Splitting fields for $x^3-3$ and $x^5-1$












1












$begingroup$


I'm looking for the splitting fields of



(a) $x^3-3$



(b) $x^5-1$.




EDIT:



(a) Thanks to all the hints and suggestions, the three roots are



$x_1=3^{frac{1}{3}}$, $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$, $x_3=e^{frac{4 pi i}{3}}3^{frac{1}{3}}$



Now, the question doesn't specify the field over which these polynomials are defined, I'll take a guess and say $Q$. Now, all the roots can be generated from $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$ over the rationals, so is the answer $Q(e^{frac{2 pi i}{3}}3^{frac{1}{3}})$ correct?




(b) Again, the roots are the 5 complex roots of unity, all of which can be generated by the root $x_1=e^{frac{2 pi i}{5}}$. So would the correct answer now be $Q(e^{frac{2 pi i}{5}})$






proof-verification polynomials field-theory galois-theory extension-field







share|cite|improve this question











$endgroup$












  • $begingroup$
    Splitting fields over which ground field?
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:45










  • $begingroup$
    @TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals.
    $endgroup$
    – Mike
    Dec 2 '18 at 23:46












  • $begingroup$
    Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:49






  • 1




    $begingroup$
    Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:54






  • 1




    $begingroup$
    See math.stackexchange.com/questions/1597326/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:14


















1












$begingroup$


I'm looking for the splitting fields of



(a) $x^3-3$



(b) $x^5-1$.




EDIT:



(a) Thanks to all the hints and suggestions, the three roots are



$x_1=3^{frac{1}{3}}$, $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$, $x_3=e^{frac{4 pi i}{3}}3^{frac{1}{3}}$



Now, the question doesn't specify the field over which these polynomials are defined, I'll take a guess and say $Q$. Now, all the roots can be generated from $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$ over the rationals, so is the answer $Q(e^{frac{2 pi i}{3}}3^{frac{1}{3}})$ correct?




(b) Again, the roots are the 5 complex roots of unity, all of which can be generated by the root $x_1=e^{frac{2 pi i}{5}}$. So would the correct answer now be $Q(e^{frac{2 pi i}{5}})$






proof-verification polynomials field-theory galois-theory extension-field







share|cite|improve this question











$endgroup$












  • $begingroup$
    Splitting fields over which ground field?
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:45










  • $begingroup$
    @TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals.
    $endgroup$
    – Mike
    Dec 2 '18 at 23:46












  • $begingroup$
    Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:49






  • 1




    $begingroup$
    Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:54






  • 1




    $begingroup$
    See math.stackexchange.com/questions/1597326/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:14
















1












1








1





$begingroup$


I'm looking for the splitting fields of



(a) $x^3-3$



(b) $x^5-1$.




EDIT:



(a) Thanks to all the hints and suggestions, the three roots are



$x_1=3^{frac{1}{3}}$, $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$, $x_3=e^{frac{4 pi i}{3}}3^{frac{1}{3}}$



Now, the question doesn't specify the field over which these polynomials are defined, I'll take a guess and say $Q$. Now, all the roots can be generated from $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$ over the rationals, so is the answer $Q(e^{frac{2 pi i}{3}}3^{frac{1}{3}})$ correct?




(b) Again, the roots are the 5 complex roots of unity, all of which can be generated by the root $x_1=e^{frac{2 pi i}{5}}$. So would the correct answer now be $Q(e^{frac{2 pi i}{5}})$






proof-verification polynomials field-theory galois-theory extension-field







share|cite|improve this question











$endgroup$




I'm looking for the splitting fields of



(a) $x^3-3$



(b) $x^5-1$.




EDIT:



(a) Thanks to all the hints and suggestions, the three roots are



$x_1=3^{frac{1}{3}}$, $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$, $x_3=e^{frac{4 pi i}{3}}3^{frac{1}{3}}$



Now, the question doesn't specify the field over which these polynomials are defined, I'll take a guess and say $Q$. Now, all the roots can be generated from $x_2=e^{frac{2 pi i}{3}}3^{frac{1}{3}}$ over the rationals, so is the answer $Q(e^{frac{2 pi i}{3}}3^{frac{1}{3}})$ correct?




(b) Again, the roots are the 5 complex roots of unity, all of which can be generated by the root $x_1=e^{frac{2 pi i}{5}}$. So would the correct answer now be $Q(e^{frac{2 pi i}{5}})$






proof-verification polynomials field-theory galois-theory extension-field




proof-verification polynomials field-theory galois-theory extension-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 0:12







Mike

















asked Dec 2 '18 at 23:40









MikeMike

700414




700414












  • $begingroup$
    Splitting fields over which ground field?
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:45










  • $begingroup$
    @TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals.
    $endgroup$
    – Mike
    Dec 2 '18 at 23:46












  • $begingroup$
    Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:49






  • 1




    $begingroup$
    Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:54






  • 1




    $begingroup$
    See math.stackexchange.com/questions/1597326/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:14




















  • $begingroup$
    Splitting fields over which ground field?
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:45










  • $begingroup$
    @TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals.
    $endgroup$
    – Mike
    Dec 2 '18 at 23:46












  • $begingroup$
    Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384
    $endgroup$
    – Torsten Schoeneberg
    Dec 2 '18 at 23:49






  • 1




    $begingroup$
    Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:54






  • 1




    $begingroup$
    See math.stackexchange.com/questions/1597326/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:14


















$begingroup$
Splitting fields over which ground field?
$endgroup$
– Torsten Schoeneberg
Dec 2 '18 at 23:45




$begingroup$
Splitting fields over which ground field?
$endgroup$
– Torsten Schoeneberg
Dec 2 '18 at 23:45












$begingroup$
@TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals.
$endgroup$
– Mike
Dec 2 '18 at 23:46






$begingroup$
@TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals.
$endgroup$
– Mike
Dec 2 '18 at 23:46














$begingroup$
Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384
$endgroup$
– Torsten Schoeneberg
Dec 2 '18 at 23:49




$begingroup$
Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384
$endgroup$
– Torsten Schoeneberg
Dec 2 '18 at 23:49




1




1




$begingroup$
Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$.
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 23:54




$begingroup$
Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$.
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 23:54




1




1




$begingroup$
See math.stackexchange.com/questions/1597326/…
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:14






$begingroup$
See math.stackexchange.com/questions/1597326/…
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:14












1 Answer
1






active

oldest

votes


















2












$begingroup$

Hints:




  • A real number has $3$ cube roots in $mathbf C$. One is the standard real cube root, he other two are this real cube root, multiplied by one of the complex cube roots of unity.

  • For $x^5-1$, solve it in the form $mathrm e^{itheta}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:57










  • $begingroup$
    @Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:59










  • $begingroup$
    Correct. ${}{}{}$
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:02










  • $begingroup$
    @Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
    $endgroup$
    – Mike
    Dec 3 '18 at 0:08











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Hints:




  • A real number has $3$ cube roots in $mathbf C$. One is the standard real cube root, he other two are this real cube root, multiplied by one of the complex cube roots of unity.

  • For $x^5-1$, solve it in the form $mathrm e^{itheta}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:57










  • $begingroup$
    @Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:59










  • $begingroup$
    Correct. ${}{}{}$
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:02










  • $begingroup$
    @Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
    $endgroup$
    – Mike
    Dec 3 '18 at 0:08
















2












$begingroup$

Hints:




  • A real number has $3$ cube roots in $mathbf C$. One is the standard real cube root, he other two are this real cube root, multiplied by one of the complex cube roots of unity.

  • For $x^5-1$, solve it in the form $mathrm e^{itheta}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:57










  • $begingroup$
    @Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:59










  • $begingroup$
    Correct. ${}{}{}$
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:02










  • $begingroup$
    @Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
    $endgroup$
    – Mike
    Dec 3 '18 at 0:08














2












2








2





$begingroup$

Hints:




  • A real number has $3$ cube roots in $mathbf C$. One is the standard real cube root, he other two are this real cube root, multiplied by one of the complex cube roots of unity.

  • For $x^5-1$, solve it in the form $mathrm e^{itheta}$.






share|cite|improve this answer









$endgroup$



Hints:




  • A real number has $3$ cube roots in $mathbf C$. One is the standard real cube root, he other two are this real cube root, multiplied by one of the complex cube roots of unity.

  • For $x^5-1$, solve it in the form $mathrm e^{itheta}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 23:55









BernardBernard

119k639113




119k639113












  • $begingroup$
    That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:57










  • $begingroup$
    @Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:59










  • $begingroup$
    Correct. ${}{}{}$
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:02










  • $begingroup$
    @Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
    $endgroup$
    – Mike
    Dec 3 '18 at 0:08


















  • $begingroup$
    That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 23:57










  • $begingroup$
    @Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
    $endgroup$
    – Bernard
    Dec 2 '18 at 23:59










  • $begingroup$
    Correct. ${}{}{}$
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 0:02










  • $begingroup$
    @Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
    $endgroup$
    – Mike
    Dec 3 '18 at 0:08
















$begingroup$
That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 23:57




$begingroup$
That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica.
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 23:57












$begingroup$
@Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
$endgroup$
– Bernard
Dec 2 '18 at 23:59




$begingroup$
@Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title.
$endgroup$
– Bernard
Dec 2 '18 at 23:59












$begingroup$
Correct. ${}{}{}$
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:02




$begingroup$
Correct. ${}{}{}$
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:02












$begingroup$
@Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
$endgroup$
– Mike
Dec 3 '18 at 0:08




$begingroup$
@Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer.
$endgroup$
– Mike
Dec 3 '18 at 0:08


















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