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
asked Dec 2 '18 at 23:40
MikeMike
700414
$endgroup$
|
show 4 more comments
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
asked Dec 2 '18 at 23:40
MikeMike
700414
$endgroup$
$begingroup$
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
|
show 4 more comments
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
asked Dec 2 '18 at 23:40
MikeMike
700414
$endgroup$
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
asked Dec 2 '18 at 23:40
MikeMike
700414
asked Dec 2 '18 at 23:40
MikeMike
700414
edited Dec 3 '18 at 0:12
Mike
asked Dec 2 '18 at 23:40
MikeMike
700414
asked Dec 2 '18 at 23:40
MikeMike
700414
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
1 Answer
1
active
oldest
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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 $mathrm e^{itheta}$.
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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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 $mathrm e^{itheta}$.
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
$endgroup$
$begingroup$
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
add a comment |
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 $mathrm e^{itheta}$.
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
$endgroup$
$begingroup$
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
add a comment |
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 $mathrm e^{itheta}$.
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
$endgroup$
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 $mathrm e^{itheta}$.
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
answered Dec 2 '18 at 23:55
BernardBernard
119k639113
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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