Justification of an isomorphism between $mathbb{Z}[t]/(t,3)$ and $mathbb{Z}/3$
up vote
2
down vote
favorite
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
add a comment |
up vote
2
down vote
favorite
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 2 at 9:24
Asaf Karagila♦
301k32422753
301k32422753
asked Dec 2 at 3:23
J. Wang
524
524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
up vote
2
down vote
It's legitimate and a consequence of the isomorphism theorems.
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',
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%2f3022187%2fjustification-of-an-isomorphism-between-mathbbzt-t-3-and-mathbbz-3%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
up vote
3
down vote
accepted
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
up vote
3
down vote
accepted
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
edited Dec 3 at 3:57
answered Dec 2 at 4:23
Indrajit Ghosh
1,0491718
1,0491718
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
up vote
2
down vote
It's legitimate and a consequence of the isomorphism theorems.
add a comment |
up vote
2
down vote
It's legitimate and a consequence of the isomorphism theorems.
add a comment |
up vote
2
down vote
up vote
2
down vote
It's legitimate and a consequence of the isomorphism theorems.
It's legitimate and a consequence of the isomorphism theorems.
answered Dec 2 at 4:24
CyclotomicField
2,1541312
2,1541312
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%2f3022187%2fjustification-of-an-isomorphism-between-mathbbzt-t-3-and-mathbbz-3%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