Why can't Eisenstein Criterion be used for certain polynomials (to show that it's irreducible over...












1












$begingroup$


Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?



I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.



A detailed explanation would be great! Thanks.






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    @WillJagy but why would you want to do that when you have a degree as big as 10?
    $endgroup$
    – dean3794
    May 13 '15 at 21:14






  • 1




    $begingroup$
    Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
    $endgroup$
    – Mariano Suárez-Álvarez
    May 13 '15 at 21:39










  • $begingroup$
    @MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
    $endgroup$
    – dean3794
    May 13 '15 at 21:41










  • $begingroup$
    I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
    $endgroup$
    – James
    May 13 '15 at 21:44
















1












$begingroup$


Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?



I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.



A detailed explanation would be great! Thanks.






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    @WillJagy but why would you want to do that when you have a degree as big as 10?
    $endgroup$
    – dean3794
    May 13 '15 at 21:14






  • 1




    $begingroup$
    Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
    $endgroup$
    – Mariano Suárez-Álvarez
    May 13 '15 at 21:39










  • $begingroup$
    @MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
    $endgroup$
    – dean3794
    May 13 '15 at 21:41










  • $begingroup$
    I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
    $endgroup$
    – James
    May 13 '15 at 21:44














1












1








1


1



$begingroup$


Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?



I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.



A detailed explanation would be great! Thanks.






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question











$endgroup$




Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?



I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.



A detailed explanation would be great! Thanks.






abstract-algebra field-theory irreducible-polynomials




abstract-algebra field-theory irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 13 '15 at 21:35









Ken

3,63151728




3,63151728










asked May 13 '15 at 21:07









dean3794dean3794

788




788












  • $begingroup$
    @WillJagy but why would you want to do that when you have a degree as big as 10?
    $endgroup$
    – dean3794
    May 13 '15 at 21:14






  • 1




    $begingroup$
    Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
    $endgroup$
    – Mariano Suárez-Álvarez
    May 13 '15 at 21:39










  • $begingroup$
    @MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
    $endgroup$
    – dean3794
    May 13 '15 at 21:41










  • $begingroup$
    I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
    $endgroup$
    – James
    May 13 '15 at 21:44


















  • $begingroup$
    @WillJagy but why would you want to do that when you have a degree as big as 10?
    $endgroup$
    – dean3794
    May 13 '15 at 21:14






  • 1




    $begingroup$
    Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
    $endgroup$
    – Mariano Suárez-Álvarez
    May 13 '15 at 21:39










  • $begingroup$
    @MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
    $endgroup$
    – dean3794
    May 13 '15 at 21:41










  • $begingroup$
    I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
    $endgroup$
    – James
    May 13 '15 at 21:44
















$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14




$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14




1




1




$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39




$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39












$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41




$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41












$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44




$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44










2 Answers
2






active

oldest

votes


















2












$begingroup$

If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
    $endgroup$
    – Henry
    Dec 3 '18 at 16:44












  • $begingroup$
    @Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
    $endgroup$
    – KCd
    Dec 3 '18 at 19:58



















1












$begingroup$

The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
    $endgroup$
    – k1.M
    May 13 '15 at 21:23










  • $begingroup$
    Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
    $endgroup$
    – William Stagner
    May 13 '15 at 21:28






  • 1




    $begingroup$
    So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
    $endgroup$
    – k1.M
    May 13 '15 at 21:30











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
    $endgroup$
    – Henry
    Dec 3 '18 at 16:44












  • $begingroup$
    @Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
    $endgroup$
    – KCd
    Dec 3 '18 at 19:58
















2












$begingroup$

If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
    $endgroup$
    – Henry
    Dec 3 '18 at 16:44












  • $begingroup$
    @Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
    $endgroup$
    – KCd
    Dec 3 '18 at 19:58














2












2








2





$begingroup$

If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.






share|cite|improve this answer









$endgroup$



If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 21:06









KCdKCd

16.7k4075




16.7k4075












  • $begingroup$
    Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
    $endgroup$
    – Henry
    Dec 3 '18 at 16:44












  • $begingroup$
    @Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
    $endgroup$
    – KCd
    Dec 3 '18 at 19:58


















  • $begingroup$
    Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
    $endgroup$
    – Henry
    Dec 3 '18 at 16:44












  • $begingroup$
    @Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
    $endgroup$
    – KCd
    Dec 3 '18 at 19:58
















$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44






$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44














$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58




$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58











1












$begingroup$

The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
    $endgroup$
    – k1.M
    May 13 '15 at 21:23










  • $begingroup$
    Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
    $endgroup$
    – William Stagner
    May 13 '15 at 21:28






  • 1




    $begingroup$
    So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
    $endgroup$
    – k1.M
    May 13 '15 at 21:30
















1












$begingroup$

The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
    $endgroup$
    – k1.M
    May 13 '15 at 21:23










  • $begingroup$
    Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
    $endgroup$
    – William Stagner
    May 13 '15 at 21:28






  • 1




    $begingroup$
    So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
    $endgroup$
    – k1.M
    May 13 '15 at 21:30














1












1








1





$begingroup$

The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.






share|cite|improve this answer









$endgroup$



The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 13 '15 at 21:11









William StagnerWilliam Stagner

3,5831027




3,5831027








  • 2




    $begingroup$
    The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
    $endgroup$
    – k1.M
    May 13 '15 at 21:23










  • $begingroup$
    Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
    $endgroup$
    – William Stagner
    May 13 '15 at 21:28






  • 1




    $begingroup$
    So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
    $endgroup$
    – k1.M
    May 13 '15 at 21:30














  • 2




    $begingroup$
    The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
    $endgroup$
    – k1.M
    May 13 '15 at 21:23










  • $begingroup$
    Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
    $endgroup$
    – William Stagner
    May 13 '15 at 21:28






  • 1




    $begingroup$
    So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
    $endgroup$
    – k1.M
    May 13 '15 at 21:30








2




2




$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23




$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23












$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28




$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28




1




1




$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30




$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30


















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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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