Discriminant of $Bbb Q(sqrt[3]2)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Easy way to show that $mathbbZ[sqrt[3]2]$ is the ring of integers of $mathbbQ[sqrt[3]2]$The discriminant of a cubic extensionShow that $mathbbQ(6^1/3)$ and $mathbbQ(12^1/3)$ have the same degree and discriminant but are not isomorphic.All number fields with absolute value of discriminant $le 20$Primes corresponding to embeddings of a number fieldThe discriminant of a cubic extension$Bbb Q (sqrt-535, sqrt 5)$ is unramified over $Bbb Q (sqrt -535)$Absolute values induced by embeddings (after Lang's “Algebraic number theory”)What is the discriminant of $R:=mathbb Z[sqrt2,sqrt3]$?Archimedean places of a number fieldDoes a full-rank set of appropriate discriminant always generate an algebraic number ring?Is the conductor of an L-function F the absolute value of the discriminant ofsome number field related to F?
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Discriminant of $Bbb Q(sqrt[3]2)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Easy way to show that $mathbbZ[sqrt[3]2]$ is the ring of integers of $mathbbQ[sqrt[3]2]$The discriminant of a cubic extensionShow that $mathbbQ(6^1/3)$ and $mathbbQ(12^1/3)$ have the same degree and discriminant but are not isomorphic.All number fields with absolute value of discriminant $le 20$Primes corresponding to embeddings of a number fieldThe discriminant of a cubic extension$Bbb Q (sqrt-535, sqrt 5)$ is unramified over $Bbb Q (sqrt -535)$Absolute values induced by embeddings (after Lang's “Algebraic number theory”)What is the discriminant of $R:=mathbb Z[sqrt2,sqrt3]$?Archimedean places of a number fieldDoes a full-rank set of appropriate discriminant always generate an algebraic number ring?Is the conductor of an L-function F the absolute value of the discriminant ofsome number field related to F?
$begingroup$
I want to understand a way of computing the discriminant of the number field $K=mathbbQ(sqrt[3]2)$. The degree of $K|mathbbQ$ is $n=3$ and we have $3=1+2cdot 1$, so there are one real and two complex embeddings.
Now my teacher concludes that the absolute value of the discriminant is equal to $2^2cdot 3^3$. Why that?
abstract-algebra algebraic-number-theory
$endgroup$
add a comment |
$begingroup$
I want to understand a way of computing the discriminant of the number field $K=mathbbQ(sqrt[3]2)$. The degree of $K|mathbbQ$ is $n=3$ and we have $3=1+2cdot 1$, so there are one real and two complex embeddings.
Now my teacher concludes that the absolute value of the discriminant is equal to $2^2cdot 3^3$. Why that?
abstract-algebra algebraic-number-theory
$endgroup$
1
$begingroup$
Is the class using a textbook or did the teacher just pull that out of thin air?
$endgroup$
– Robert Soupe
Mar 11 '17 at 18:59
add a comment |
$begingroup$
I want to understand a way of computing the discriminant of the number field $K=mathbbQ(sqrt[3]2)$. The degree of $K|mathbbQ$ is $n=3$ and we have $3=1+2cdot 1$, so there are one real and two complex embeddings.
Now my teacher concludes that the absolute value of the discriminant is equal to $2^2cdot 3^3$. Why that?
abstract-algebra algebraic-number-theory
$endgroup$
I want to understand a way of computing the discriminant of the number field $K=mathbbQ(sqrt[3]2)$. The degree of $K|mathbbQ$ is $n=3$ and we have $3=1+2cdot 1$, so there are one real and two complex embeddings.
Now my teacher concludes that the absolute value of the discriminant is equal to $2^2cdot 3^3$. Why that?
abstract-algebra algebraic-number-theory
abstract-algebra algebraic-number-theory
edited Mar 11 '17 at 17:25
asked Mar 11 '17 at 17:24
user404105
1
$begingroup$
Is the class using a textbook or did the teacher just pull that out of thin air?
$endgroup$
– Robert Soupe
Mar 11 '17 at 18:59
add a comment |
1
$begingroup$
Is the class using a textbook or did the teacher just pull that out of thin air?
$endgroup$
– Robert Soupe
Mar 11 '17 at 18:59
1
1
$begingroup$
Is the class using a textbook or did the teacher just pull that out of thin air?
$endgroup$
– Robert Soupe
Mar 11 '17 at 18:59
$begingroup$
Is the class using a textbook or did the teacher just pull that out of thin air?
$endgroup$
– Robert Soupe
Mar 11 '17 at 18:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.
Let $f in mathbbQ[x]$ monic of degree $n$ be the minimum polynomial of $alpha$ and let $K=mathbbQ(alpha)$.
- The discriminant of $mathbbZ[alpha]$ is $(-1)^frac12n(n-1)N^K_mathbbQ(f'(alpha))$.
- The field norm $N^K_mathbbQ$ is multiplicative.
- $N^K_mathbbQ(b) = b^n$ for $b in mathbbQ$.
- $N^K_mathbbQ(alpha)$ is $(-1)^n$ times the constant term of $f$.
Putting these together for $f = x^3 - 2$ (using $mathcalO_K = mathbbZ[alpha]$ here, which is nontrivial) yields
The absolute value of the discriminant of $mathbbQ(sqrt[3]2)$ is $N(3alpha^2) = N(3)N(alpha)^2 = 3^3 cdot 2^2$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
$endgroup$
1
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
3
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
add a comment |
$begingroup$
Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(sigma_i(x_j))^2_i,j$ gives the discriminant, where $x_1,x_2,ldots, x_n$ is an integral basis of the number field $K$ over $mathbbQ$, and the $sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
$endgroup$
3
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
add a comment |
$begingroup$
Let $L=mathbbQ(sqrt[3]2)$ and let $f(x)=x^3-2$ be the minimal polynomial of $sqrt[3]2$ over $mathbbQ$. Consider $e_1,e_2,e_3:=1,x,x^2$ be a basis of $L$ over $mathbbQ$. Let $theta_1 = sqrt[3]2,theta_2 = wsqrt[3]2$ and $theta_3=w^2sqrt[3]2$ be the roots of $f$ (in $mathbbC$ as a splitting field of $f$) where $w$ is a complex cube root of $1$. Then the discriminant is the square of the determinant of the matrix $big(sigma_i(e_j)big)_ij$ where $sigma_i$ are $mathbbQ$-embeddings.
We know that the $i$-th embedding takes $x$ to $theta_i$, i.e. $sigma_i(1)=1$, $sigma_i(x)=theta_i$ and $sigma_i(x^2)=theta_i^2$. Clearly the matrix $big(sigma_i(e_j)big)_ij = big(theta_i^j-1big)$ is Vandermonde whose determinant satisfies
beginequation
detbig(theta_i^j-1big)^2 = (theta_1-theta_2)^2(theta_1-theta_3)^2(theta_2-theta_3)^2 = -108.
endequation
Now we obtain the discriminant.
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.
Let $f in mathbbQ[x]$ monic of degree $n$ be the minimum polynomial of $alpha$ and let $K=mathbbQ(alpha)$.
- The discriminant of $mathbbZ[alpha]$ is $(-1)^frac12n(n-1)N^K_mathbbQ(f'(alpha))$.
- The field norm $N^K_mathbbQ$ is multiplicative.
- $N^K_mathbbQ(b) = b^n$ for $b in mathbbQ$.
- $N^K_mathbbQ(alpha)$ is $(-1)^n$ times the constant term of $f$.
Putting these together for $f = x^3 - 2$ (using $mathcalO_K = mathbbZ[alpha]$ here, which is nontrivial) yields
The absolute value of the discriminant of $mathbbQ(sqrt[3]2)$ is $N(3alpha^2) = N(3)N(alpha)^2 = 3^3 cdot 2^2$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
$endgroup$
1
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
3
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
add a comment |
$begingroup$
I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.
Let $f in mathbbQ[x]$ monic of degree $n$ be the minimum polynomial of $alpha$ and let $K=mathbbQ(alpha)$.
- The discriminant of $mathbbZ[alpha]$ is $(-1)^frac12n(n-1)N^K_mathbbQ(f'(alpha))$.
- The field norm $N^K_mathbbQ$ is multiplicative.
- $N^K_mathbbQ(b) = b^n$ for $b in mathbbQ$.
- $N^K_mathbbQ(alpha)$ is $(-1)^n$ times the constant term of $f$.
Putting these together for $f = x^3 - 2$ (using $mathcalO_K = mathbbZ[alpha]$ here, which is nontrivial) yields
The absolute value of the discriminant of $mathbbQ(sqrt[3]2)$ is $N(3alpha^2) = N(3)N(alpha)^2 = 3^3 cdot 2^2$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
$endgroup$
1
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
3
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
add a comment |
$begingroup$
I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.
Let $f in mathbbQ[x]$ monic of degree $n$ be the minimum polynomial of $alpha$ and let $K=mathbbQ(alpha)$.
- The discriminant of $mathbbZ[alpha]$ is $(-1)^frac12n(n-1)N^K_mathbbQ(f'(alpha))$.
- The field norm $N^K_mathbbQ$ is multiplicative.
- $N^K_mathbbQ(b) = b^n$ for $b in mathbbQ$.
- $N^K_mathbbQ(alpha)$ is $(-1)^n$ times the constant term of $f$.
Putting these together for $f = x^3 - 2$ (using $mathcalO_K = mathbbZ[alpha]$ here, which is nontrivial) yields
The absolute value of the discriminant of $mathbbQ(sqrt[3]2)$ is $N(3alpha^2) = N(3)N(alpha)^2 = 3^3 cdot 2^2$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
$endgroup$
I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.
Let $f in mathbbQ[x]$ monic of degree $n$ be the minimum polynomial of $alpha$ and let $K=mathbbQ(alpha)$.
- The discriminant of $mathbbZ[alpha]$ is $(-1)^frac12n(n-1)N^K_mathbbQ(f'(alpha))$.
- The field norm $N^K_mathbbQ$ is multiplicative.
- $N^K_mathbbQ(b) = b^n$ for $b in mathbbQ$.
- $N^K_mathbbQ(alpha)$ is $(-1)^n$ times the constant term of $f$.
Putting these together for $f = x^3 - 2$ (using $mathcalO_K = mathbbZ[alpha]$ here, which is nontrivial) yields
The absolute value of the discriminant of $mathbbQ(sqrt[3]2)$ is $N(3alpha^2) = N(3)N(alpha)^2 = 3^3 cdot 2^2$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
answered Mar 15 '17 at 13:51
Ricardo BuringRicardo Buring
1,90621437
1,90621437
1
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
3
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
add a comment |
1
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
3
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
1
1
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
$begingroup$
what's the discriminant of $Bbb Q(sqrt[3] 4)$ ?
$endgroup$
– mercio
Mar 15 '17 at 14:01
3
3
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
This formula works only if powers of $alpha$ form an integral basis. In case you didn't get mercio's hint :-)
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:07
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
$begingroup$
Thanks for the correction; I edited my answer accordingly.
$endgroup$
– Ricardo Buring
Mar 17 '17 at 17:05
add a comment |
$begingroup$
Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(sigma_i(x_j))^2_i,j$ gives the discriminant, where $x_1,x_2,ldots, x_n$ is an integral basis of the number field $K$ over $mathbbQ$, and the $sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
$endgroup$
3
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
add a comment |
$begingroup$
Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(sigma_i(x_j))^2_i,j$ gives the discriminant, where $x_1,x_2,ldots, x_n$ is an integral basis of the number field $K$ over $mathbbQ$, and the $sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
$endgroup$
3
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
add a comment |
$begingroup$
Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(sigma_i(x_j))^2_i,j$ gives the discriminant, where $x_1,x_2,ldots, x_n$ is an integral basis of the number field $K$ over $mathbbQ$, and the $sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
$endgroup$
Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(sigma_i(x_j))^2_i,j$ gives the discriminant, where $x_1,x_2,ldots, x_n$ is an integral basis of the number field $K$ over $mathbbQ$, and the $sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
answered Mar 11 '17 at 19:45
Dietrich BurdeDietrich Burde
82.3k649107
82.3k649107
3
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
add a comment |
3
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
3
3
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
$begingroup$
Emphasis: $x_1,x_2,ldots,x_n$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral.
$endgroup$
– Jyrki Lahtonen
Mar 15 '17 at 14:06
add a comment |
$begingroup$
Let $L=mathbbQ(sqrt[3]2)$ and let $f(x)=x^3-2$ be the minimal polynomial of $sqrt[3]2$ over $mathbbQ$. Consider $e_1,e_2,e_3:=1,x,x^2$ be a basis of $L$ over $mathbbQ$. Let $theta_1 = sqrt[3]2,theta_2 = wsqrt[3]2$ and $theta_3=w^2sqrt[3]2$ be the roots of $f$ (in $mathbbC$ as a splitting field of $f$) where $w$ is a complex cube root of $1$. Then the discriminant is the square of the determinant of the matrix $big(sigma_i(e_j)big)_ij$ where $sigma_i$ are $mathbbQ$-embeddings.
We know that the $i$-th embedding takes $x$ to $theta_i$, i.e. $sigma_i(1)=1$, $sigma_i(x)=theta_i$ and $sigma_i(x^2)=theta_i^2$. Clearly the matrix $big(sigma_i(e_j)big)_ij = big(theta_i^j-1big)$ is Vandermonde whose determinant satisfies
beginequation
detbig(theta_i^j-1big)^2 = (theta_1-theta_2)^2(theta_1-theta_3)^2(theta_2-theta_3)^2 = -108.
endequation
Now we obtain the discriminant.
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
$endgroup$
add a comment |
$begingroup$
Let $L=mathbbQ(sqrt[3]2)$ and let $f(x)=x^3-2$ be the minimal polynomial of $sqrt[3]2$ over $mathbbQ$. Consider $e_1,e_2,e_3:=1,x,x^2$ be a basis of $L$ over $mathbbQ$. Let $theta_1 = sqrt[3]2,theta_2 = wsqrt[3]2$ and $theta_3=w^2sqrt[3]2$ be the roots of $f$ (in $mathbbC$ as a splitting field of $f$) where $w$ is a complex cube root of $1$. Then the discriminant is the square of the determinant of the matrix $big(sigma_i(e_j)big)_ij$ where $sigma_i$ are $mathbbQ$-embeddings.
We know that the $i$-th embedding takes $x$ to $theta_i$, i.e. $sigma_i(1)=1$, $sigma_i(x)=theta_i$ and $sigma_i(x^2)=theta_i^2$. Clearly the matrix $big(sigma_i(e_j)big)_ij = big(theta_i^j-1big)$ is Vandermonde whose determinant satisfies
beginequation
detbig(theta_i^j-1big)^2 = (theta_1-theta_2)^2(theta_1-theta_3)^2(theta_2-theta_3)^2 = -108.
endequation
Now we obtain the discriminant.
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
$endgroup$
add a comment |
$begingroup$
Let $L=mathbbQ(sqrt[3]2)$ and let $f(x)=x^3-2$ be the minimal polynomial of $sqrt[3]2$ over $mathbbQ$. Consider $e_1,e_2,e_3:=1,x,x^2$ be a basis of $L$ over $mathbbQ$. Let $theta_1 = sqrt[3]2,theta_2 = wsqrt[3]2$ and $theta_3=w^2sqrt[3]2$ be the roots of $f$ (in $mathbbC$ as a splitting field of $f$) where $w$ is a complex cube root of $1$. Then the discriminant is the square of the determinant of the matrix $big(sigma_i(e_j)big)_ij$ where $sigma_i$ are $mathbbQ$-embeddings.
We know that the $i$-th embedding takes $x$ to $theta_i$, i.e. $sigma_i(1)=1$, $sigma_i(x)=theta_i$ and $sigma_i(x^2)=theta_i^2$. Clearly the matrix $big(sigma_i(e_j)big)_ij = big(theta_i^j-1big)$ is Vandermonde whose determinant satisfies
beginequation
detbig(theta_i^j-1big)^2 = (theta_1-theta_2)^2(theta_1-theta_3)^2(theta_2-theta_3)^2 = -108.
endequation
Now we obtain the discriminant.
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
$endgroup$
Let $L=mathbbQ(sqrt[3]2)$ and let $f(x)=x^3-2$ be the minimal polynomial of $sqrt[3]2$ over $mathbbQ$. Consider $e_1,e_2,e_3:=1,x,x^2$ be a basis of $L$ over $mathbbQ$. Let $theta_1 = sqrt[3]2,theta_2 = wsqrt[3]2$ and $theta_3=w^2sqrt[3]2$ be the roots of $f$ (in $mathbbC$ as a splitting field of $f$) where $w$ is a complex cube root of $1$. Then the discriminant is the square of the determinant of the matrix $big(sigma_i(e_j)big)_ij$ where $sigma_i$ are $mathbbQ$-embeddings.
We know that the $i$-th embedding takes $x$ to $theta_i$, i.e. $sigma_i(1)=1$, $sigma_i(x)=theta_i$ and $sigma_i(x^2)=theta_i^2$. Clearly the matrix $big(sigma_i(e_j)big)_ij = big(theta_i^j-1big)$ is Vandermonde whose determinant satisfies
beginequation
detbig(theta_i^j-1big)^2 = (theta_1-theta_2)^2(theta_1-theta_3)^2(theta_2-theta_3)^2 = -108.
endequation
Now we obtain the discriminant.
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
edited Mar 27 at 16:02
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
answered Mar 27 at 13:24
Yibo ZhangYibo Zhang
11
11
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$begingroup$
Is the class using a textbook or did the teacher just pull that out of thin air?
$endgroup$
– Robert Soupe
Mar 11 '17 at 18:59