Trouble understanding proof regarding elements of an integral domain which are not product of irreducible...
$begingroup$
I'm having trouble understanding the proof of a proposition regarding elements which are not product of irreducible elements in integral domains. The proposition is the following:
Let $A$ be an integral domain and $ain A$ different of 0 and not a unit. If $a$ is not product of irreducible elements then there exists a sequence ${a_n}_{nin N}$ of elements of $A$ such that $a_{n+1}$ is a proper divisor of $a_n$ for every natural $n$.
The proof that has been given to me is this:
Let's build inductively a sequence with this property. Let $a_0 = a$ and suppose $a_0,dots,a_n$, $ngeq0$ are already built. Since $a_n$ is not a product of irreducibles, it must be composite, thus $a_n = bc$ with $b,c$ proper divisors of $a_n$. Clearly at least one of the two factors must not be a product of irreducibles. We define $a_{n+1}$ as this factor.
The problem I have understanding this proof is that I don't get why not being product of irreducibles implies it is composite. Wouldn't such elements be irreducibles themselves? I mean if it is composite, couldn't we keep decomposing its factors until all of them are irreducible?
Thanks for your help.
ring-theory integral-domain unique-factorization-domains
edited Jan 15 at 23:42
Bernard
125k743119
asked Jan 15 at 23:38
GenisGenis
485
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding the proof of a proposition regarding elements which are not product of irreducible elements in integral domains. The proposition is the following:
Let $A$ be an integral domain and $ain A$ different of 0 and not a unit. If $a$ is not product of irreducible elements then there exists a sequence ${a_n}_{nin N}$ of elements of $A$ such that $a_{n+1}$ is a proper divisor of $a_n$ for every natural $n$.
The proof that has been given to me is this:
Let's build inductively a sequence with this property. Let $a_0 = a$ and suppose $a_0,dots,a_n$, $ngeq0$ are already built. Since $a_n$ is not a product of irreducibles, it must be composite, thus $a_n = bc$ with $b,c$ proper divisors of $a_n$. Clearly at least one of the two factors must not be a product of irreducibles. We define $a_{n+1}$ as this factor.
The problem I have understanding this proof is that I don't get why not being product of irreducibles implies it is composite. Wouldn't such elements be irreducibles themselves? I mean if it is composite, couldn't we keep decomposing its factors until all of them are irreducible?
Thanks for your help.
ring-theory integral-domain unique-factorization-domains
edited Jan 15 at 23:42
Bernard
125k743119
asked Jan 15 at 23:38
GenisGenis
485
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding the proof of a proposition regarding elements which are not product of irreducible elements in integral domains. The proposition is the following:
Let $A$ be an integral domain and $ain A$ different of 0 and not a unit. If $a$ is not product of irreducible elements then there exists a sequence ${a_n}_{nin N}$ of elements of $A$ such that $a_{n+1}$ is a proper divisor of $a_n$ for every natural $n$.
The proof that has been given to me is this:
Let's build inductively a sequence with this property. Let $a_0 = a$ and suppose $a_0,dots,a_n$, $ngeq0$ are already built. Since $a_n$ is not a product of irreducibles, it must be composite, thus $a_n = bc$ with $b,c$ proper divisors of $a_n$. Clearly at least one of the two factors must not be a product of irreducibles. We define $a_{n+1}$ as this factor.
The problem I have understanding this proof is that I don't get why not being product of irreducibles implies it is composite. Wouldn't such elements be irreducibles themselves? I mean if it is composite, couldn't we keep decomposing its factors until all of them are irreducible?
Thanks for your help.
ring-theory integral-domain unique-factorization-domains
edited Jan 15 at 23:42
Bernard
125k743119
asked Jan 15 at 23:38
GenisGenis
485
$endgroup$
I'm having trouble understanding the proof of a proposition regarding elements which are not product of irreducible elements in integral domains. The proposition is the following:
Let $A$ be an integral domain and $ain A$ different of 0 and not a unit. If $a$ is not product of irreducible elements then there exists a sequence ${a_n}_{nin N}$ of elements of $A$ such that $a_{n+1}$ is a proper divisor of $a_n$ for every natural $n$.
The proof that has been given to me is this:
Let's build inductively a sequence with this property. Let $a_0 = a$ and suppose $a_0,dots,a_n$, $ngeq0$ are already built. Since $a_n$ is not a product of irreducibles, it must be composite, thus $a_n = bc$ with $b,c$ proper divisors of $a_n$. Clearly at least one of the two factors must not be a product of irreducibles. We define $a_{n+1}$ as this factor.
The problem I have understanding this proof is that I don't get why not being product of irreducibles implies it is composite. Wouldn't such elements be irreducibles themselves? I mean if it is composite, couldn't we keep decomposing its factors until all of them are irreducible?
Thanks for your help.
ring-theory integral-domain unique-factorization-domains
ring-theory integral-domain unique-factorization-domains
edited Jan 15 at 23:42
Bernard
125k743119
asked Jan 15 at 23:38
GenisGenis
485
edited Jan 15 at 23:42
Bernard
125k743119
asked Jan 15 at 23:38
GenisGenis
485
edited Jan 15 at 23:42
Bernard
125k743119
edited Jan 15 at 23:42
Bernard
125k743119
edited Jan 15 at 23:42
Bernard
125k743119
125k743119
asked Jan 15 at 23:38
GenisGenis
485
asked Jan 15 at 23:38
GenisGenis
485
asked Jan 15 at 23:38
GenisGenis
485
485
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
"Composite" here just means "not irreducible (and not a unit)". Since $a_n$ is not a product of irreducibles, it in particular is not irreducible (since then it would be a product of $1$ irreducible, itself). So, by definition of "irreducible", this means there exist $b$ and $c$ which are not units such that $a_n=bc$.
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
$endgroup$
add a comment |
Your Answer
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%2f3075122%2ftrouble-understanding-proof-regarding-elements-of-an-integral-domain-which-are-n%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
$begingroup$
"Composite" here just means "not irreducible (and not a unit)". Since $a_n$ is not a product of irreducibles, it in particular is not irreducible (since then it would be a product of $1$ irreducible, itself). So, by definition of "irreducible", this means there exist $b$ and $c$ which are not units such that $a_n=bc$.
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
$endgroup$
add a comment |
$begingroup$
"Composite" here just means "not irreducible (and not a unit)". Since $a_n$ is not a product of irreducibles, it in particular is not irreducible (since then it would be a product of $1$ irreducible, itself). So, by definition of "irreducible", this means there exist $b$ and $c$ which are not units such that $a_n=bc$.
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
$endgroup$
add a comment |
$begingroup$
"Composite" here just means "not irreducible (and not a unit)". Since $a_n$ is not a product of irreducibles, it in particular is not irreducible (since then it would be a product of $1$ irreducible, itself). So, by definition of "irreducible", this means there exist $b$ and $c$ which are not units such that $a_n=bc$.
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
$endgroup$
"Composite" here just means "not irreducible (and not a unit)". Since $a_n$ is not a product of irreducibles, it in particular is not irreducible (since then it would be a product of $1$ irreducible, itself). So, by definition of "irreducible", this means there exist $b$ and $c$ which are not units such that $a_n=bc$.
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
answered Jan 15 at 23:41
Eric WofseyEric Wofsey
194k14222353
194k14222353
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%2f3075122%2ftrouble-understanding-proof-regarding-elements-of-an-integral-domain-which-are-n%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
