Prime ideals in quadratic ring $mathbb{Z}[sqrt{-5}]$
Consider quadratic ring $mathbb{Z}[sqrt{-5}]$. For each of the following elements tell whether or not the principal ideal $langle xrangle$ generated by $x$ is a prime ideal.
$x=29,11.$
My approach: Firstly, I proved that $29$ is reducible element and $11$ is irreducible element in this ring.
1) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 29rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 29rangle cong mathbb{Z}_{29}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{29}[sqrt{-5}]$ is not integral domain because $(3+2sqrt{-5})(3-2sqrt{-5})=0.$ Hence, an ideal $langle 29rangle$ is NOT prime.
2) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 11rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 11rangle cong mathbb{Z}_{11}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{11}[sqrt{-5}]$ is integral domain.
How did I show that it is an integral domain? Suppose $(a+bsqrt{-5})(c+dsqrt{-5})=0$, where $a,b,c,din mathbb{Z}_{11}$. Multiplying by conjugates we get that $(a^2+5b^2)(c^2+5d^2)=0$ in $mathbb{Z}_{29}$. But since it is field then at least one of them is zero. WLOG suppose that $a^2+5b^2=0$. And after some trial-error method with $pmod {29}$ I have found that $a=b=0$ in $mathbb{Z}_{11}$.
1) Is my solution and approach correct?
2) Is there is another one more shorter I'd be happy to see it.
3) Also how to apply such method for ideal generated by $1+2sqrt{-5}$?
ring-theory
asked Dec 7 at 21:48
ZFR
4,95631338
add a comment |
Consider quadratic ring $mathbb{Z}[sqrt{-5}]$. For each of the following elements tell whether or not the principal ideal $langle xrangle$ generated by $x$ is a prime ideal.
$x=29,11.$
My approach: Firstly, I proved that $29$ is reducible element and $11$ is irreducible element in this ring.
1) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 29rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 29rangle cong mathbb{Z}_{29}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{29}[sqrt{-5}]$ is not integral domain because $(3+2sqrt{-5})(3-2sqrt{-5})=0.$ Hence, an ideal $langle 29rangle$ is NOT prime.
2) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 11rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 11rangle cong mathbb{Z}_{11}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{11}[sqrt{-5}]$ is integral domain.
How did I show that it is an integral domain? Suppose $(a+bsqrt{-5})(c+dsqrt{-5})=0$, where $a,b,c,din mathbb{Z}_{11}$. Multiplying by conjugates we get that $(a^2+5b^2)(c^2+5d^2)=0$ in $mathbb{Z}_{29}$. But since it is field then at least one of them is zero. WLOG suppose that $a^2+5b^2=0$. And after some trial-error method with $pmod {29}$ I have found that $a=b=0$ in $mathbb{Z}_{11}$.
1) Is my solution and approach correct?
2) Is there is another one more shorter I'd be happy to see it.
3) Also how to apply such method for ideal generated by $1+2sqrt{-5}$?
ring-theory
asked Dec 7 at 21:48
ZFR
4,95631338
1
If you haven't already, having a look at Dedekind's factorisation criterion might help.
– ODF
Dec 7 at 23:52
add a comment |
Consider quadratic ring $mathbb{Z}[sqrt{-5}]$. For each of the following elements tell whether or not the principal ideal $langle xrangle$ generated by $x$ is a prime ideal.
$x=29,11.$
My approach: Firstly, I proved that $29$ is reducible element and $11$ is irreducible element in this ring.
1) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 29rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 29rangle cong mathbb{Z}_{29}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{29}[sqrt{-5}]$ is not integral domain because $(3+2sqrt{-5})(3-2sqrt{-5})=0.$ Hence, an ideal $langle 29rangle$ is NOT prime.
2) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 11rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 11rangle cong mathbb{Z}_{11}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{11}[sqrt{-5}]$ is integral domain.
How did I show that it is an integral domain? Suppose $(a+bsqrt{-5})(c+dsqrt{-5})=0$, where $a,b,c,din mathbb{Z}_{11}$. Multiplying by conjugates we get that $(a^2+5b^2)(c^2+5d^2)=0$ in $mathbb{Z}_{29}$. But since it is field then at least one of them is zero. WLOG suppose that $a^2+5b^2=0$. And after some trial-error method with $pmod {29}$ I have found that $a=b=0$ in $mathbb{Z}_{11}$.
1) Is my solution and approach correct?
2) Is there is another one more shorter I'd be happy to see it.
3) Also how to apply such method for ideal generated by $1+2sqrt{-5}$?
ring-theory
asked Dec 7 at 21:48
ZFR
4,95631338
Consider quadratic ring $mathbb{Z}[sqrt{-5}]$. For each of the following elements tell whether or not the principal ideal $langle xrangle$ generated by $x$ is a prime ideal.
$x=29,11.$
My approach: Firstly, I proved that $29$ is reducible element and $11$ is irreducible element in this ring.
1) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 29rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 29rangle cong mathbb{Z}_{29}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{29}[sqrt{-5}]$ is not integral domain because $(3+2sqrt{-5})(3-2sqrt{-5})=0.$ Hence, an ideal $langle 29rangle$ is NOT prime.
2) Consider the quotient-ring $mathbb{Z}[sqrt{-5}]/langle 11rangle$ and we can show that $mathbb{Z}[sqrt{-5}]/langle 11rangle cong mathbb{Z}_{11}[sqrt{-5}]$. After trial-error I've found that $mathbb{Z}_{11}[sqrt{-5}]$ is integral domain.
How did I show that it is an integral domain? Suppose $(a+bsqrt{-5})(c+dsqrt{-5})=0$, where $a,b,c,din mathbb{Z}_{11}$. Multiplying by conjugates we get that $(a^2+5b^2)(c^2+5d^2)=0$ in $mathbb{Z}_{29}$. But since it is field then at least one of them is zero. WLOG suppose that $a^2+5b^2=0$. And after some trial-error method with $pmod {29}$ I have found that $a=b=0$ in $mathbb{Z}_{11}$.
1) Is my solution and approach correct?
2) Is there is another one more shorter I'd be happy to see it.
3) Also how to apply such method for ideal generated by $1+2sqrt{-5}$?
ring-theory
ring-theory
asked Dec 7 at 21:48
ZFR
4,95631338
asked Dec 7 at 21:48
ZFR
4,95631338
asked Dec 7 at 21:48
ZFR
4,95631338
asked Dec 7 at 21:48
ZFR
4,95631338
asked Dec 7 at 21:48
ZFR
4,95631338
4,95631338
1
If you haven't already, having a look at Dedekind's factorisation criterion might help.
– ODF
Dec 7 at 23:52
add a comment |
1
If you haven't already, having a look at Dedekind's factorisation criterion might help.
– ODF
Dec 7 at 23:52
1
1
If you haven't already, having a look at Dedekind's factorisation criterion might help.
– ODF
Dec 7 at 23:52
If you haven't already, having a look at Dedekind's factorisation criterion might help.
– ODF
Dec 7 at 23:52
add a comment |
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%2f3030399%2fprime-ideals-in-quadratic-ring-mathbbz-sqrt-5%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
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%2f3030399%2fprime-ideals-in-quadratic-ring-mathbbz-sqrt-5%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
If you haven't already, having a look at Dedekind's factorisation criterion might help.
– ODF
Dec 7 at 23:52