Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?












4















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







share|cite|improve this question




















  • 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
















4















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







share|cite|improve this question




















  • 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














4












4








4








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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 4:04

























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














  • 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










1 Answer
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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.






share|cite|improve this answer























  • 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











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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.






share|cite|improve this answer























  • 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
















4














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.






share|cite|improve this answer























  • 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














4












4








4






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 0:13









egreg

177k1484200




177k1484200










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


















  • 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


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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