Nil radical of $(n)$ in $mathbb{Z}$
$begingroup$
I did not understand the following proof, from time to time the questions will be inserted at the critical point.
In the ring $mathbb{Z}$, let us show that if $n=p_1^{k_1}p_2^{k_2}cdots p_r^{k_r}$ is a factorization of the positive integer $nne 1$ into distinct primes $p_j$, then $$sqrt(n)=(p_1cdots p_r).$$
Indeed, if the integer $a=p_1cdots p_r$ and $k=max{k_1,k_2dots,k_r}$, then we have $a^kin (n)$ (Why?); this makes clear that $(p_1p_2cdots p_r)subseteq sqrt (n)$ (Why?).
On the other hand, if some positive integral power of the integer $m$ is divisible by $n$ (that is, if $minsqrt (n)$), then $m$ itself must be divisible by each of the primes $p_1, p_2,dots, p_r$ (why?) and, hence, a member of the ideal $$(p_1)cap (p_2)capcdotscap(p_r)=(p_1p_2cdots p_r)quadtext{Why?}.$$
Thanks for your patience!
abstract-algebra proof-explanation
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
$endgroup$
add a comment |
$begingroup$
I did not understand the following proof, from time to time the questions will be inserted at the critical point.
In the ring $mathbb{Z}$, let us show that if $n=p_1^{k_1}p_2^{k_2}cdots p_r^{k_r}$ is a factorization of the positive integer $nne 1$ into distinct primes $p_j$, then $$sqrt(n)=(p_1cdots p_r).$$
Indeed, if the integer $a=p_1cdots p_r$ and $k=max{k_1,k_2dots,k_r}$, then we have $a^kin (n)$ (Why?); this makes clear that $(p_1p_2cdots p_r)subseteq sqrt (n)$ (Why?).
On the other hand, if some positive integral power of the integer $m$ is divisible by $n$ (that is, if $minsqrt (n)$), then $m$ itself must be divisible by each of the primes $p_1, p_2,dots, p_r$ (why?) and, hence, a member of the ideal $$(p_1)cap (p_2)capcdotscap(p_r)=(p_1p_2cdots p_r)quadtext{Why?}.$$
Thanks for your patience!
abstract-algebra proof-explanation
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
$endgroup$
add a comment |
$begingroup$
I did not understand the following proof, from time to time the questions will be inserted at the critical point.
In the ring $mathbb{Z}$, let us show that if $n=p_1^{k_1}p_2^{k_2}cdots p_r^{k_r}$ is a factorization of the positive integer $nne 1$ into distinct primes $p_j$, then $$sqrt(n)=(p_1cdots p_r).$$
Indeed, if the integer $a=p_1cdots p_r$ and $k=max{k_1,k_2dots,k_r}$, then we have $a^kin (n)$ (Why?); this makes clear that $(p_1p_2cdots p_r)subseteq sqrt (n)$ (Why?).
On the other hand, if some positive integral power of the integer $m$ is divisible by $n$ (that is, if $minsqrt (n)$), then $m$ itself must be divisible by each of the primes $p_1, p_2,dots, p_r$ (why?) and, hence, a member of the ideal $$(p_1)cap (p_2)capcdotscap(p_r)=(p_1p_2cdots p_r)quadtext{Why?}.$$
Thanks for your patience!
abstract-algebra proof-explanation
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
$endgroup$
I did not understand the following proof, from time to time the questions will be inserted at the critical point.
In the ring $mathbb{Z}$, let us show that if $n=p_1^{k_1}p_2^{k_2}cdots p_r^{k_r}$ is a factorization of the positive integer $nne 1$ into distinct primes $p_j$, then $$sqrt(n)=(p_1cdots p_r).$$
Indeed, if the integer $a=p_1cdots p_r$ and $k=max{k_1,k_2dots,k_r}$, then we have $a^kin (n)$ (Why?); this makes clear that $(p_1p_2cdots p_r)subseteq sqrt (n)$ (Why?).
On the other hand, if some positive integral power of the integer $m$ is divisible by $n$ (that is, if $minsqrt (n)$), then $m$ itself must be divisible by each of the primes $p_1, p_2,dots, p_r$ (why?) and, hence, a member of the ideal $$(p_1)cap (p_2)capcdotscap(p_r)=(p_1p_2cdots p_r)quadtext{Why?}.$$
Thanks for your patience!
abstract-algebra proof-explanation
abstract-algebra proof-explanation
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
asked Dec 3 '18 at 16:25
Jack J.Jack J.
4701419
4701419
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Remember the definition of $$sqrt{(n)}:={rin Bbb Z : r^sin (n), sin Bbb Z (squad text{varies with}quad r}.$$To show that $(p_1p_2cdots p_r)subseteq sqrt{(n)}$ you can prove that $p_1p_2cdots p_rin sqrt{(n)}$ and this follows by definition of $sqrt{(n)}$. On the other hand, if $min sqrt {(n)}$ then exist $sin Bbb Z$ such that $m^sin (n)$. This implies that $n | m$ , $p_1 |m cdots p_r | m$ and therefore $min (p_1)capcdotscap (p_r)$. I believe that you can prove the last “small proposition”.
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
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%2f3024297%2fnil-radical-of-n-in-mathbbz%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
$begingroup$
Remember the definition of $$sqrt{(n)}:={rin Bbb Z : r^sin (n), sin Bbb Z (squad text{varies with}quad r}.$$To show that $(p_1p_2cdots p_r)subseteq sqrt{(n)}$ you can prove that $p_1p_2cdots p_rin sqrt{(n)}$ and this follows by definition of $sqrt{(n)}$. On the other hand, if $min sqrt {(n)}$ then exist $sin Bbb Z$ such that $m^sin (n)$. This implies that $n | m$ , $p_1 |m cdots p_r | m$ and therefore $min (p_1)capcdotscap (p_r)$. I believe that you can prove the last “small proposition”.
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
Remember the definition of $$sqrt{(n)}:={rin Bbb Z : r^sin (n), sin Bbb Z (squad text{varies with}quad r}.$$To show that $(p_1p_2cdots p_r)subseteq sqrt{(n)}$ you can prove that $p_1p_2cdots p_rin sqrt{(n)}$ and this follows by definition of $sqrt{(n)}$. On the other hand, if $min sqrt {(n)}$ then exist $sin Bbb Z$ such that $m^sin (n)$. This implies that $n | m$ , $p_1 |m cdots p_r | m$ and therefore $min (p_1)capcdotscap (p_r)$. I believe that you can prove the last “small proposition”.
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
Remember the definition of $$sqrt{(n)}:={rin Bbb Z : r^sin (n), sin Bbb Z (squad text{varies with}quad r}.$$To show that $(p_1p_2cdots p_r)subseteq sqrt{(n)}$ you can prove that $p_1p_2cdots p_rin sqrt{(n)}$ and this follows by definition of $sqrt{(n)}$. On the other hand, if $min sqrt {(n)}$ then exist $sin Bbb Z$ such that $m^sin (n)$. This implies that $n | m$ , $p_1 |m cdots p_r | m$ and therefore $min (p_1)capcdotscap (p_r)$. I believe that you can prove the last “small proposition”.
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
Remember the definition of $$sqrt{(n)}:={rin Bbb Z : r^sin (n), sin Bbb Z (squad text{varies with}quad r}.$$To show that $(p_1p_2cdots p_r)subseteq sqrt{(n)}$ you can prove that $p_1p_2cdots p_rin sqrt{(n)}$ and this follows by definition of $sqrt{(n)}$. On the other hand, if $min sqrt {(n)}$ then exist $sin Bbb Z$ such that $m^sin (n)$. This implies that $n | m$ , $p_1 |m cdots p_r | m$ and therefore $min (p_1)capcdotscap (p_r)$. I believe that you can prove the last “small proposition”.
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
answered Dec 3 '18 at 20:08
Domenico VuonoDomenico Vuono
2,3161523
2,3161523
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.
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%2f3024297%2fnil-radical-of-n-in-mathbbz%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
