Find a factorisation in a cubic field.
Take $alpha$ a root of $x^3-7$ and consider $mathbb{Q}(alpha)$. This is a pure cubic field. I am interested in the ideals which ramify. Since the discriminant here is $27cdot7^2$, I have to look at the ideals generated by 3 and 7 in $mathbb{Z}[alpha]$, which is the ring of integers. Surely $(7)=(alpha)^3$.
Now we consider $(3)$. I think I have to find $a,b,cin mathbb{Z}$ such that $$(3)=(3,alpha+a)(3,alpha+b)(3,alpha+c).$$ Computing, I get a system of modular equation which solution is $a,b,c=2$. Thus $$(3)supset(3,alpha+2)(3,alpha+2)(3,alpha+2).$$ For the other inclusion, is there a better way or have I to find 3 in this product by hand? I think that not always is simple. And more, I am always sure that the ideals in the factorisation are of the form $(3,alpha+a)$? Why not, for example, $(3,alpha^2+a)$?
algebraic-number-theory extension-field
asked Nov 29 at 19:57
Yecabel
1547
add a comment |
Take $alpha$ a root of $x^3-7$ and consider $mathbb{Q}(alpha)$. This is a pure cubic field. I am interested in the ideals which ramify. Since the discriminant here is $27cdot7^2$, I have to look at the ideals generated by 3 and 7 in $mathbb{Z}[alpha]$, which is the ring of integers. Surely $(7)=(alpha)^3$.
Now we consider $(3)$. I think I have to find $a,b,cin mathbb{Z}$ such that $$(3)=(3,alpha+a)(3,alpha+b)(3,alpha+c).$$ Computing, I get a system of modular equation which solution is $a,b,c=2$. Thus $$(3)supset(3,alpha+2)(3,alpha+2)(3,alpha+2).$$ For the other inclusion, is there a better way or have I to find 3 in this product by hand? I think that not always is simple. And more, I am always sure that the ideals in the factorisation are of the form $(3,alpha+a)$? Why not, for example, $(3,alpha^2+a)$?
algebraic-number-theory extension-field
asked Nov 29 at 19:57
Yecabel
1547
add a comment |
Take $alpha$ a root of $x^3-7$ and consider $mathbb{Q}(alpha)$. This is a pure cubic field. I am interested in the ideals which ramify. Since the discriminant here is $27cdot7^2$, I have to look at the ideals generated by 3 and 7 in $mathbb{Z}[alpha]$, which is the ring of integers. Surely $(7)=(alpha)^3$.
Now we consider $(3)$. I think I have to find $a,b,cin mathbb{Z}$ such that $$(3)=(3,alpha+a)(3,alpha+b)(3,alpha+c).$$ Computing, I get a system of modular equation which solution is $a,b,c=2$. Thus $$(3)supset(3,alpha+2)(3,alpha+2)(3,alpha+2).$$ For the other inclusion, is there a better way or have I to find 3 in this product by hand? I think that not always is simple. And more, I am always sure that the ideals in the factorisation are of the form $(3,alpha+a)$? Why not, for example, $(3,alpha^2+a)$?
algebraic-number-theory extension-field
asked Nov 29 at 19:57
Yecabel
1547
Take $alpha$ a root of $x^3-7$ and consider $mathbb{Q}(alpha)$. This is a pure cubic field. I am interested in the ideals which ramify. Since the discriminant here is $27cdot7^2$, I have to look at the ideals generated by 3 and 7 in $mathbb{Z}[alpha]$, which is the ring of integers. Surely $(7)=(alpha)^3$.
Now we consider $(3)$. I think I have to find $a,b,cin mathbb{Z}$ such that $$(3)=(3,alpha+a)(3,alpha+b)(3,alpha+c).$$ Computing, I get a system of modular equation which solution is $a,b,c=2$. Thus $$(3)supset(3,alpha+2)(3,alpha+2)(3,alpha+2).$$ For the other inclusion, is there a better way or have I to find 3 in this product by hand? I think that not always is simple. And more, I am always sure that the ideals in the factorisation are of the form $(3,alpha+a)$? Why not, for example, $(3,alpha^2+a)$?
algebraic-number-theory extension-field
algebraic-number-theory extension-field
asked Nov 29 at 19:57
Yecabel
1547
asked Nov 29 at 19:57
Yecabel
1547
asked Nov 29 at 19:57
Yecabel
1547
asked Nov 29 at 19:57
Yecabel
1547
asked Nov 29 at 19:57
Yecabel
1547
1547
add a comment |
add a comment |
1 Answer
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Are you aware of Kummer's factorization theorem? This gives you the answer instantly. Since $x^2-7$ is the minimal polynomial of $alpha$ over $Bbb Q$, we can take its reduction mod $3$ and reduce it into a product of irreducibles to get the prime factorization of $(3)$ in $Bbb Q(alpha)$: as polynomials in $Bbb Z/3Bbb Z$ we have
$$x^3-7=x^3+2=(x+2)^3$$
and therefore $(3)=mathfrak p^3$ where $mathfrak p=(3,alpha+2)$.
Edit: You could also look at the "norm" function, it sends an ideal $IsubsetBbb Z[alpha]$ to the cardinality of $Bbb Z[alpha]/I$ (this is always finite if $I$ is nonzero since $Bbb Z[alpha]$ is the ring of integers of a number field). This norm is multilpicative with respect to products of ideals, and as a result, in general we have that if
$$Isubseteq Jtext{ and } N(I)=N(J)implies I=J.$$
Applied to our case, if you could show that
$$N((3))=N((3,alpha+2)^3)=(N(3,alpha+2))^3$$
(which is the same as showing that $N(3,alpha+2)=3$), then it follows that the two ideals are equal.
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
1
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
1
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
|
show 1 more comment
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Are you aware of Kummer's factorization theorem? This gives you the answer instantly. Since $x^2-7$ is the minimal polynomial of $alpha$ over $Bbb Q$, we can take its reduction mod $3$ and reduce it into a product of irreducibles to get the prime factorization of $(3)$ in $Bbb Q(alpha)$: as polynomials in $Bbb Z/3Bbb Z$ we have
$$x^3-7=x^3+2=(x+2)^3$$
and therefore $(3)=mathfrak p^3$ where $mathfrak p=(3,alpha+2)$.
Edit: You could also look at the "norm" function, it sends an ideal $IsubsetBbb Z[alpha]$ to the cardinality of $Bbb Z[alpha]/I$ (this is always finite if $I$ is nonzero since $Bbb Z[alpha]$ is the ring of integers of a number field). This norm is multilpicative with respect to products of ideals, and as a result, in general we have that if
$$Isubseteq Jtext{ and } N(I)=N(J)implies I=J.$$
Applied to our case, if you could show that
$$N((3))=N((3,alpha+2)^3)=(N(3,alpha+2))^3$$
(which is the same as showing that $N(3,alpha+2)=3$), then it follows that the two ideals are equal.
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
1
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
1
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
|
show 1 more comment
Are you aware of Kummer's factorization theorem? This gives you the answer instantly. Since $x^2-7$ is the minimal polynomial of $alpha$ over $Bbb Q$, we can take its reduction mod $3$ and reduce it into a product of irreducibles to get the prime factorization of $(3)$ in $Bbb Q(alpha)$: as polynomials in $Bbb Z/3Bbb Z$ we have
$$x^3-7=x^3+2=(x+2)^3$$
and therefore $(3)=mathfrak p^3$ where $mathfrak p=(3,alpha+2)$.
Edit: You could also look at the "norm" function, it sends an ideal $IsubsetBbb Z[alpha]$ to the cardinality of $Bbb Z[alpha]/I$ (this is always finite if $I$ is nonzero since $Bbb Z[alpha]$ is the ring of integers of a number field). This norm is multilpicative with respect to products of ideals, and as a result, in general we have that if
$$Isubseteq Jtext{ and } N(I)=N(J)implies I=J.$$
Applied to our case, if you could show that
$$N((3))=N((3,alpha+2)^3)=(N(3,alpha+2))^3$$
(which is the same as showing that $N(3,alpha+2)=3$), then it follows that the two ideals are equal.
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
1
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
1
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
|
show 1 more comment
Are you aware of Kummer's factorization theorem? This gives you the answer instantly. Since $x^2-7$ is the minimal polynomial of $alpha$ over $Bbb Q$, we can take its reduction mod $3$ and reduce it into a product of irreducibles to get the prime factorization of $(3)$ in $Bbb Q(alpha)$: as polynomials in $Bbb Z/3Bbb Z$ we have
$$x^3-7=x^3+2=(x+2)^3$$
and therefore $(3)=mathfrak p^3$ where $mathfrak p=(3,alpha+2)$.
Edit: You could also look at the "norm" function, it sends an ideal $IsubsetBbb Z[alpha]$ to the cardinality of $Bbb Z[alpha]/I$ (this is always finite if $I$ is nonzero since $Bbb Z[alpha]$ is the ring of integers of a number field). This norm is multilpicative with respect to products of ideals, and as a result, in general we have that if
$$Isubseteq Jtext{ and } N(I)=N(J)implies I=J.$$
Applied to our case, if you could show that
$$N((3))=N((3,alpha+2)^3)=(N(3,alpha+2))^3$$
(which is the same as showing that $N(3,alpha+2)=3$), then it follows that the two ideals are equal.
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
Are you aware of Kummer's factorization theorem? This gives you the answer instantly. Since $x^2-7$ is the minimal polynomial of $alpha$ over $Bbb Q$, we can take its reduction mod $3$ and reduce it into a product of irreducibles to get the prime factorization of $(3)$ in $Bbb Q(alpha)$: as polynomials in $Bbb Z/3Bbb Z$ we have
$$x^3-7=x^3+2=(x+2)^3$$
and therefore $(3)=mathfrak p^3$ where $mathfrak p=(3,alpha+2)$.
Edit: You could also look at the "norm" function, it sends an ideal $IsubsetBbb Z[alpha]$ to the cardinality of $Bbb Z[alpha]/I$ (this is always finite if $I$ is nonzero since $Bbb Z[alpha]$ is the ring of integers of a number field). This norm is multilpicative with respect to products of ideals, and as a result, in general we have that if
$$Isubseteq Jtext{ and } N(I)=N(J)implies I=J.$$
Applied to our case, if you could show that
$$N((3))=N((3,alpha+2)^3)=(N(3,alpha+2))^3$$
(which is the same as showing that $N(3,alpha+2)=3$), then it follows that the two ideals are equal.
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
edited Nov 29 at 20:24
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
answered Nov 29 at 20:09
Alex Mathers
10.7k21344
10.7k21344
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
1
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
1
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
|
show 1 more comment
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
1
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
1
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
No, I was not aware of this theorem, thank you! Are there other techniques to show it?
– Yecabel
Nov 29 at 20:17
1
1
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
@Yecabel See my edit
– Alex Mathers
Nov 29 at 20:24
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
We get $N(3,alpha+2)=|mathbb{Z}[alpha]/(3,alpha+2)|=|mathbb{Z}[x]/(x^3-7,3,x+2)|=|mathbb{Z}_3[x]/(x+2)|=3$. Is this correct?
– Yecabel
Nov 29 at 20:38
1
1
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
@Yecabel Looks good, nice!
– Alex Mathers
Nov 29 at 20:40
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
You need to be a bit more careful. Modular factorization of the minimal polynomial of $alpha$ works as prescribed if one the conditions A) the factors are all simple, B) the powers of $alpha$ form an integral basis.
– Jyrki Lahtonen
Dec 1 at 20:27
|
show 1 more comment
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