Mixing
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
asked Nov 21 '18 at 1:22
lomber
777220
add a comment |
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
asked Nov 21 '18 at 1:22
lomber
777220
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 '18 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 '18 at 1:47
@RobArthan thanks u
– lomber
Nov 21 '18 at 9:31
@ThomasShelby okss
– lomber
Nov 21 '18 at 9:31
add a comment |
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
asked Nov 21 '18 at 1:22
lomber
777220
find the number of Ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
My answer : one
let $f$ be a ring homorphism from $mathbb{Z_2} times mathbb{Z_2}$ to $mathbb{Z_4}$
we know that $mathbb{Z_2} times mathbb{Z_2}= K_4$ where $K_4$ is Klien Group
$K_4 = {a^2 =b^2 =c^2 =e text{and} c=ab}$
There is only one idempotent element $e$ and there are $3$ nilpotent elements $a,b,c$ .
Now in $mathbb{Z_4} $ we have
$0^2=0,$
$1^2=1,$
$2^2=0,$
$3^2= 1$
so, $0$ and $1$ are idempotent elements
we know that in a ring Homomorphism idempotent map to idempotent and nilpotent map to nilpotents.
so here I get only one ring homorphism by the diagram given below
$K_4 mathbb{Z_4}$
$e rightarrow 0$
$ 1$
$(a,b,c) rightarrow 2$
$ 3$
Is its correct ?
Is their any mistake in my solution?
thanks u
abstract-algebra
abstract-algebra
asked Nov 21 '18 at 1:22
lomber
777220
asked Nov 21 '18 at 1:22
lomber
777220
asked Nov 21 '18 at 1:22
lomber
777220
asked Nov 21 '18 at 1:22
lomber
777220
asked Nov 21 '18 at 1:22
lomber
777220
777220
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 '18 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 '18 at 1:47
@RobArthan thanks u
– lomber
Nov 21 '18 at 9:31
@ThomasShelby okss
– lomber
Nov 21 '18 at 9:31
add a comment |
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 '18 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 '18 at 1:47
@RobArthan thanks u
– lomber
Nov 21 '18 at 9:31
@ThomasShelby okss
– lomber
Nov 21 '18 at 9:31
1
1
What about trivial homomorphism?
– Thomas Shelby
Nov 21 '18 at 1:47
What about trivial homomorphism?
– Thomas Shelby
Nov 21 '18 at 1:47
1
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 '18 at 1:47
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 '18 at 1:47
@RobArthan thanks u
– lomber
Nov 21 '18 at 9:31
@RobArthan thanks u
– lomber
Nov 21 '18 at 9:31
@ThomasShelby okss
– lomber
Nov 21 '18 at 9:31
@ThomasShelby okss
– lomber
Nov 21 '18 at 9:31
add a comment |
0
active
oldest
votes
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%2f3007133%2ffind-the-number-of-ring-homorphism-from-mathbbz-2-times-mathbbz-2-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3007133%2ffind-the-number-of-ring-homorphism-from-mathbbz-2-times-mathbbz-2-to%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
What about trivial homomorphism?
– Thomas Shelby
Nov 21 '18 at 1:47
1
You are confusing the additive structure and the multiplicative structure in $Bbb{Z}_2timesBbb{Z}_2$: it has no nilpotent elements and all its elements are idempotent. The answer to your question depends on whether you are working with rings with $1$ or not. If you are working with rngs there are are some homomorphisms; if you are working with rings with $1$ there are not (or only the trivial homomorphism).
– Rob Arthan
Nov 21 '18 at 1:47
@RobArthan thanks u
– lomber
Nov 21 '18 at 9:31
@ThomasShelby okss
– lomber
Nov 21 '18 at 9:31