Mixing
Galois Group of $(x^2-p_1)cdots(x^2-p_n)$
$begingroup$
For distinct prime numbers $p_1,...,p_n$, what is the Galois group of $(x^2-p_1)cdots(x^2-p_n)$ over $mathbb{Q}$?
This problem appears to be quite common, however my understanding of Galois theory is quite poor, and I have no idea how to do this problem.
abstract-algebra field-theory galois-theory
edited Dec 20 '15 at 15:59
user26857
39.3k124083
asked Dec 10 '15 at 9:44
BobBob
8719
$endgroup$
add a comment |
$begingroup$
For distinct prime numbers $p_1,...,p_n$, what is the Galois group of $(x^2-p_1)cdots(x^2-p_n)$ over $mathbb{Q}$?
This problem appears to be quite common, however my understanding of Galois theory is quite poor, and I have no idea how to do this problem.
abstract-algebra field-theory galois-theory
edited Dec 20 '15 at 15:59
user26857
39.3k124083
asked Dec 10 '15 at 9:44
BobBob
8719
$endgroup$
$begingroup$
Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging.
$endgroup$
– Mathmo123
Dec 10 '15 at 10:45
1
$begingroup$
See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this.
$endgroup$
– Jyrki Lahtonen
Dec 10 '15 at 11:14
$begingroup$
Possible duplicate of Generating Elements of Galois Group
$endgroup$
– Watson
Aug 17 '16 at 22:06
add a comment |
$begingroup$
For distinct prime numbers $p_1,...,p_n$, what is the Galois group of $(x^2-p_1)cdots(x^2-p_n)$ over $mathbb{Q}$?
This problem appears to be quite common, however my understanding of Galois theory is quite poor, and I have no idea how to do this problem.
abstract-algebra field-theory galois-theory
edited Dec 20 '15 at 15:59
user26857
39.3k124083
asked Dec 10 '15 at 9:44
BobBob
8719
$endgroup$
For distinct prime numbers $p_1,...,p_n$, what is the Galois group of $(x^2-p_1)cdots(x^2-p_n)$ over $mathbb{Q}$?
This problem appears to be quite common, however my understanding of Galois theory is quite poor, and I have no idea how to do this problem.
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
edited Dec 20 '15 at 15:59
user26857
39.3k124083
asked Dec 10 '15 at 9:44
BobBob
8719
edited Dec 20 '15 at 15:59
user26857
39.3k124083
asked Dec 10 '15 at 9:44
BobBob
8719
edited Dec 20 '15 at 15:59
user26857
39.3k124083
edited Dec 20 '15 at 15:59
user26857
39.3k124083
edited Dec 20 '15 at 15:59
user26857
39.3k124083
39.3k124083
asked Dec 10 '15 at 9:44
BobBob
8719
asked Dec 10 '15 at 9:44
BobBob
8719
asked Dec 10 '15 at 9:44
BobBob
8719
8719
$begingroup$
Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging.
$endgroup$
– Mathmo123
Dec 10 '15 at 10:45
1
$begingroup$
See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this.
$endgroup$
– Jyrki Lahtonen
Dec 10 '15 at 11:14
$begingroup$
Possible duplicate of Generating Elements of Galois Group
$endgroup$
– Watson
Aug 17 '16 at 22:06
add a comment |
$begingroup$
Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging.
$endgroup$
– Mathmo123
Dec 10 '15 at 10:45
1
$begingroup$
See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this.
$endgroup$
– Jyrki Lahtonen
Dec 10 '15 at 11:14
$begingroup$
Possible duplicate of Generating Elements of Galois Group
$endgroup$
– Watson
Aug 17 '16 at 22:06
$begingroup$
Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging.
$endgroup$
– Mathmo123
Dec 10 '15 at 10:45
$begingroup$
Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging.
$endgroup$
– Mathmo123
Dec 10 '15 at 10:45
1
1
$begingroup$
See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this.
$endgroup$
– Jyrki Lahtonen
Dec 10 '15 at 11:14
$begingroup$
See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this.
$endgroup$
– Jyrki Lahtonen
Dec 10 '15 at 11:14
$begingroup$
Possible duplicate of Generating Elements of Galois Group
$endgroup$
– Watson
Aug 17 '16 at 22:06
$begingroup$
Possible duplicate of Generating Elements of Galois Group
$endgroup$
– Watson
Aug 17 '16 at 22:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This question has been asked many times, and been given many answers (see the links, especially those given by Jykri). I propose here a very simple proof based on Kummer theory. I recall the setting of this theory (which can be found in any course on Galois theory): let $m$ be a fixed integer, $K$ a field of characteristic not dividing $m$, containing the group $W_m$ of all $m$th roots of $1$; let $A$ be a subgroup of $K^ast$ containing $K^{ast m}$ , and let $L = K(A^{1/m})$, the field obtained by adding to $K$ all the $m$th roots of all the elements of $A$. Then $L/K$ is Galois, with abelian group $G$ of exponent $m$, isomorphic to $operatorname{Hom}(A/K^{ast m}, W_m)$.
Here we take $K = Bbb Q$, $m = 2$, $A_n$ = the (multiplicative) subgroup generated by $p_1$ , ..., $p_n$ and $Bbb Q^{ast 2}$. We want to show that $Bbb Q((A_n)^{1/2})/Bbb Q$ has Galois group isomorphic to $underbrace{Bbb Z/2Bbb Z times cdots Bbb Z/2Bbb Z}_{text{$n$ times}}$. By Kummer theory, this amounts to show that $A_n mod{Bbb Q^{ast 2}}$ has the same description. Here we shift perspectives and use elementary linear algebra: all the previous multiplicative groups are of exponent 2, hence can be viewed as vector spaces over the field $Bbb Z/2Bbb Z$. Let us show that $p_1 mod {Bbb Q^{ast 2}}$, ..., $p_n mod {Bbb Q^{ast 2}}$ form a basis of $A_n mod {Bbb Q^{ast 2}}$. Any relation of linear dependence between them, written multiplicatively, would be of the form:
A product of distinct $p_i$'s is equal to an element of $Bbb Q^{ast 2}$.
This is impossible by the unicity of decomposition into primes in $Bbb Z$. Thus we have shown that $A_n mod{Bbb Q^{ast 2}}$, as a vector space over $Bbb Z/2Bbb Z$ (written multiplicatively), has dimension $n$. QED
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
$endgroup$
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
add a comment |
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$begingroup$
This question has been asked many times, and been given many answers (see the links, especially those given by Jykri). I propose here a very simple proof based on Kummer theory. I recall the setting of this theory (which can be found in any course on Galois theory): let $m$ be a fixed integer, $K$ a field of characteristic not dividing $m$, containing the group $W_m$ of all $m$th roots of $1$; let $A$ be a subgroup of $K^ast$ containing $K^{ast m}$ , and let $L = K(A^{1/m})$, the field obtained by adding to $K$ all the $m$th roots of all the elements of $A$. Then $L/K$ is Galois, with abelian group $G$ of exponent $m$, isomorphic to $operatorname{Hom}(A/K^{ast m}, W_m)$.
Here we take $K = Bbb Q$, $m = 2$, $A_n$ = the (multiplicative) subgroup generated by $p_1$ , ..., $p_n$ and $Bbb Q^{ast 2}$. We want to show that $Bbb Q((A_n)^{1/2})/Bbb Q$ has Galois group isomorphic to $underbrace{Bbb Z/2Bbb Z times cdots Bbb Z/2Bbb Z}_{text{$n$ times}}$. By Kummer theory, this amounts to show that $A_n mod{Bbb Q^{ast 2}}$ has the same description. Here we shift perspectives and use elementary linear algebra: all the previous multiplicative groups are of exponent 2, hence can be viewed as vector spaces over the field $Bbb Z/2Bbb Z$. Let us show that $p_1 mod {Bbb Q^{ast 2}}$, ..., $p_n mod {Bbb Q^{ast 2}}$ form a basis of $A_n mod {Bbb Q^{ast 2}}$. Any relation of linear dependence between them, written multiplicatively, would be of the form:
A product of distinct $p_i$'s is equal to an element of $Bbb Q^{ast 2}$.
This is impossible by the unicity of decomposition into primes in $Bbb Z$. Thus we have shown that $A_n mod{Bbb Q^{ast 2}}$, as a vector space over $Bbb Z/2Bbb Z$ (written multiplicatively), has dimension $n$. QED
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
$endgroup$
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
add a comment |
$begingroup$
This question has been asked many times, and been given many answers (see the links, especially those given by Jykri). I propose here a very simple proof based on Kummer theory. I recall the setting of this theory (which can be found in any course on Galois theory): let $m$ be a fixed integer, $K$ a field of characteristic not dividing $m$, containing the group $W_m$ of all $m$th roots of $1$; let $A$ be a subgroup of $K^ast$ containing $K^{ast m}$ , and let $L = K(A^{1/m})$, the field obtained by adding to $K$ all the $m$th roots of all the elements of $A$. Then $L/K$ is Galois, with abelian group $G$ of exponent $m$, isomorphic to $operatorname{Hom}(A/K^{ast m}, W_m)$.
Here we take $K = Bbb Q$, $m = 2$, $A_n$ = the (multiplicative) subgroup generated by $p_1$ , ..., $p_n$ and $Bbb Q^{ast 2}$. We want to show that $Bbb Q((A_n)^{1/2})/Bbb Q$ has Galois group isomorphic to $underbrace{Bbb Z/2Bbb Z times cdots Bbb Z/2Bbb Z}_{text{$n$ times}}$. By Kummer theory, this amounts to show that $A_n mod{Bbb Q^{ast 2}}$ has the same description. Here we shift perspectives and use elementary linear algebra: all the previous multiplicative groups are of exponent 2, hence can be viewed as vector spaces over the field $Bbb Z/2Bbb Z$. Let us show that $p_1 mod {Bbb Q^{ast 2}}$, ..., $p_n mod {Bbb Q^{ast 2}}$ form a basis of $A_n mod {Bbb Q^{ast 2}}$. Any relation of linear dependence between them, written multiplicatively, would be of the form:
A product of distinct $p_i$'s is equal to an element of $Bbb Q^{ast 2}$.
This is impossible by the unicity of decomposition into primes in $Bbb Z$. Thus we have shown that $A_n mod{Bbb Q^{ast 2}}$, as a vector space over $Bbb Z/2Bbb Z$ (written multiplicatively), has dimension $n$. QED
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
$endgroup$
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
add a comment |
$begingroup$
This question has been asked many times, and been given many answers (see the links, especially those given by Jykri). I propose here a very simple proof based on Kummer theory. I recall the setting of this theory (which can be found in any course on Galois theory): let $m$ be a fixed integer, $K$ a field of characteristic not dividing $m$, containing the group $W_m$ of all $m$th roots of $1$; let $A$ be a subgroup of $K^ast$ containing $K^{ast m}$ , and let $L = K(A^{1/m})$, the field obtained by adding to $K$ all the $m$th roots of all the elements of $A$. Then $L/K$ is Galois, with abelian group $G$ of exponent $m$, isomorphic to $operatorname{Hom}(A/K^{ast m}, W_m)$.
Here we take $K = Bbb Q$, $m = 2$, $A_n$ = the (multiplicative) subgroup generated by $p_1$ , ..., $p_n$ and $Bbb Q^{ast 2}$. We want to show that $Bbb Q((A_n)^{1/2})/Bbb Q$ has Galois group isomorphic to $underbrace{Bbb Z/2Bbb Z times cdots Bbb Z/2Bbb Z}_{text{$n$ times}}$. By Kummer theory, this amounts to show that $A_n mod{Bbb Q^{ast 2}}$ has the same description. Here we shift perspectives and use elementary linear algebra: all the previous multiplicative groups are of exponent 2, hence can be viewed as vector spaces over the field $Bbb Z/2Bbb Z$. Let us show that $p_1 mod {Bbb Q^{ast 2}}$, ..., $p_n mod {Bbb Q^{ast 2}}$ form a basis of $A_n mod {Bbb Q^{ast 2}}$. Any relation of linear dependence between them, written multiplicatively, would be of the form:
A product of distinct $p_i$'s is equal to an element of $Bbb Q^{ast 2}$.
This is impossible by the unicity of decomposition into primes in $Bbb Z$. Thus we have shown that $A_n mod{Bbb Q^{ast 2}}$, as a vector space over $Bbb Z/2Bbb Z$ (written multiplicatively), has dimension $n$. QED
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
$endgroup$
This question has been asked many times, and been given many answers (see the links, especially those given by Jykri). I propose here a very simple proof based on Kummer theory. I recall the setting of this theory (which can be found in any course on Galois theory): let $m$ be a fixed integer, $K$ a field of characteristic not dividing $m$, containing the group $W_m$ of all $m$th roots of $1$; let $A$ be a subgroup of $K^ast$ containing $K^{ast m}$ , and let $L = K(A^{1/m})$, the field obtained by adding to $K$ all the $m$th roots of all the elements of $A$. Then $L/K$ is Galois, with abelian group $G$ of exponent $m$, isomorphic to $operatorname{Hom}(A/K^{ast m}, W_m)$.
Here we take $K = Bbb Q$, $m = 2$, $A_n$ = the (multiplicative) subgroup generated by $p_1$ , ..., $p_n$ and $Bbb Q^{ast 2}$. We want to show that $Bbb Q((A_n)^{1/2})/Bbb Q$ has Galois group isomorphic to $underbrace{Bbb Z/2Bbb Z times cdots Bbb Z/2Bbb Z}_{text{$n$ times}}$. By Kummer theory, this amounts to show that $A_n mod{Bbb Q^{ast 2}}$ has the same description. Here we shift perspectives and use elementary linear algebra: all the previous multiplicative groups are of exponent 2, hence can be viewed as vector spaces over the field $Bbb Z/2Bbb Z$. Let us show that $p_1 mod {Bbb Q^{ast 2}}$, ..., $p_n mod {Bbb Q^{ast 2}}$ form a basis of $A_n mod {Bbb Q^{ast 2}}$. Any relation of linear dependence between them, written multiplicatively, would be of the form:
A product of distinct $p_i$'s is equal to an element of $Bbb Q^{ast 2}$.
This is impossible by the unicity of decomposition into primes in $Bbb Z$. Thus we have shown that $A_n mod{Bbb Q^{ast 2}}$, as a vector space over $Bbb Z/2Bbb Z$ (written multiplicatively), has dimension $n$. QED
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
edited Jan 2 at 1:03
Kenny Lau
19.9k2159
19.9k2159
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
answered Jan 12 '16 at 7:47
nguyen quang donguyen quang do
8,4611723
8,4611723
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
add a comment |
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
What is MathJax ?
$endgroup$
– nguyen quang do
Jan 12 '16 at 20:31
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
Where can I find it ? I'm kind of "illiterate" concerning Tex
$endgroup$
– nguyen quang do
Jan 13 '16 at 7:07
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
$begingroup$
math.stackexchange.com/help/notation
$endgroup$
– Morgan Rodgers
Jan 13 '16 at 7:43
add a comment |
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$begingroup$
Can you deal with the case where $n=1$? What about $n=2$. Once you've tried the first few cases, you may see a pattern emerging.
$endgroup$
– Mathmo123
Dec 10 '15 at 10:45
1
$begingroup$
See this thread for IMO a good explanation as to why the dimension (and hence also the order of the Galois group) is $2^n$. With that out of the way it is easy to prove that the Galois group is an $n$-fold cartesian power of $C_2$. See also this and this.
$endgroup$
– Jyrki Lahtonen
Dec 10 '15 at 11:14
$begingroup$
Possible duplicate of Generating Elements of Galois Group
$endgroup$
– Watson
Aug 17 '16 at 22:06