Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?
Question: Let $F$ be a commutative ring with identity.
Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?
If $F$ is a field, clearly one can do division algorithm to prove that $F[X]$ is an Euclidean domain.
What puzzles me is the converse, that is, if $F[X]$ is an Euclidean domain, can we conclude that $F$ is a field?
My attempt:
Let $uin F$ be a nonzero element.
Consider $1,uin F[X].$
Since $F[X]$ is an ED, there exist $f(X),r(X)in F[X]$ such that
$$1 = uf(X) + r(X)$$
where $deg(r(X))<deg(f(X)).$
Since the constant polynomial $1$ has degree $0,$ it follows that $deg(f(X)) = 0.$
Denote $f(X) = vin F.$
Since $deg(r(X))<deg(f(X)) = 0,$ it implies that $r(X) = 0.$
Therefore, we have
$$1 = uv.$$
Hence, $u$ is a unit and thus $F$ is a field.
Is my attempt above correct?
abstract-algebra proof-verification polynomials field-theory
asked Nov 24 at 3:52
Idonknow
2,292749112
add a comment |
Question: Let $F$ be a commutative ring with identity.
Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?
If $F$ is a field, clearly one can do division algorithm to prove that $F[X]$ is an Euclidean domain.
What puzzles me is the converse, that is, if $F[X]$ is an Euclidean domain, can we conclude that $F$ is a field?
My attempt:
Let $uin F$ be a nonzero element.
Consider $1,uin F[X].$
Since $F[X]$ is an ED, there exist $f(X),r(X)in F[X]$ such that
$$1 = uf(X) + r(X)$$
where $deg(r(X))<deg(f(X)).$
Since the constant polynomial $1$ has degree $0,$ it follows that $deg(f(X)) = 0.$
Denote $f(X) = vin F.$
Since $deg(r(X))<deg(f(X)) = 0,$ it implies that $r(X) = 0.$
Therefore, we have
$$1 = uv.$$
Hence, $u$ is a unit and thus $F$ is a field.
Is my attempt above correct?
abstract-algebra proof-verification polynomials field-theory
asked Nov 24 at 3:52
Idonknow
2,292749112
5
You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$.
– Lord Shark the Unknown
Nov 24 at 4:11
@LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $fin F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit.
– Idonknow
Nov 24 at 5:01
1
There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment...
– KCd
Nov 24 at 6:37
Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :)
– darij grinberg
Nov 26 at 0:53
@darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that.
– KCd
Nov 26 at 0:58
add a comment |
Question: Let $F$ be a commutative ring with identity.
Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?
If $F$ is a field, clearly one can do division algorithm to prove that $F[X]$ is an Euclidean domain.
What puzzles me is the converse, that is, if $F[X]$ is an Euclidean domain, can we conclude that $F$ is a field?
My attempt:
Let $uin F$ be a nonzero element.
Consider $1,uin F[X].$
Since $F[X]$ is an ED, there exist $f(X),r(X)in F[X]$ such that
$$1 = uf(X) + r(X)$$
where $deg(r(X))<deg(f(X)).$
Since the constant polynomial $1$ has degree $0,$ it follows that $deg(f(X)) = 0.$
Denote $f(X) = vin F.$
Since $deg(r(X))<deg(f(X)) = 0,$ it implies that $r(X) = 0.$
Therefore, we have
$$1 = uv.$$
Hence, $u$ is a unit and thus $F$ is a field.
Is my attempt above correct?
abstract-algebra proof-verification polynomials field-theory
asked Nov 24 at 3:52
Idonknow
2,292749112
Question: Let $F$ be a commutative ring with identity.
Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?
If $F$ is a field, clearly one can do division algorithm to prove that $F[X]$ is an Euclidean domain.
What puzzles me is the converse, that is, if $F[X]$ is an Euclidean domain, can we conclude that $F$ is a field?
My attempt:
Let $uin F$ be a nonzero element.
Consider $1,uin F[X].$
Since $F[X]$ is an ED, there exist $f(X),r(X)in F[X]$ such that
$$1 = uf(X) + r(X)$$
where $deg(r(X))<deg(f(X)).$
Since the constant polynomial $1$ has degree $0,$ it follows that $deg(f(X)) = 0.$
Denote $f(X) = vin F.$
Since $deg(r(X))<deg(f(X)) = 0,$ it implies that $r(X) = 0.$
Therefore, we have
$$1 = uv.$$
Hence, $u$ is a unit and thus $F$ is a field.
Is my attempt above correct?
abstract-algebra proof-verification polynomials field-theory
abstract-algebra proof-verification polynomials field-theory
asked Nov 24 at 3:52
Idonknow
2,292749112
asked Nov 24 at 3:52
Idonknow
2,292749112
edited Nov 24 at 4:04
asked Nov 24 at 3:52
Idonknow
2,292749112
asked Nov 24 at 3:52
Idonknow
2,292749112
asked Nov 24 at 3:52
Idonknow
2,292749112
2,292749112
5
You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$.
– Lord Shark the Unknown
Nov 24 at 4:11
@LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $fin F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit.
– Idonknow
Nov 24 at 5:01
1
There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment...
– KCd
Nov 24 at 6:37
Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :)
– darij grinberg
Nov 26 at 0:53
@darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that.
– KCd
Nov 26 at 0:58
add a comment |
5
You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$.
– Lord Shark the Unknown
Nov 24 at 4:11
@LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $fin F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit.
– Idonknow
Nov 24 at 5:01
1
There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment...
– KCd
Nov 24 at 6:37
Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :)
– darij grinberg
Nov 26 at 0:53
@darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that.
– KCd
Nov 26 at 0:58
5
5
You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$.
– Lord Shark the Unknown
Nov 24 at 4:11
You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$.
– Lord Shark the Unknown
Nov 24 at 4:11
@LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $fin F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit.
– Idonknow
Nov 24 at 5:01
@LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $fin F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit.
– Idonknow
Nov 24 at 5:01
1
1
There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment...
– KCd
Nov 24 at 6:37
There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment...
– KCd
Nov 24 at 6:37
Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :)
– darij grinberg
Nov 26 at 0:53
Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :)
– darij grinberg
Nov 26 at 0:53
@darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that.
– KCd
Nov 26 at 0:58
@darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that.
– KCd
Nov 26 at 0:58
add a comment |
1 Answer
1
active
oldest
votes
No, that's not right. You're using Euclidean Division wrong. For one thing, you would have to conclude that $deg(r(x)) < deg(u)$, not $deg(f(x))$. This would actually imply that $r(x) = 0$, and thus that $f(x)$ was an inverse for $u$, and thus would have to be in $F$, implying that $F$ is a field. But, as Lord Shark the Unknown points out, this depends on assuming that the degree is the Euclidean function. Since we can't do this, we have to find some other approach.
It's actually easier to prove a stronger version of the statement, namely that if $F[x]$ is a PID then $F$ is a field. If $F[x]$ is a PID, $F$ must be an integral domain as well, and so $F[x]/langle x rangle cong F$ implies that $langle x rangle$ is prime. Since nonzero prime ideals are maximal in a PID, it follows that the quotient $F[x]/langle x rangle cong F$ is actually a field, and so we are done.
edited Nov 26 at 0:13
egreg
177k1484200
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
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%2f3011161%2fis-it-true-that-f-is-a-field-if-and-only-if-fx-is-an-euclidean-domain%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
No, that's not right. You're using Euclidean Division wrong. For one thing, you would have to conclude that $deg(r(x)) < deg(u)$, not $deg(f(x))$. This would actually imply that $r(x) = 0$, and thus that $f(x)$ was an inverse for $u$, and thus would have to be in $F$, implying that $F$ is a field. But, as Lord Shark the Unknown points out, this depends on assuming that the degree is the Euclidean function. Since we can't do this, we have to find some other approach.
It's actually easier to prove a stronger version of the statement, namely that if $F[x]$ is a PID then $F$ is a field. If $F[x]$ is a PID, $F$ must be an integral domain as well, and so $F[x]/langle x rangle cong F$ implies that $langle x rangle$ is prime. Since nonzero prime ideals are maximal in a PID, it follows that the quotient $F[x]/langle x rangle cong F$ is actually a field, and so we are done.
edited Nov 26 at 0:13
egreg
177k1484200
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
add a comment |
No, that's not right. You're using Euclidean Division wrong. For one thing, you would have to conclude that $deg(r(x)) < deg(u)$, not $deg(f(x))$. This would actually imply that $r(x) = 0$, and thus that $f(x)$ was an inverse for $u$, and thus would have to be in $F$, implying that $F$ is a field. But, as Lord Shark the Unknown points out, this depends on assuming that the degree is the Euclidean function. Since we can't do this, we have to find some other approach.
It's actually easier to prove a stronger version of the statement, namely that if $F[x]$ is a PID then $F$ is a field. If $F[x]$ is a PID, $F$ must be an integral domain as well, and so $F[x]/langle x rangle cong F$ implies that $langle x rangle$ is prime. Since nonzero prime ideals are maximal in a PID, it follows that the quotient $F[x]/langle x rangle cong F$ is actually a field, and so we are done.
edited Nov 26 at 0:13
egreg
177k1484200
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
add a comment |
No, that's not right. You're using Euclidean Division wrong. For one thing, you would have to conclude that $deg(r(x)) < deg(u)$, not $deg(f(x))$. This would actually imply that $r(x) = 0$, and thus that $f(x)$ was an inverse for $u$, and thus would have to be in $F$, implying that $F$ is a field. But, as Lord Shark the Unknown points out, this depends on assuming that the degree is the Euclidean function. Since we can't do this, we have to find some other approach.
It's actually easier to prove a stronger version of the statement, namely that if $F[x]$ is a PID then $F$ is a field. If $F[x]$ is a PID, $F$ must be an integral domain as well, and so $F[x]/langle x rangle cong F$ implies that $langle x rangle$ is prime. Since nonzero prime ideals are maximal in a PID, it follows that the quotient $F[x]/langle x rangle cong F$ is actually a field, and so we are done.
edited Nov 26 at 0:13
egreg
177k1484200
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
No, that's not right. You're using Euclidean Division wrong. For one thing, you would have to conclude that $deg(r(x)) < deg(u)$, not $deg(f(x))$. This would actually imply that $r(x) = 0$, and thus that $f(x)$ was an inverse for $u$, and thus would have to be in $F$, implying that $F$ is a field. But, as Lord Shark the Unknown points out, this depends on assuming that the degree is the Euclidean function. Since we can't do this, we have to find some other approach.
It's actually easier to prove a stronger version of the statement, namely that if $F[x]$ is a PID then $F$ is a field. If $F[x]$ is a PID, $F$ must be an integral domain as well, and so $F[x]/langle x rangle cong F$ implies that $langle x rangle$ is prime. Since nonzero prime ideals are maximal in a PID, it follows that the quotient $F[x]/langle x rangle cong F$ is actually a field, and so we are done.
edited Nov 26 at 0:13
egreg
177k1484200
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
edited Nov 26 at 0:13
egreg
177k1484200
edited Nov 26 at 0:13
egreg
177k1484200
edited Nov 26 at 0:13
egreg
177k1484200
177k1484200
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
answered Nov 24 at 5:05
Monstrous Moonshiner
2,25511337
2,25511337
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
add a comment |
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind...
– David C. Ullrich
Nov 25 at 14:17
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
@David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly.
– Monstrous Moonshiner
Nov 25 at 22:56
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
$F[x]$ is a domain, hence $F$ is a domain; since $F[x]/langle xranglecong F$, it follows that $langle xrangle$ is a prime ideal.
– egreg
Nov 25 at 23:13
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
@egreg You’re right, that is a bit slicker. The answer has been updated.
– Monstrous Moonshiner
Nov 25 at 23:59
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%2f3011161%2fis-it-true-that-f-is-a-field-if-and-only-if-fx-is-an-euclidean-domain%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
5
You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$.
– Lord Shark the Unknown
Nov 24 at 4:11
@LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $fin F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit.
– Idonknow
Nov 24 at 5:01
1
There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment...
– KCd
Nov 24 at 6:37
Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :)
– darij grinberg
Nov 26 at 0:53
@darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that.
– KCd
Nov 26 at 0:58