Mixing
Order of Galois group of $X^n-a$ over $mathbb{Q}$
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The order of a Galois group, $G$, of a polynomial of the type $X^n-a$ over $mathbb{Q}$, with $a, n in mathbb{N}$ is $|G|=n cdot varphi(n)$ when $n$ and $varphi(n)$ are coprime (although I m not sure about this condition). For instance, if $f(X)=X^5-2$, the Galois group has $5 cdotvarphi(5)=5 cdot 4 = 20$ elements$. This comes in handy when deriving the corresponding Galois group.
Can this expression be generalized for the case when $n$ and $varphi(n)$ are not coprime? If so, when does $a$ play a role in the number of elements?
I have checked that if $n$ divides 8, then the right expression seems to be $|G|=frac{1}{2}n cdot varphi(n)$, but I don't know how to proceed with other cases (eg, $4, 6, 10, etc.)$
abstract-algebra galois-theory
asked Jan 9 at 22:32
pendermathpendermath
55912
$endgroup$
|
show 5 more comments
$begingroup$
The order of a Galois group, $G$, of a polynomial of the type $X^n-a$ over $mathbb{Q}$, with $a, n in mathbb{N}$ is $|G|=n cdot varphi(n)$ when $n$ and $varphi(n)$ are coprime (although I m not sure about this condition). For instance, if $f(X)=X^5-2$, the Galois group has $5 cdotvarphi(5)=5 cdot 4 = 20$ elements$. This comes in handy when deriving the corresponding Galois group.
Can this expression be generalized for the case when $n$ and $varphi(n)$ are not coprime? If so, when does $a$ play a role in the number of elements?
I have checked that if $n$ divides 8, then the right expression seems to be $|G|=frac{1}{2}n cdot varphi(n)$, but I don't know how to proceed with other cases (eg, $4, 6, 10, etc.)$
abstract-algebra galois-theory
asked Jan 9 at 22:32
pendermathpendermath
55912
$endgroup$
$begingroup$
You might find useful Chapter IV.9 of Lang's Algebra that treats irreducibility of $x^n-a$ over general fields.
$endgroup$
– Kolja
Jan 9 at 22:45
$begingroup$
You mean chapter VI, right?
$endgroup$
– pendermath
Jan 9 at 22:51
$begingroup$
Indeed, chapter VI.
$endgroup$
– Kolja
Jan 9 at 22:54
1
$begingroup$
Well you know that $G=operatorname{Aut}(mathbb{K}/mathbb{Q})$ where $mathbb{K}=mathbb{Q}[sqrt[n]{a},xi]$ and $xi$ is a primitive $n$'th root of unity. This is exactly the reason why $|G|=n cdot varphi(n)$ when $(n,varphi(n))=1$.
$endgroup$
– Kolja
Jan 9 at 23:20
1
$begingroup$
You need some condition on $a$ as well (not a $d$-th power for some $d|n$ maybe, square-free is going to be cleanest) though.
$endgroup$
– Mindlack
Jan 9 at 23:27
|
show 5 more comments
$begingroup$
The order of a Galois group, $G$, of a polynomial of the type $X^n-a$ over $mathbb{Q}$, with $a, n in mathbb{N}$ is $|G|=n cdot varphi(n)$ when $n$ and $varphi(n)$ are coprime (although I m not sure about this condition). For instance, if $f(X)=X^5-2$, the Galois group has $5 cdotvarphi(5)=5 cdot 4 = 20$ elements$. This comes in handy when deriving the corresponding Galois group.
Can this expression be generalized for the case when $n$ and $varphi(n)$ are not coprime? If so, when does $a$ play a role in the number of elements?
I have checked that if $n$ divides 8, then the right expression seems to be $|G|=frac{1}{2}n cdot varphi(n)$, but I don't know how to proceed with other cases (eg, $4, 6, 10, etc.)$
abstract-algebra galois-theory
asked Jan 9 at 22:32
pendermathpendermath
55912
$endgroup$
The order of a Galois group, $G$, of a polynomial of the type $X^n-a$ over $mathbb{Q}$, with $a, n in mathbb{N}$ is $|G|=n cdot varphi(n)$ when $n$ and $varphi(n)$ are coprime (although I m not sure about this condition). For instance, if $f(X)=X^5-2$, the Galois group has $5 cdotvarphi(5)=5 cdot 4 = 20$ elements$. This comes in handy when deriving the corresponding Galois group.
Can this expression be generalized for the case when $n$ and $varphi(n)$ are not coprime? If so, when does $a$ play a role in the number of elements?
I have checked that if $n$ divides 8, then the right expression seems to be $|G|=frac{1}{2}n cdot varphi(n)$, but I don't know how to proceed with other cases (eg, $4, 6, 10, etc.)$
abstract-algebra galois-theory
abstract-algebra galois-theory
asked Jan 9 at 22:32
pendermathpendermath
55912
asked Jan 9 at 22:32
pendermathpendermath
55912
edited Jan 9 at 23:08
pendermath
asked Jan 9 at 22:32
pendermathpendermath
55912
asked Jan 9 at 22:32
pendermathpendermath
55912
asked Jan 9 at 22:32
pendermathpendermath
55912
55912
$begingroup$
You might find useful Chapter IV.9 of Lang's Algebra that treats irreducibility of $x^n-a$ over general fields.
$endgroup$
– Kolja
Jan 9 at 22:45
$begingroup$
You mean chapter VI, right?
$endgroup$
– pendermath
Jan 9 at 22:51
$begingroup$
Indeed, chapter VI.
$endgroup$
– Kolja
Jan 9 at 22:54
1
$begingroup$
Well you know that $G=operatorname{Aut}(mathbb{K}/mathbb{Q})$ where $mathbb{K}=mathbb{Q}[sqrt[n]{a},xi]$ and $xi$ is a primitive $n$'th root of unity. This is exactly the reason why $|G|=n cdot varphi(n)$ when $(n,varphi(n))=1$.
$endgroup$
– Kolja
Jan 9 at 23:20
1
$begingroup$
You need some condition on $a$ as well (not a $d$-th power for some $d|n$ maybe, square-free is going to be cleanest) though.
$endgroup$
– Mindlack
Jan 9 at 23:27
|
show 5 more comments
$begingroup$
You might find useful Chapter IV.9 of Lang's Algebra that treats irreducibility of $x^n-a$ over general fields.
$endgroup$
– Kolja
Jan 9 at 22:45
$begingroup$
You mean chapter VI, right?
$endgroup$
– pendermath
Jan 9 at 22:51
$begingroup$
Indeed, chapter VI.
$endgroup$
– Kolja
Jan 9 at 22:54
1
$begingroup$
Well you know that $G=operatorname{Aut}(mathbb{K}/mathbb{Q})$ where $mathbb{K}=mathbb{Q}[sqrt[n]{a},xi]$ and $xi$ is a primitive $n$'th root of unity. This is exactly the reason why $|G|=n cdot varphi(n)$ when $(n,varphi(n))=1$.
$endgroup$
– Kolja
Jan 9 at 23:20
1
$begingroup$
You need some condition on $a$ as well (not a $d$-th power for some $d|n$ maybe, square-free is going to be cleanest) though.
$endgroup$
– Mindlack
Jan 9 at 23:27
$begingroup$
You might find useful Chapter IV.9 of Lang's Algebra that treats irreducibility of $x^n-a$ over general fields.
$endgroup$
– Kolja
Jan 9 at 22:45
$begingroup$
You might find useful Chapter IV.9 of Lang's Algebra that treats irreducibility of $x^n-a$ over general fields.
$endgroup$
– Kolja
Jan 9 at 22:45
$begingroup$
You mean chapter VI, right?
$endgroup$
– pendermath
Jan 9 at 22:51
$begingroup$
You mean chapter VI, right?
$endgroup$
– pendermath
Jan 9 at 22:51
$begingroup$
Indeed, chapter VI.
$endgroup$
– Kolja
Jan 9 at 22:54
$begingroup$
Indeed, chapter VI.
$endgroup$
– Kolja
Jan 9 at 22:54
1
1
$begingroup$
Well you know that $G=operatorname{Aut}(mathbb{K}/mathbb{Q})$ where $mathbb{K}=mathbb{Q}[sqrt[n]{a},xi]$ and $xi$ is a primitive $n$'th root of unity. This is exactly the reason why $|G|=n cdot varphi(n)$ when $(n,varphi(n))=1$.
$endgroup$
– Kolja
Jan 9 at 23:20
$begingroup$
Well you know that $G=operatorname{Aut}(mathbb{K}/mathbb{Q})$ where $mathbb{K}=mathbb{Q}[sqrt[n]{a},xi]$ and $xi$ is a primitive $n$'th root of unity. This is exactly the reason why $|G|=n cdot varphi(n)$ when $(n,varphi(n))=1$.
$endgroup$
– Kolja
Jan 9 at 23:20
1
1
$begingroup$
You need some condition on $a$ as well (not a $d$-th power for some $d|n$ maybe, square-free is going to be cleanest) though.
$endgroup$
– Mindlack
Jan 9 at 23:27
$begingroup$
You need some condition on $a$ as well (not a $d$-th power for some $d|n$ maybe, square-free is going to be cleanest) though.
$endgroup$
– Mindlack
Jan 9 at 23:27
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
As noted, the splitting field is $mathbb{Q}(zeta_n, sqrt[n]{a})$ where $zeta_n$ is a primitive $n$-th root of unity. Now it's known that $[mathbb{Q}(zeta_n):mathbb{Q}]=varphi(n)$ so $varphi(n)mid |G|$.
For the remaining part, we need to know a bit more about $a$. For example if $a$ is a $n$-th power already, then the remaining extension is trivial so $a$ does play a role. In fact, the claim about $n$ and $varphi(n)$ being coprime is not actually sufficient due to this: the polynomial $x^2-1$ is completely reducible so has trivial Galois group but $1 neq varphi(2) times 2$.
Morally, the remaining condition to check is whether $a$ is a $d$-th power for any $d mid n$. However, as @Jyrki points out, there are slight problems such as $2$ being a square in $mathbb{Q}(zeta_8)$ but not $mathbb{Q}$. In general, $a$ is always a square in $mathbb{Q}(zeta_{4a})$ (and potentially even in $mathbb{Q}(zeta_a)$ e.g. if $a$ is prime congruent to $1 bmod{4}$). Fortunately, it is only square roots where this problem occurs as noted in the comments.
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
$endgroup$
1
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
add a comment |
$begingroup$
"Can anyone refine @Matt B contribution?"
To the best of my knowledge, the most general result (for any field $K$) on the degree of $K(sqrt [n] a)/K$ does not come from Kummer's theory, but from the following theorems:
1) Let $ain K$ and $m,n$ two coprime integers. Then $X^{mn}-a$ is irreducible over $K$ iff both $X^m-a$ and $X^n-a$ are. This allows to reduce the problem to prime power exponents.
2) Let $p$ be a prime. Suppose that $X^p-a$ is irreducible over $K$, and let $alpha$ be a root. Then: (i) If $pneq2$, or $p=2$ and char$(K)=2$, $a$ is not a $p$-th power in $K(alpha)$; (ii) If $p=2$ and char$(K)neq 2$, $alpha$ is a square in $K(alpha)$ iff $-4a$ is a $4$-th power in $K$. The proof goes through the "Kummer detour".
3) Assume that $a$ is not a $p$-th power in $K$. Then: (i) If $pneq2, X^{p^n}-a$ is irreducible over $K$ for any $n$ ; (ii) If $p=2$ and char$(K)=2, X^{2^n}-a$ is irreducible over $K$ for any $n$ ; (iii) If $p=2, char(K)neq 2$ and $nge 2, X^{2^n}-a$ is irreducible over $K$ iff $-4a$ is not a $4$-th power. The proof relies of course on (1).
Addendum. I have just realized that the OP question was about the degree of the splitting field of $X^n-a$ (I was confused by the expression "Galois group of a polynomial"), so I must go on further, taking $mathbf Q$ as the base field. The final answer depends of course on data given about the element $a$. Let us first concentrate on the $p$-primary case, i.e. $n=p^m$, with $p$ odd. Throughout, suppose that $a$ is not a $p$-th power in $mathbf Q$, and write $alpha_r$ for a $p^r$-th root of $a$. According to 3i), the extension $F=mathbf Q(alpha_m)$ admits a tower of subextensions $F_r=mathbf Q(alpha_r)$ of degree $p^r$ over $mathbf Q , rle m$. On the other hand, consider the cyclotomic field $L=mathbf Q(zeta_{p^m})$ and its canonical tower of cyclic subextensions $L_r=mathbf Q(zeta_{p^r}), rle m$. For $p$ odd, the extensions $F$ and $L_1$ are linearly disjoint, and $a$ is not a $p$-th power in $L_r$ for any $r$ (this comes from linear disjointness for $r=1$, and from the cyclicity of $L_r/L_{r-1}$ for $r>1$). So we can apply again 3i) to construct the tower of extensions $F_r.L_r, rle m$, which culminates at the splitting field $F_m.L_m$, of degree $p^mphi (p^m)$.
The passage from a $p$-primary degree to a composite degree $n$ is easy as long as $n$ is odd ./.
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
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add a comment |
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A very incomplete answer: assume that $a$ is neither a square nor a cube and $n=6$.
Then the splitting field is $mathbb{Q}(a^{1/6})(isqrt{3})$ which does have degree $6*2=12$.
answered Jan 9 at 23:35
MindlackMindlack
3,53717
$endgroup$
$begingroup$
So the condition about coprimality doesn’t work...
$endgroup$
– pendermath
Jan 9 at 23:38
1
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
add a comment |
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3 Answers
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$begingroup$
As noted, the splitting field is $mathbb{Q}(zeta_n, sqrt[n]{a})$ where $zeta_n$ is a primitive $n$-th root of unity. Now it's known that $[mathbb{Q}(zeta_n):mathbb{Q}]=varphi(n)$ so $varphi(n)mid |G|$.
For the remaining part, we need to know a bit more about $a$. For example if $a$ is a $n$-th power already, then the remaining extension is trivial so $a$ does play a role. In fact, the claim about $n$ and $varphi(n)$ being coprime is not actually sufficient due to this: the polynomial $x^2-1$ is completely reducible so has trivial Galois group but $1 neq varphi(2) times 2$.
Morally, the remaining condition to check is whether $a$ is a $d$-th power for any $d mid n$. However, as @Jyrki points out, there are slight problems such as $2$ being a square in $mathbb{Q}(zeta_8)$ but not $mathbb{Q}$. In general, $a$ is always a square in $mathbb{Q}(zeta_{4a})$ (and potentially even in $mathbb{Q}(zeta_a)$ e.g. if $a$ is prime congruent to $1 bmod{4}$). Fortunately, it is only square roots where this problem occurs as noted in the comments.
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
$endgroup$
1
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
add a comment |
$begingroup$
As noted, the splitting field is $mathbb{Q}(zeta_n, sqrt[n]{a})$ where $zeta_n$ is a primitive $n$-th root of unity. Now it's known that $[mathbb{Q}(zeta_n):mathbb{Q}]=varphi(n)$ so $varphi(n)mid |G|$.
For the remaining part, we need to know a bit more about $a$. For example if $a$ is a $n$-th power already, then the remaining extension is trivial so $a$ does play a role. In fact, the claim about $n$ and $varphi(n)$ being coprime is not actually sufficient due to this: the polynomial $x^2-1$ is completely reducible so has trivial Galois group but $1 neq varphi(2) times 2$.
Morally, the remaining condition to check is whether $a$ is a $d$-th power for any $d mid n$. However, as @Jyrki points out, there are slight problems such as $2$ being a square in $mathbb{Q}(zeta_8)$ but not $mathbb{Q}$. In general, $a$ is always a square in $mathbb{Q}(zeta_{4a})$ (and potentially even in $mathbb{Q}(zeta_a)$ e.g. if $a$ is prime congruent to $1 bmod{4}$). Fortunately, it is only square roots where this problem occurs as noted in the comments.
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
$endgroup$
1
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
add a comment |
$begingroup$
As noted, the splitting field is $mathbb{Q}(zeta_n, sqrt[n]{a})$ where $zeta_n$ is a primitive $n$-th root of unity. Now it's known that $[mathbb{Q}(zeta_n):mathbb{Q}]=varphi(n)$ so $varphi(n)mid |G|$.
For the remaining part, we need to know a bit more about $a$. For example if $a$ is a $n$-th power already, then the remaining extension is trivial so $a$ does play a role. In fact, the claim about $n$ and $varphi(n)$ being coprime is not actually sufficient due to this: the polynomial $x^2-1$ is completely reducible so has trivial Galois group but $1 neq varphi(2) times 2$.
Morally, the remaining condition to check is whether $a$ is a $d$-th power for any $d mid n$. However, as @Jyrki points out, there are slight problems such as $2$ being a square in $mathbb{Q}(zeta_8)$ but not $mathbb{Q}$. In general, $a$ is always a square in $mathbb{Q}(zeta_{4a})$ (and potentially even in $mathbb{Q}(zeta_a)$ e.g. if $a$ is prime congruent to $1 bmod{4}$). Fortunately, it is only square roots where this problem occurs as noted in the comments.
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
$endgroup$
As noted, the splitting field is $mathbb{Q}(zeta_n, sqrt[n]{a})$ where $zeta_n$ is a primitive $n$-th root of unity. Now it's known that $[mathbb{Q}(zeta_n):mathbb{Q}]=varphi(n)$ so $varphi(n)mid |G|$.
For the remaining part, we need to know a bit more about $a$. For example if $a$ is a $n$-th power already, then the remaining extension is trivial so $a$ does play a role. In fact, the claim about $n$ and $varphi(n)$ being coprime is not actually sufficient due to this: the polynomial $x^2-1$ is completely reducible so has trivial Galois group but $1 neq varphi(2) times 2$.
Morally, the remaining condition to check is whether $a$ is a $d$-th power for any $d mid n$. However, as @Jyrki points out, there are slight problems such as $2$ being a square in $mathbb{Q}(zeta_8)$ but not $mathbb{Q}$. In general, $a$ is always a square in $mathbb{Q}(zeta_{4a})$ (and potentially even in $mathbb{Q}(zeta_a)$ e.g. if $a$ is prime congruent to $1 bmod{4}$). Fortunately, it is only square roots where this problem occurs as noted in the comments.
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
edited Jan 10 at 11:46
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
answered Jan 10 at 0:44
Matt BMatt B
2,7881615
2,7881615
1
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
add a comment |
1
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
1
1
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Re: the last paragraph, we have subtleties coming from the fact that square roots of primes belong to some cyclotomic fields. $sqrt2inBbb{Q}(zeta_8)$, $sqrt3inBbb{Q}(zeta_{12})$, $sqrt5inBbb{Q}(zeta_5)$ etc. This may effect the extension degree when $n$ is even. Irrational cube roots (or higher) of integers never belong to a cyclotomic field because all subfields of cyclotomic fields are Galois, hence normal.
$endgroup$
– Jyrki Lahtonen
Jan 10 at 6:45
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
$begingroup$
Can anyone refine @Matt B contribution?
$endgroup$
– pendermath
Jan 10 at 10:39
add a comment |
$begingroup$
"Can anyone refine @Matt B contribution?"
To the best of my knowledge, the most general result (for any field $K$) on the degree of $K(sqrt [n] a)/K$ does not come from Kummer's theory, but from the following theorems:
1) Let $ain K$ and $m,n$ two coprime integers. Then $X^{mn}-a$ is irreducible over $K$ iff both $X^m-a$ and $X^n-a$ are. This allows to reduce the problem to prime power exponents.
2) Let $p$ be a prime. Suppose that $X^p-a$ is irreducible over $K$, and let $alpha$ be a root. Then: (i) If $pneq2$, or $p=2$ and char$(K)=2$, $a$ is not a $p$-th power in $K(alpha)$; (ii) If $p=2$ and char$(K)neq 2$, $alpha$ is a square in $K(alpha)$ iff $-4a$ is a $4$-th power in $K$. The proof goes through the "Kummer detour".
3) Assume that $a$ is not a $p$-th power in $K$. Then: (i) If $pneq2, X^{p^n}-a$ is irreducible over $K$ for any $n$ ; (ii) If $p=2$ and char$(K)=2, X^{2^n}-a$ is irreducible over $K$ for any $n$ ; (iii) If $p=2, char(K)neq 2$ and $nge 2, X^{2^n}-a$ is irreducible over $K$ iff $-4a$ is not a $4$-th power. The proof relies of course on (1).
Addendum. I have just realized that the OP question was about the degree of the splitting field of $X^n-a$ (I was confused by the expression "Galois group of a polynomial"), so I must go on further, taking $mathbf Q$ as the base field. The final answer depends of course on data given about the element $a$. Let us first concentrate on the $p$-primary case, i.e. $n=p^m$, with $p$ odd. Throughout, suppose that $a$ is not a $p$-th power in $mathbf Q$, and write $alpha_r$ for a $p^r$-th root of $a$. According to 3i), the extension $F=mathbf Q(alpha_m)$ admits a tower of subextensions $F_r=mathbf Q(alpha_r)$ of degree $p^r$ over $mathbf Q , rle m$. On the other hand, consider the cyclotomic field $L=mathbf Q(zeta_{p^m})$ and its canonical tower of cyclic subextensions $L_r=mathbf Q(zeta_{p^r}), rle m$. For $p$ odd, the extensions $F$ and $L_1$ are linearly disjoint, and $a$ is not a $p$-th power in $L_r$ for any $r$ (this comes from linear disjointness for $r=1$, and from the cyclicity of $L_r/L_{r-1}$ for $r>1$). So we can apply again 3i) to construct the tower of extensions $F_r.L_r, rle m$, which culminates at the splitting field $F_m.L_m$, of degree $p^mphi (p^m)$.
The passage from a $p$-primary degree to a composite degree $n$ is easy as long as $n$ is odd ./.
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
$endgroup$
add a comment |
$begingroup$
"Can anyone refine @Matt B contribution?"
To the best of my knowledge, the most general result (for any field $K$) on the degree of $K(sqrt [n] a)/K$ does not come from Kummer's theory, but from the following theorems:
1) Let $ain K$ and $m,n$ two coprime integers. Then $X^{mn}-a$ is irreducible over $K$ iff both $X^m-a$ and $X^n-a$ are. This allows to reduce the problem to prime power exponents.
2) Let $p$ be a prime. Suppose that $X^p-a$ is irreducible over $K$, and let $alpha$ be a root. Then: (i) If $pneq2$, or $p=2$ and char$(K)=2$, $a$ is not a $p$-th power in $K(alpha)$; (ii) If $p=2$ and char$(K)neq 2$, $alpha$ is a square in $K(alpha)$ iff $-4a$ is a $4$-th power in $K$. The proof goes through the "Kummer detour".
3) Assume that $a$ is not a $p$-th power in $K$. Then: (i) If $pneq2, X^{p^n}-a$ is irreducible over $K$ for any $n$ ; (ii) If $p=2$ and char$(K)=2, X^{2^n}-a$ is irreducible over $K$ for any $n$ ; (iii) If $p=2, char(K)neq 2$ and $nge 2, X^{2^n}-a$ is irreducible over $K$ iff $-4a$ is not a $4$-th power. The proof relies of course on (1).
Addendum. I have just realized that the OP question was about the degree of the splitting field of $X^n-a$ (I was confused by the expression "Galois group of a polynomial"), so I must go on further, taking $mathbf Q$ as the base field. The final answer depends of course on data given about the element $a$. Let us first concentrate on the $p$-primary case, i.e. $n=p^m$, with $p$ odd. Throughout, suppose that $a$ is not a $p$-th power in $mathbf Q$, and write $alpha_r$ for a $p^r$-th root of $a$. According to 3i), the extension $F=mathbf Q(alpha_m)$ admits a tower of subextensions $F_r=mathbf Q(alpha_r)$ of degree $p^r$ over $mathbf Q , rle m$. On the other hand, consider the cyclotomic field $L=mathbf Q(zeta_{p^m})$ and its canonical tower of cyclic subextensions $L_r=mathbf Q(zeta_{p^r}), rle m$. For $p$ odd, the extensions $F$ and $L_1$ are linearly disjoint, and $a$ is not a $p$-th power in $L_r$ for any $r$ (this comes from linear disjointness for $r=1$, and from the cyclicity of $L_r/L_{r-1}$ for $r>1$). So we can apply again 3i) to construct the tower of extensions $F_r.L_r, rle m$, which culminates at the splitting field $F_m.L_m$, of degree $p^mphi (p^m)$.
The passage from a $p$-primary degree to a composite degree $n$ is easy as long as $n$ is odd ./.
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
$endgroup$
add a comment |
$begingroup$
"Can anyone refine @Matt B contribution?"
To the best of my knowledge, the most general result (for any field $K$) on the degree of $K(sqrt [n] a)/K$ does not come from Kummer's theory, but from the following theorems:
1) Let $ain K$ and $m,n$ two coprime integers. Then $X^{mn}-a$ is irreducible over $K$ iff both $X^m-a$ and $X^n-a$ are. This allows to reduce the problem to prime power exponents.
2) Let $p$ be a prime. Suppose that $X^p-a$ is irreducible over $K$, and let $alpha$ be a root. Then: (i) If $pneq2$, or $p=2$ and char$(K)=2$, $a$ is not a $p$-th power in $K(alpha)$; (ii) If $p=2$ and char$(K)neq 2$, $alpha$ is a square in $K(alpha)$ iff $-4a$ is a $4$-th power in $K$. The proof goes through the "Kummer detour".
3) Assume that $a$ is not a $p$-th power in $K$. Then: (i) If $pneq2, X^{p^n}-a$ is irreducible over $K$ for any $n$ ; (ii) If $p=2$ and char$(K)=2, X^{2^n}-a$ is irreducible over $K$ for any $n$ ; (iii) If $p=2, char(K)neq 2$ and $nge 2, X^{2^n}-a$ is irreducible over $K$ iff $-4a$ is not a $4$-th power. The proof relies of course on (1).
Addendum. I have just realized that the OP question was about the degree of the splitting field of $X^n-a$ (I was confused by the expression "Galois group of a polynomial"), so I must go on further, taking $mathbf Q$ as the base field. The final answer depends of course on data given about the element $a$. Let us first concentrate on the $p$-primary case, i.e. $n=p^m$, with $p$ odd. Throughout, suppose that $a$ is not a $p$-th power in $mathbf Q$, and write $alpha_r$ for a $p^r$-th root of $a$. According to 3i), the extension $F=mathbf Q(alpha_m)$ admits a tower of subextensions $F_r=mathbf Q(alpha_r)$ of degree $p^r$ over $mathbf Q , rle m$. On the other hand, consider the cyclotomic field $L=mathbf Q(zeta_{p^m})$ and its canonical tower of cyclic subextensions $L_r=mathbf Q(zeta_{p^r}), rle m$. For $p$ odd, the extensions $F$ and $L_1$ are linearly disjoint, and $a$ is not a $p$-th power in $L_r$ for any $r$ (this comes from linear disjointness for $r=1$, and from the cyclicity of $L_r/L_{r-1}$ for $r>1$). So we can apply again 3i) to construct the tower of extensions $F_r.L_r, rle m$, which culminates at the splitting field $F_m.L_m$, of degree $p^mphi (p^m)$.
The passage from a $p$-primary degree to a composite degree $n$ is easy as long as $n$ is odd ./.
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
$endgroup$
"Can anyone refine @Matt B contribution?"
To the best of my knowledge, the most general result (for any field $K$) on the degree of $K(sqrt [n] a)/K$ does not come from Kummer's theory, but from the following theorems:
1) Let $ain K$ and $m,n$ two coprime integers. Then $X^{mn}-a$ is irreducible over $K$ iff both $X^m-a$ and $X^n-a$ are. This allows to reduce the problem to prime power exponents.
2) Let $p$ be a prime. Suppose that $X^p-a$ is irreducible over $K$, and let $alpha$ be a root. Then: (i) If $pneq2$, or $p=2$ and char$(K)=2$, $a$ is not a $p$-th power in $K(alpha)$; (ii) If $p=2$ and char$(K)neq 2$, $alpha$ is a square in $K(alpha)$ iff $-4a$ is a $4$-th power in $K$. The proof goes through the "Kummer detour".
3) Assume that $a$ is not a $p$-th power in $K$. Then: (i) If $pneq2, X^{p^n}-a$ is irreducible over $K$ for any $n$ ; (ii) If $p=2$ and char$(K)=2, X^{2^n}-a$ is irreducible over $K$ for any $n$ ; (iii) If $p=2, char(K)neq 2$ and $nge 2, X^{2^n}-a$ is irreducible over $K$ iff $-4a$ is not a $4$-th power. The proof relies of course on (1).
Addendum. I have just realized that the OP question was about the degree of the splitting field of $X^n-a$ (I was confused by the expression "Galois group of a polynomial"), so I must go on further, taking $mathbf Q$ as the base field. The final answer depends of course on data given about the element $a$. Let us first concentrate on the $p$-primary case, i.e. $n=p^m$, with $p$ odd. Throughout, suppose that $a$ is not a $p$-th power in $mathbf Q$, and write $alpha_r$ for a $p^r$-th root of $a$. According to 3i), the extension $F=mathbf Q(alpha_m)$ admits a tower of subextensions $F_r=mathbf Q(alpha_r)$ of degree $p^r$ over $mathbf Q , rle m$. On the other hand, consider the cyclotomic field $L=mathbf Q(zeta_{p^m})$ and its canonical tower of cyclic subextensions $L_r=mathbf Q(zeta_{p^r}), rle m$. For $p$ odd, the extensions $F$ and $L_1$ are linearly disjoint, and $a$ is not a $p$-th power in $L_r$ for any $r$ (this comes from linear disjointness for $r=1$, and from the cyclicity of $L_r/L_{r-1}$ for $r>1$). So we can apply again 3i) to construct the tower of extensions $F_r.L_r, rle m$, which culminates at the splitting field $F_m.L_m$, of degree $p^mphi (p^m)$.
The passage from a $p$-primary degree to a composite degree $n$ is easy as long as $n$ is odd ./.
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
edited Jan 12 at 9:13
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
answered Jan 11 at 21:24
nguyen quang donguyen quang do
8,6341723
8,6341723
add a comment |
add a comment |
$begingroup$
A very incomplete answer: assume that $a$ is neither a square nor a cube and $n=6$.
Then the splitting field is $mathbb{Q}(a^{1/6})(isqrt{3})$ which does have degree $6*2=12$.
answered Jan 9 at 23:35
MindlackMindlack
3,53717
$endgroup$
$begingroup$
So the condition about coprimality doesn’t work...
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– pendermath
Jan 9 at 23:38
1
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
add a comment |
$begingroup$
A very incomplete answer: assume that $a$ is neither a square nor a cube and $n=6$.
Then the splitting field is $mathbb{Q}(a^{1/6})(isqrt{3})$ which does have degree $6*2=12$.
answered Jan 9 at 23:35
MindlackMindlack
3,53717
$endgroup$
$begingroup$
So the condition about coprimality doesn’t work...
$endgroup$
– pendermath
Jan 9 at 23:38
1
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
add a comment |
$begingroup$
A very incomplete answer: assume that $a$ is neither a square nor a cube and $n=6$.
Then the splitting field is $mathbb{Q}(a^{1/6})(isqrt{3})$ which does have degree $6*2=12$.
answered Jan 9 at 23:35
MindlackMindlack
3,53717
$endgroup$
A very incomplete answer: assume that $a$ is neither a square nor a cube and $n=6$.
Then the splitting field is $mathbb{Q}(a^{1/6})(isqrt{3})$ which does have degree $6*2=12$.
answered Jan 9 at 23:35
MindlackMindlack
3,53717
answered Jan 9 at 23:35
MindlackMindlack
3,53717
answered Jan 9 at 23:35
MindlackMindlack
3,53717
answered Jan 9 at 23:35
MindlackMindlack
3,53717
3,53717
$begingroup$
So the condition about coprimality doesn’t work...
$endgroup$
– pendermath
Jan 9 at 23:38
1
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
add a comment |
$begingroup$
So the condition about coprimality doesn’t work...
$endgroup$
– pendermath
Jan 9 at 23:38
1
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
$begingroup$
So the condition about coprimality doesn’t work...
$endgroup$
– pendermath
Jan 9 at 23:38
$begingroup$
So the condition about coprimality doesn’t work...
$endgroup$
– pendermath
Jan 9 at 23:38
1
1
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
$begingroup$
To be more accurate, it is sufficient, not necessary.
$endgroup$
– Mindlack
Jan 9 at 23:41
add a comment |
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You might find useful Chapter IV.9 of Lang's Algebra that treats irreducibility of $x^n-a$ over general fields.
$endgroup$
– Kolja
Jan 9 at 22:45
$begingroup$
You mean chapter VI, right?
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– pendermath
Jan 9 at 22:51
$begingroup$
Indeed, chapter VI.
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– Kolja
Jan 9 at 22:54
1
$begingroup$
Well you know that $G=operatorname{Aut}(mathbb{K}/mathbb{Q})$ where $mathbb{K}=mathbb{Q}[sqrt[n]{a},xi]$ and $xi$ is a primitive $n$'th root of unity. This is exactly the reason why $|G|=n cdot varphi(n)$ when $(n,varphi(n))=1$.
$endgroup$
– Kolja
Jan 9 at 23:20
1
$begingroup$
You need some condition on $a$ as well (not a $d$-th power for some $d|n$ maybe, square-free is going to be cleanest) though.
$endgroup$
– Mindlack
Jan 9 at 23:27