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Confusion about the splitting field of $x^3-7$
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I'm asked to find the degree of the splitting field of $x^3-7$ over the rationals.
The roots are $sqrt[3] 7e^frac2pi ik3, k=0,1,2$. Explicitly, $$x_1=sqrt[3] 7,\ x_2=sqrt[3] 7 bigg(-frac12+ifracsqrt 32bigg),\ x_3=sqrt[3] 7 bigg(-frac12-ifracsqrt 32bigg).$$
The splitting field is an extension that contains all roots. I see at least two essentially different possibilities of building it. First, I could adjoin $sqrt[3] 7, i,sqrt 3$. Such an extension will contain all roots. On the other hand, I could adjoin $sqrt[3] 7, isqrt 3$. This extension also contains all the roots. I believe the two extensions have different degrees. So how should I understand which degree I should find to begin with? Which extension is the splitting field?
abstract-algebra field-theory galois-theory extension-field splitting-field
edited Jul 17 at 1:57
Key Flex
4,396425
asked Jul 16 at 18:38
user437309
556212
 |Â
show 4 more comments
up vote
4
down vote
favorite
I'm asked to find the degree of the splitting field of $x^3-7$ over the rationals.
The roots are $sqrt[3] 7e^frac2pi ik3, k=0,1,2$. Explicitly, $$x_1=sqrt[3] 7,\ x_2=sqrt[3] 7 bigg(-frac12+ifracsqrt 32bigg),\ x_3=sqrt[3] 7 bigg(-frac12-ifracsqrt 32bigg).$$
The splitting field is an extension that contains all roots. I see at least two essentially different possibilities of building it. First, I could adjoin $sqrt[3] 7, i,sqrt 3$. Such an extension will contain all roots. On the other hand, I could adjoin $sqrt[3] 7, isqrt 3$. This extension also contains all the roots. I believe the two extensions have different degrees. So how should I understand which degree I should find to begin with? Which extension is the splitting field?
abstract-algebra field-theory galois-theory extension-field splitting-field
edited Jul 17 at 1:57
Key Flex
4,396425
asked Jul 16 at 18:38
user437309
556212
2
The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree?
– Batominovski
Jul 16 at 18:39
@Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree.
– user437309
Jul 16 at 18:41
3
Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+isqrt3)/2$ is certainly in the splitting field. Because $BbbQ$ is a subset of the splitting field you can then deduce that $isqrt3$ must be in. But, it stops there, right?
– Jyrki Lahtonen
Jul 16 at 18:42
1
On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors."
– Batominovski
Jul 16 at 18:42
@Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki?
– user437309
Jul 16 at 18:43
 |Â
show 4 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm asked to find the degree of the splitting field of $x^3-7$ over the rationals.
The roots are $sqrt[3] 7e^frac2pi ik3, k=0,1,2$. Explicitly, $$x_1=sqrt[3] 7,\ x_2=sqrt[3] 7 bigg(-frac12+ifracsqrt 32bigg),\ x_3=sqrt[3] 7 bigg(-frac12-ifracsqrt 32bigg).$$
The splitting field is an extension that contains all roots. I see at least two essentially different possibilities of building it. First, I could adjoin $sqrt[3] 7, i,sqrt 3$. Such an extension will contain all roots. On the other hand, I could adjoin $sqrt[3] 7, isqrt 3$. This extension also contains all the roots. I believe the two extensions have different degrees. So how should I understand which degree I should find to begin with? Which extension is the splitting field?
abstract-algebra field-theory galois-theory extension-field splitting-field
edited Jul 17 at 1:57
Key Flex
4,396425
asked Jul 16 at 18:38
user437309
556212
I'm asked to find the degree of the splitting field of $x^3-7$ over the rationals.
The roots are $sqrt[3] 7e^frac2pi ik3, k=0,1,2$. Explicitly, $$x_1=sqrt[3] 7,\ x_2=sqrt[3] 7 bigg(-frac12+ifracsqrt 32bigg),\ x_3=sqrt[3] 7 bigg(-frac12-ifracsqrt 32bigg).$$
The splitting field is an extension that contains all roots. I see at least two essentially different possibilities of building it. First, I could adjoin $sqrt[3] 7, i,sqrt 3$. Such an extension will contain all roots. On the other hand, I could adjoin $sqrt[3] 7, isqrt 3$. This extension also contains all the roots. I believe the two extensions have different degrees. So how should I understand which degree I should find to begin with? Which extension is the splitting field?
abstract-algebra field-theory galois-theory extension-field splitting-field
edited Jul 17 at 1:57
Key Flex
4,396425
asked Jul 16 at 18:38
user437309
556212
edited Jul 17 at 1:57
Key Flex
4,396425
edited Jul 17 at 1:57
Key Flex
4,396425
edited Jul 17 at 1:57
Key Flex
4,396425
4,396425
asked Jul 16 at 18:38
user437309
556212
asked Jul 16 at 18:38
user437309
556212
asked Jul 16 at 18:38
user437309
556212
556212
2
The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree?
– Batominovski
Jul 16 at 18:39
@Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree.
– user437309
Jul 16 at 18:41
3
Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+isqrt3)/2$ is certainly in the splitting field. Because $BbbQ$ is a subset of the splitting field you can then deduce that $isqrt3$ must be in. But, it stops there, right?
– Jyrki Lahtonen
Jul 16 at 18:42
1
On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors."
– Batominovski
Jul 16 at 18:42
@Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki?
– user437309
Jul 16 at 18:43
 |Â
show 4 more comments
2
The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree?
– Batominovski
Jul 16 at 18:39
@Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree.
– user437309
Jul 16 at 18:41
3
Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+isqrt3)/2$ is certainly in the splitting field. Because $BbbQ$ is a subset of the splitting field you can then deduce that $isqrt3$ must be in. But, it stops there, right?
– Jyrki Lahtonen
Jul 16 at 18:42
1
On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors."
– Batominovski
Jul 16 at 18:42
@Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki?
– user437309
Jul 16 at 18:43
2
2
The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree?
– Batominovski
Jul 16 at 18:39
The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree?
– Batominovski
Jul 16 at 18:39
@Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree.
– user437309
Jul 16 at 18:41
@Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree.
– user437309
Jul 16 at 18:41
3
3
Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+isqrt3)/2$ is certainly in the splitting field. Because $BbbQ$ is a subset of the splitting field you can then deduce that $isqrt3$ must be in. But, it stops there, right?
– Jyrki Lahtonen
Jul 16 at 18:42
Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+isqrt3)/2$ is certainly in the splitting field. Because $BbbQ$ is a subset of the splitting field you can then deduce that $isqrt3$ must be in. But, it stops there, right?
– Jyrki Lahtonen
Jul 16 at 18:42
1
1
On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors."
– Batominovski
Jul 16 at 18:42
On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors."
– Batominovski
Jul 16 at 18:42
@Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki?
– user437309
Jul 16 at 18:43
@Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki?
– user437309
Jul 16 at 18:43
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The splitting field $K$ of $x^3-7$ is $mathbb Q(sqrt[3]7, omega)$, where $omega=-frac12+ifracsqrt 32$ is a primitive cubic root of the unity, a root of $x^2+x+1$.
There is no need to decompose $omega$ into $i$ and $sqrt 3$.
It is true that $L =mathbb Q(sqrt[3]7, i, sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.
answered Jul 16 at 22:44
lhf
156k9160367
add a comment |Â
up vote
1
down vote
You have $x^3-7$ and $alpha=sqrt [ 3 ] 7 $ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-alpha)(x-alphazeta )(x-alphazeta ^2)$ where $zeta $ is a complex cube root of unity. $zeta ^2+zeta +1=0$ and $zeta notinmathbbR$ hence $notinmathbbQ(alpha)$, so splitting the field has the degree $3cdot2=6$. In fact the splitting field is $mathbbQ(alpha,zeta)$.
answered Jul 16 at 23:12
Key Flex
4,396425
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The splitting field $K$ of $x^3-7$ is $mathbb Q(sqrt[3]7, omega)$, where $omega=-frac12+ifracsqrt 32$ is a primitive cubic root of the unity, a root of $x^2+x+1$.
There is no need to decompose $omega$ into $i$ and $sqrt 3$.
It is true that $L =mathbb Q(sqrt[3]7, i, sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.
answered Jul 16 at 22:44
lhf
156k9160367
add a comment |Â
up vote
1
down vote
accepted
The splitting field $K$ of $x^3-7$ is $mathbb Q(sqrt[3]7, omega)$, where $omega=-frac12+ifracsqrt 32$ is a primitive cubic root of the unity, a root of $x^2+x+1$.
There is no need to decompose $omega$ into $i$ and $sqrt 3$.
It is true that $L =mathbb Q(sqrt[3]7, i, sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.
answered Jul 16 at 22:44
lhf
156k9160367
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The splitting field $K$ of $x^3-7$ is $mathbb Q(sqrt[3]7, omega)$, where $omega=-frac12+ifracsqrt 32$ is a primitive cubic root of the unity, a root of $x^2+x+1$.
There is no need to decompose $omega$ into $i$ and $sqrt 3$.
It is true that $L =mathbb Q(sqrt[3]7, i, sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.
answered Jul 16 at 22:44
lhf
156k9160367
The splitting field $K$ of $x^3-7$ is $mathbb Q(sqrt[3]7, omega)$, where $omega=-frac12+ifracsqrt 32$ is a primitive cubic root of the unity, a root of $x^2+x+1$.
There is no need to decompose $omega$ into $i$ and $sqrt 3$.
It is true that $L =mathbb Q(sqrt[3]7, i, sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.
answered Jul 16 at 22:44
lhf
156k9160367
edited Jul 17 at 1:46
answered Jul 16 at 22:44
lhf
156k9160367
answered Jul 16 at 22:44
lhf
156k9160367
answered Jul 16 at 22:44
lhf
156k9160367
156k9160367
add a comment |Â
add a comment |Â
up vote
1
down vote
You have $x^3-7$ and $alpha=sqrt [ 3 ] 7 $ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-alpha)(x-alphazeta )(x-alphazeta ^2)$ where $zeta $ is a complex cube root of unity. $zeta ^2+zeta +1=0$ and $zeta notinmathbbR$ hence $notinmathbbQ(alpha)$, so splitting the field has the degree $3cdot2=6$. In fact the splitting field is $mathbbQ(alpha,zeta)$.
answered Jul 16 at 23:12
Key Flex
4,396425
add a comment |Â
up vote
1
down vote
You have $x^3-7$ and $alpha=sqrt [ 3 ] 7 $ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-alpha)(x-alphazeta )(x-alphazeta ^2)$ where $zeta $ is a complex cube root of unity. $zeta ^2+zeta +1=0$ and $zeta notinmathbbR$ hence $notinmathbbQ(alpha)$, so splitting the field has the degree $3cdot2=6$. In fact the splitting field is $mathbbQ(alpha,zeta)$.
answered Jul 16 at 23:12
Key Flex
4,396425
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have $x^3-7$ and $alpha=sqrt [ 3 ] 7 $ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-alpha)(x-alphazeta )(x-alphazeta ^2)$ where $zeta $ is a complex cube root of unity. $zeta ^2+zeta +1=0$ and $zeta notinmathbbR$ hence $notinmathbbQ(alpha)$, so splitting the field has the degree $3cdot2=6$. In fact the splitting field is $mathbbQ(alpha,zeta)$.
answered Jul 16 at 23:12
Key Flex
4,396425
You have $x^3-7$ and $alpha=sqrt [ 3 ] 7 $ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-alpha)(x-alphazeta )(x-alphazeta ^2)$ where $zeta $ is a complex cube root of unity. $zeta ^2+zeta +1=0$ and $zeta notinmathbbR$ hence $notinmathbbQ(alpha)$, so splitting the field has the degree $3cdot2=6$. In fact the splitting field is $mathbbQ(alpha,zeta)$.
answered Jul 16 at 23:12
Key Flex
4,396425
answered Jul 16 at 23:12
Key Flex
4,396425
answered Jul 16 at 23:12
Key Flex
4,396425
answered Jul 16 at 23:12
Key Flex
4,396425
4,396425
add a comment |Â
add a comment |Â
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2
The splitting field should be the one with lowest degree of extension, shouldn't it? Which field of the two has a smaller degree?
– Batominovski
Jul 16 at 18:39
@Batominovski For example in the Wikipedia definition, I don't see the requirement on the degree.
– user437309
Jul 16 at 18:41
3
Yes, the degrees of the two extension are different. So one of them is not the splitting field. You can think about it as follows. The number $x_2/x_1=(-1+isqrt3)/2$ is certainly in the splitting field. Because $BbbQ$ is a subset of the splitting field you can then deduce that $isqrt3$ must be in. But, it stops there, right?
– Jyrki Lahtonen
Jul 16 at 18:42
1
On the wiki page, that is said in literally the first sentence of the page. "In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial splits or decomposes into linear factors."
– Batominovski
Jul 16 at 18:42
@Batominovski I looked in the "Definition" section. Is it implied by some condition in the formal definition stated on wiki?
– user437309
Jul 16 at 18:43