome Don't I get cosets as entries then yes, and $R/I$ is a set of cosets. What's unexpected about this? :)
– rschwieb
Oct 15 '14 at 18:28
Example: let $n = 2$, $R=mathbb{Z}$, and $I= 2mathbb{Z}$, then for each 2x2 matrix, take the map that sends each entry to $mathbb{Z} /2mathbb{Z}$. That is, sends entries divisible by $2$ to $0$ and all other entries to $1$.
– Kevin Johnson
Oct 27 '14 at 6:23
2
2
Apply $Rto R/I$ to each entry.
– Hagen von Eitzen
Oct 15 '14 at 16:36
Apply $Rto R/I$ to each entry.
– Hagen von Eitzen
Oct 15 '14 at 16:36
@HagenvonEitzen That'd make a good hint-answer. If you really hate rep you can always CW it.
– rschwieb
Oct 15 '14 at 16:52
@HagenvonEitzen That'd make a good hint-answer. If you really hate rep you can always CW it.
– rschwieb
Oct 15 '14 at 16:52
Don't I get cosets as entries then? I don't understand that...
– Noome
Oct 15 '14 at 18:01
Don't I get cosets as entries then? I don't understand that...
– Noome
Oct 15 '14 at 18:01
@Noome Don't I get cosets as entries then yes, and $R/I$ is a set of cosets. What's unexpected about this? :)
– rschwieb
Oct 15 '14 at 18:28
@Noome Don't I get cosets as entries then yes, and $R/I$ is a set of cosets. What's unexpected about this? :)
– rschwieb
Oct 15 '14 at 18:28
Example: let $n = 2$, $R=mathbb{Z}$, and $I= 2mathbb{Z}$, then for each 2x2 matrix, take the map that sends each entry to $mathbb{Z} /2mathbb{Z}$. That is, sends entries divisible by $2$ to $0$ and all other entries to $1$.
– Kevin Johnson
Oct 27 '14 at 6:23
Example: let $n = 2$, $R=mathbb{Z}$, and $I= 2mathbb{Z}$, then for each 2x2 matrix, take the map that sends each entry to $mathbb{Z} /2mathbb{Z}$. That is, sends entries divisible by $2$ to $0$ and all other entries to $1$.
– Kevin Johnson
Oct 27 '14 at 6:23
add a comment |
1 Answer
1
active
oldest
votes
If $n > 0$, we have a functor $M_n : operatorname{Ring} to operatorname{Ring}$ where $M_n(R)$ is the ring of $n times n$ matrices with entries in $R$. If $f : R to S$ is a ring homomorphism, then $M_n(f) : M_n(R) to M_n(S)$ is a ring homomorphism defined by $$M_n(f)Bigg( begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} f(a_{11}) & cdots & f(a_{1n}) \ vdots & & vdots \ f(a_{n1}) & cdots & f(a_{nn})end{bmatrix}.$$
In our specific case, we let $R$ be a ring and $I$ an ideal, so we can consider the ring $R/I$ and the quotient homomorphism: $f : R to R/I$ defined by $f(r) = overline r = r + I$. Notice that our homormorphism is clearly surjective.
Let's pick an arbitrary element in $M_n(R/I)$: $$begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix} = begin{bmatrix} a_{11} + I & cdots & a_{1n} + I \ vdots & & vdots \ a_{n1} + I & cdots & a_{nn} + Iend{bmatrix}$$ where we can choose the obvious representatives: for $overline{a_{ij}}$ we pick $a_{ij} in R$ and notice that $f(a_{ij}) = overline{a_{ij}} = a_{ij} + I$.
It should be clear that $$begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}$$ is an element such that $$M_n(f)Bigg(begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix}.$$
Hence the surjective homomorphism you requested should be $M_n(f)$ where $f$ is the canonical quotient map.
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
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%2f975219%2fsurjective-ring-homomorphism-from-m-nr-to-m-nr-i-where-r-is-a-ring-and%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
If $n > 0$, we have a functor $M_n : operatorname{Ring} to operatorname{Ring}$ where $M_n(R)$ is the ring of $n times n$ matrices with entries in $R$. If $f : R to S$ is a ring homomorphism, then $M_n(f) : M_n(R) to M_n(S)$ is a ring homomorphism defined by $$M_n(f)Bigg( begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} f(a_{11}) & cdots & f(a_{1n}) \ vdots & & vdots \ f(a_{n1}) & cdots & f(a_{nn})end{bmatrix}.$$
In our specific case, we let $R$ be a ring and $I$ an ideal, so we can consider the ring $R/I$ and the quotient homomorphism: $f : R to R/I$ defined by $f(r) = overline r = r + I$. Notice that our homormorphism is clearly surjective.
Let's pick an arbitrary element in $M_n(R/I)$: $$begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix} = begin{bmatrix} a_{11} + I & cdots & a_{1n} + I \ vdots & & vdots \ a_{n1} + I & cdots & a_{nn} + Iend{bmatrix}$$ where we can choose the obvious representatives: for $overline{a_{ij}}$ we pick $a_{ij} in R$ and notice that $f(a_{ij}) = overline{a_{ij}} = a_{ij} + I$.
It should be clear that $$begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}$$ is an element such that $$M_n(f)Bigg(begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix}.$$
Hence the surjective homomorphism you requested should be $M_n(f)$ where $f$ is the canonical quotient map.
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
add a comment |
If $n > 0$, we have a functor $M_n : operatorname{Ring} to operatorname{Ring}$ where $M_n(R)$ is the ring of $n times n$ matrices with entries in $R$. If $f : R to S$ is a ring homomorphism, then $M_n(f) : M_n(R) to M_n(S)$ is a ring homomorphism defined by $$M_n(f)Bigg( begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} f(a_{11}) & cdots & f(a_{1n}) \ vdots & & vdots \ f(a_{n1}) & cdots & f(a_{nn})end{bmatrix}.$$
In our specific case, we let $R$ be a ring and $I$ an ideal, so we can consider the ring $R/I$ and the quotient homomorphism: $f : R to R/I$ defined by $f(r) = overline r = r + I$. Notice that our homormorphism is clearly surjective.
Let's pick an arbitrary element in $M_n(R/I)$: $$begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix} = begin{bmatrix} a_{11} + I & cdots & a_{1n} + I \ vdots & & vdots \ a_{n1} + I & cdots & a_{nn} + Iend{bmatrix}$$ where we can choose the obvious representatives: for $overline{a_{ij}}$ we pick $a_{ij} in R$ and notice that $f(a_{ij}) = overline{a_{ij}} = a_{ij} + I$.
It should be clear that $$begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}$$ is an element such that $$M_n(f)Bigg(begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix}.$$
Hence the surjective homomorphism you requested should be $M_n(f)$ where $f$ is the canonical quotient map.
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
add a comment |
If $n > 0$, we have a functor $M_n : operatorname{Ring} to operatorname{Ring}$ where $M_n(R)$ is the ring of $n times n$ matrices with entries in $R$. If $f : R to S$ is a ring homomorphism, then $M_n(f) : M_n(R) to M_n(S)$ is a ring homomorphism defined by $$M_n(f)Bigg( begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} f(a_{11}) & cdots & f(a_{1n}) \ vdots & & vdots \ f(a_{n1}) & cdots & f(a_{nn})end{bmatrix}.$$
In our specific case, we let $R$ be a ring and $I$ an ideal, so we can consider the ring $R/I$ and the quotient homomorphism: $f : R to R/I$ defined by $f(r) = overline r = r + I$. Notice that our homormorphism is clearly surjective.
Let's pick an arbitrary element in $M_n(R/I)$: $$begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix} = begin{bmatrix} a_{11} + I & cdots & a_{1n} + I \ vdots & & vdots \ a_{n1} + I & cdots & a_{nn} + Iend{bmatrix}$$ where we can choose the obvious representatives: for $overline{a_{ij}}$ we pick $a_{ij} in R$ and notice that $f(a_{ij}) = overline{a_{ij}} = a_{ij} + I$.
It should be clear that $$begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}$$ is an element such that $$M_n(f)Bigg(begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix}.$$
Hence the surjective homomorphism you requested should be $M_n(f)$ where $f$ is the canonical quotient map.
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
If $n > 0$, we have a functor $M_n : operatorname{Ring} to operatorname{Ring}$ where $M_n(R)$ is the ring of $n times n$ matrices with entries in $R$. If $f : R to S$ is a ring homomorphism, then $M_n(f) : M_n(R) to M_n(S)$ is a ring homomorphism defined by $$M_n(f)Bigg( begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} f(a_{11}) & cdots & f(a_{1n}) \ vdots & & vdots \ f(a_{n1}) & cdots & f(a_{nn})end{bmatrix}.$$
In our specific case, we let $R$ be a ring and $I$ an ideal, so we can consider the ring $R/I$ and the quotient homomorphism: $f : R to R/I$ defined by $f(r) = overline r = r + I$. Notice that our homormorphism is clearly surjective.
Let's pick an arbitrary element in $M_n(R/I)$: $$begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix} = begin{bmatrix} a_{11} + I & cdots & a_{1n} + I \ vdots & & vdots \ a_{n1} + I & cdots & a_{nn} + Iend{bmatrix}$$ where we can choose the obvious representatives: for $overline{a_{ij}}$ we pick $a_{ij} in R$ and notice that $f(a_{ij}) = overline{a_{ij}} = a_{ij} + I$.
It should be clear that $$begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}$$ is an element such that $$M_n(f)Bigg(begin{bmatrix} a_{11} & cdots & a_{1n} \ vdots & & vdots \ a_{n1} & cdots & a_{nn}end{bmatrix}Bigg) = begin{bmatrix} overline {a_{11}} & cdots &overline{ a_{1n}} \ vdots & & vdots \ overline{a_{n1}} & cdots & overline{a_{nn}}end{bmatrix}.$$
Hence the surjective homomorphism you requested should be $M_n(f)$ where $f$ is the canonical quotient map.
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
answered Feb 15 '15 at 5:31
Robert Cardona
5,23023498
5,23023498
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f975219%2fsurjective-ring-homomorphism-from-m-nr-to-m-nr-i-where-r-is-a-ring-and%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
Apply $Rto R/I$ to each entry.
– Hagen von Eitzen
Oct 15 '14 at 16:36
@HagenvonEitzen That'd make a good hint-answer. If you really hate rep you can always CW it.
– rschwieb
Oct 15 '14 at 16:52
Don't I get cosets as entries then? I don't understand that...
– Noome
Oct 15 '14 at 18:01
@Noome Don't I get cosets as entries then yes, and $R/I$ is a set of cosets. What's unexpected about this? :)
– rschwieb
Oct 15 '14 at 18:28
Example: let $n = 2$, $R=mathbb{Z}$, and $I= 2mathbb{Z}$, then for each 2x2 matrix, take the map that sends each entry to $mathbb{Z} /2mathbb{Z}$. That is, sends entries divisible by $2$ to $0$ and all other entries to $1$.
– Kevin Johnson
Oct 27 '14 at 6:23