$n$th- roots of unit in a splitting field
Let $n geq 2$ and $L$ be the splitting field of $x^n -1$ over $mathbb{Q}$. How do I show that $[L : mathbb{Q}] = phi(n)$, the Euler function of $n$. I tried considering that the set of roots of unit is a group over multiplication in $L$ and showing that it is cyclical. How do I relate this question to roots of unit in the complex numbers?
field-theory extension-field
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
add a comment |
Let $n geq 2$ and $L$ be the splitting field of $x^n -1$ over $mathbb{Q}$. How do I show that $[L : mathbb{Q}] = phi(n)$, the Euler function of $n$. I tried considering that the set of roots of unit is a group over multiplication in $L$ and showing that it is cyclical. How do I relate this question to roots of unit in the complex numbers?
field-theory extension-field
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
1
This amounts to saying that the cyclotomic polynomials are irreducible. Many textbooks have this (e.g. Washington's) but the proofs are a bit indirect.
– Lord Shark the Unknown
Nov 24 at 6:59
add a comment |
Let $n geq 2$ and $L$ be the splitting field of $x^n -1$ over $mathbb{Q}$. How do I show that $[L : mathbb{Q}] = phi(n)$, the Euler function of $n$. I tried considering that the set of roots of unit is a group over multiplication in $L$ and showing that it is cyclical. How do I relate this question to roots of unit in the complex numbers?
field-theory extension-field
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
Let $n geq 2$ and $L$ be the splitting field of $x^n -1$ over $mathbb{Q}$. How do I show that $[L : mathbb{Q}] = phi(n)$, the Euler function of $n$. I tried considering that the set of roots of unit is a group over multiplication in $L$ and showing that it is cyclical. How do I relate this question to roots of unit in the complex numbers?
field-theory extension-field
field-theory extension-field
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
asked Nov 24 at 6:19
Nuntractatuses Amável
61812
61812
1
This amounts to saying that the cyclotomic polynomials are irreducible. Many textbooks have this (e.g. Washington's) but the proofs are a bit indirect.
– Lord Shark the Unknown
Nov 24 at 6:59
add a comment |
1
This amounts to saying that the cyclotomic polynomials are irreducible. Many textbooks have this (e.g. Washington's) but the proofs are a bit indirect.
– Lord Shark the Unknown
Nov 24 at 6:59
1
1
This amounts to saying that the cyclotomic polynomials are irreducible. Many textbooks have this (e.g. Washington's) but the proofs are a bit indirect.
– Lord Shark the Unknown
Nov 24 at 6:59
This amounts to saying that the cyclotomic polynomials are irreducible. Many textbooks have this (e.g. Washington's) but the proofs are a bit indirect.
– Lord Shark the Unknown
Nov 24 at 6:59
add a comment |
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This amounts to saying that the cyclotomic polynomials are irreducible. Many textbooks have this (e.g. Washington's) but the proofs are a bit indirect.
– Lord Shark the Unknown
Nov 24 at 6:59