Mixing
$mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$
Clash Royale CLAN TAG#URR8PPP
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$textbfProblem mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$ for all $ngeq 2$.
I knew that $mathbbQ[sqrt2,sqrt3]=mathbbQ[sqrt2+sqrt3]$.
Thus, I'll prove the problem by using induction.
Suppose $mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$
It suffices to show that $mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]$ since $$ mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]subset mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1] $$ and $$[ mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]:Q[sqrt2,sqrt3,dots,sqrtn]]leq2$$ imply
$$ mathbbQ[sqrt2+sqrt3+dots+sqrtn+sqrtn+1]=mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]$$
Any help is appreciated..
Thank you!
abstract-algebra field-theory galois-theory extension-field
edited Jul 17 at 11:26
Javi
2,1631725
asked Jul 17 at 10:51
w.sdka
19918
add a comment |Â
up vote
1
down vote
favorite
$textbfProblem mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$ for all $ngeq 2$.
I knew that $mathbbQ[sqrt2,sqrt3]=mathbbQ[sqrt2+sqrt3]$.
Thus, I'll prove the problem by using induction.
Suppose $mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$
It suffices to show that $mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]$ since $$ mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]subset mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1] $$ and $$[ mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]:Q[sqrt2,sqrt3,dots,sqrtn]]leq2$$ imply
$$ mathbbQ[sqrt2+sqrt3+dots+sqrtn+sqrtn+1]=mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]$$
Any help is appreciated..
Thank you!
abstract-algebra field-theory galois-theory extension-field
edited Jul 17 at 11:26
Javi
2,1631725
asked Jul 17 at 10:51
w.sdka
19918
Hint: Look at the degrees of the extensions and what are the irreducible polynomials?
– Michael Burr
Jul 17 at 11:02
I thought the degrees of the extensions depend on $n$ and less than $2^n$. I don't know the irreducible polynomials...
– w.sdka
Jul 17 at 11:05
I don't understand one thing: is $sqrt 2+2sqrt 3in Bbb Q[sqrt 2+sqrt 3]$?
– Mostafa Ayaz
Jul 17 at 11:18
$(sqrt2+sqrt3)^-1=sqrt3-sqrt2 in Q[sqrt2+sqrt3]$
– w.sdka
Jul 17 at 11:25
FWIW: See the comments to an old answer of mine for a Galois theoretic proof. Basically, an element generates a Galois extension iff the element is not fixed by any non-trivial automorphism.
– Jyrki Lahtonen
Jul 20 at 4:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$textbfProblem mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$ for all $ngeq 2$.
I knew that $mathbbQ[sqrt2,sqrt3]=mathbbQ[sqrt2+sqrt3]$.
Thus, I'll prove the problem by using induction.
Suppose $mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$
It suffices to show that $mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]$ since $$ mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]subset mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1] $$ and $$[ mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]:Q[sqrt2,sqrt3,dots,sqrtn]]leq2$$ imply
$$ mathbbQ[sqrt2+sqrt3+dots+sqrtn+sqrtn+1]=mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]$$
Any help is appreciated..
Thank you!
abstract-algebra field-theory galois-theory extension-field
edited Jul 17 at 11:26
Javi
2,1631725
asked Jul 17 at 10:51
w.sdka
19918
$textbfProblem mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$ for all $ngeq 2$.
I knew that $mathbbQ[sqrt2,sqrt3]=mathbbQ[sqrt2+sqrt3]$.
Thus, I'll prove the problem by using induction.
Suppose $mathbbQ[sqrt2,sqrt3,dots,sqrtn]=mathbbQ[sqrt2+sqrt3+cdots+sqrtn]$
It suffices to show that $mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]$ since $$ mathbbQ[sqrt2,sqrt3,dots,sqrtn]subset mathbbQ[sqrt2+sqrt3+cdots+sqrtn+1]subset mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1] $$ and $$[ mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]:Q[sqrt2,sqrt3,dots,sqrtn]]leq2$$ imply
$$ mathbbQ[sqrt2+sqrt3+dots+sqrtn+sqrtn+1]=mathbbQ[sqrt2,sqrt3,dots,sqrtn,sqrtn+1]$$
Any help is appreciated..
Thank you!
abstract-algebra field-theory galois-theory extension-field
edited Jul 17 at 11:26
Javi
2,1631725
asked Jul 17 at 10:51
w.sdka
19918
edited Jul 17 at 11:26
Javi
2,1631725
edited Jul 17 at 11:26
Javi
2,1631725
edited Jul 17 at 11:26
Javi
2,1631725
2,1631725
asked Jul 17 at 10:51
w.sdka
19918
asked Jul 17 at 10:51
w.sdka
19918
asked Jul 17 at 10:51
w.sdka
19918
19918
Hint: Look at the degrees of the extensions and what are the irreducible polynomials?
– Michael Burr
Jul 17 at 11:02
I thought the degrees of the extensions depend on $n$ and less than $2^n$. I don't know the irreducible polynomials...
– w.sdka
Jul 17 at 11:05
I don't understand one thing: is $sqrt 2+2sqrt 3in Bbb Q[sqrt 2+sqrt 3]$?
– Mostafa Ayaz
Jul 17 at 11:18
$(sqrt2+sqrt3)^-1=sqrt3-sqrt2 in Q[sqrt2+sqrt3]$
– w.sdka
Jul 17 at 11:25
FWIW: See the comments to an old answer of mine for a Galois theoretic proof. Basically, an element generates a Galois extension iff the element is not fixed by any non-trivial automorphism.
– Jyrki Lahtonen
Jul 20 at 4:38
add a comment |Â
Hint: Look at the degrees of the extensions and what are the irreducible polynomials?
– Michael Burr
Jul 17 at 11:02
I thought the degrees of the extensions depend on $n$ and less than $2^n$. I don't know the irreducible polynomials...
– w.sdka
Jul 17 at 11:05
I don't understand one thing: is $sqrt 2+2sqrt 3in Bbb Q[sqrt 2+sqrt 3]$?
– Mostafa Ayaz
Jul 17 at 11:18
$(sqrt2+sqrt3)^-1=sqrt3-sqrt2 in Q[sqrt2+sqrt3]$
– w.sdka
Jul 17 at 11:25
FWIW: See the comments to an old answer of mine for a Galois theoretic proof. Basically, an element generates a Galois extension iff the element is not fixed by any non-trivial automorphism.
– Jyrki Lahtonen
Jul 20 at 4:38
Hint: Look at the degrees of the extensions and what are the irreducible polynomials?
– Michael Burr
Jul 17 at 11:02
Hint: Look at the degrees of the extensions and what are the irreducible polynomials?
– Michael Burr
Jul 17 at 11:02
I thought the degrees of the extensions depend on $n$ and less than $2^n$. I don't know the irreducible polynomials...
– w.sdka
Jul 17 at 11:05
I thought the degrees of the extensions depend on $n$ and less than $2^n$. I don't know the irreducible polynomials...
– w.sdka
Jul 17 at 11:05
I don't understand one thing: is $sqrt 2+2sqrt 3in Bbb Q[sqrt 2+sqrt 3]$?
– Mostafa Ayaz
Jul 17 at 11:18
I don't understand one thing: is $sqrt 2+2sqrt 3in Bbb Q[sqrt 2+sqrt 3]$?
– Mostafa Ayaz
Jul 17 at 11:18
$(sqrt2+sqrt3)^-1=sqrt3-sqrt2 in Q[sqrt2+sqrt3]$
– w.sdka
Jul 17 at 11:25
$(sqrt2+sqrt3)^-1=sqrt3-sqrt2 in Q[sqrt2+sqrt3]$
– w.sdka
Jul 17 at 11:25
FWIW: See the comments to an old answer of mine for a Galois theoretic proof. Basically, an element generates a Galois extension iff the element is not fixed by any non-trivial automorphism.
– Jyrki Lahtonen
Jul 20 at 4:38
FWIW: See the comments to an old answer of mine for a Galois theoretic proof. Basically, an element generates a Galois extension iff the element is not fixed by any non-trivial automorphism.
– Jyrki Lahtonen
Jul 20 at 4:38
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The solution to this is straightforward, but in general in any sort of such problems which have to do with simple extensions, you can use the Primitive element theorem and especially its proof. What the theorem tells you is that any finite extension $mathbb Q(a,b)$ can be made to be simple and be written as $mathbb Q(c)$. This is well known. What is perhaps less familiar and which you can see from the proof in the link is what that value c is. To be more precise, c can take the form $c=a+lambda b$ for almost all $lambda$. Looking even more carefully you can point out exactly for which $lambda$ this is not applicable. To put it simply in most of the cases (not all!) you have that $mathbb Q(a,b)=mathbb Q(a+b)$.
Returning to your example you have by induction that $ K=mathbb Q(sqrt2,....,sqrtn,sqrtn+1)=mathbb Q(sqrt2+...+sqrtn,sqrtn+1)$. If you name $a=sqrt2+....+sqrtn$ and $b=sqrtn+1$ you have that $K=mathbb Q(a,b)$ If you spend some time to understand the proof you will realize whether $lambda=1$ is applicable to your example or not in your case.
answered Jul 17 at 11:20
Foivos
35929
1
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The solution to this is straightforward, but in general in any sort of such problems which have to do with simple extensions, you can use the Primitive element theorem and especially its proof. What the theorem tells you is that any finite extension $mathbb Q(a,b)$ can be made to be simple and be written as $mathbb Q(c)$. This is well known. What is perhaps less familiar and which you can see from the proof in the link is what that value c is. To be more precise, c can take the form $c=a+lambda b$ for almost all $lambda$. Looking even more carefully you can point out exactly for which $lambda$ this is not applicable. To put it simply in most of the cases (not all!) you have that $mathbb Q(a,b)=mathbb Q(a+b)$.
Returning to your example you have by induction that $ K=mathbb Q(sqrt2,....,sqrtn,sqrtn+1)=mathbb Q(sqrt2+...+sqrtn,sqrtn+1)$. If you name $a=sqrt2+....+sqrtn$ and $b=sqrtn+1$ you have that $K=mathbb Q(a,b)$ If you spend some time to understand the proof you will realize whether $lambda=1$ is applicable to your example or not in your case.
answered Jul 17 at 11:20
Foivos
35929
1
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
add a comment |Â
up vote
3
down vote
accepted
The solution to this is straightforward, but in general in any sort of such problems which have to do with simple extensions, you can use the Primitive element theorem and especially its proof. What the theorem tells you is that any finite extension $mathbb Q(a,b)$ can be made to be simple and be written as $mathbb Q(c)$. This is well known. What is perhaps less familiar and which you can see from the proof in the link is what that value c is. To be more precise, c can take the form $c=a+lambda b$ for almost all $lambda$. Looking even more carefully you can point out exactly for which $lambda$ this is not applicable. To put it simply in most of the cases (not all!) you have that $mathbb Q(a,b)=mathbb Q(a+b)$.
Returning to your example you have by induction that $ K=mathbb Q(sqrt2,....,sqrtn,sqrtn+1)=mathbb Q(sqrt2+...+sqrtn,sqrtn+1)$. If you name $a=sqrt2+....+sqrtn$ and $b=sqrtn+1$ you have that $K=mathbb Q(a,b)$ If you spend some time to understand the proof you will realize whether $lambda=1$ is applicable to your example or not in your case.
answered Jul 17 at 11:20
Foivos
35929
1
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The solution to this is straightforward, but in general in any sort of such problems which have to do with simple extensions, you can use the Primitive element theorem and especially its proof. What the theorem tells you is that any finite extension $mathbb Q(a,b)$ can be made to be simple and be written as $mathbb Q(c)$. This is well known. What is perhaps less familiar and which you can see from the proof in the link is what that value c is. To be more precise, c can take the form $c=a+lambda b$ for almost all $lambda$. Looking even more carefully you can point out exactly for which $lambda$ this is not applicable. To put it simply in most of the cases (not all!) you have that $mathbb Q(a,b)=mathbb Q(a+b)$.
Returning to your example you have by induction that $ K=mathbb Q(sqrt2,....,sqrtn,sqrtn+1)=mathbb Q(sqrt2+...+sqrtn,sqrtn+1)$. If you name $a=sqrt2+....+sqrtn$ and $b=sqrtn+1$ you have that $K=mathbb Q(a,b)$ If you spend some time to understand the proof you will realize whether $lambda=1$ is applicable to your example or not in your case.
answered Jul 17 at 11:20
Foivos
35929
The solution to this is straightforward, but in general in any sort of such problems which have to do with simple extensions, you can use the Primitive element theorem and especially its proof. What the theorem tells you is that any finite extension $mathbb Q(a,b)$ can be made to be simple and be written as $mathbb Q(c)$. This is well known. What is perhaps less familiar and which you can see from the proof in the link is what that value c is. To be more precise, c can take the form $c=a+lambda b$ for almost all $lambda$. Looking even more carefully you can point out exactly for which $lambda$ this is not applicable. To put it simply in most of the cases (not all!) you have that $mathbb Q(a,b)=mathbb Q(a+b)$.
Returning to your example you have by induction that $ K=mathbb Q(sqrt2,....,sqrtn,sqrtn+1)=mathbb Q(sqrt2+...+sqrtn,sqrtn+1)$. If you name $a=sqrt2+....+sqrtn$ and $b=sqrtn+1$ you have that $K=mathbb Q(a,b)$ If you spend some time to understand the proof you will realize whether $lambda=1$ is applicable to your example or not in your case.
answered Jul 17 at 11:20
Foivos
35929
edited Jul 17 at 14:05
answered Jul 17 at 11:20
Foivos
35929
answered Jul 17 at 11:20
Foivos
35929
answered Jul 17 at 11:20
Foivos
35929
35929
1
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
add a comment |Â
1
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
1
1
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Note that there are many proofs to the Primitive Element theorem, some much simpler than the one I mentioned, but this proof pinpoints exactly the costruction of the simple extension which can be useful in excercises like this one.
– Foivos
Jul 17 at 11:23
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Basically, yes. But the proof of the primitive element theorem I have in mind (at least) relies on having a list of the zeros of the minimal polynomial of $a$. That may need a bit more...
– Jyrki Lahtonen
Jul 20 at 4:36
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
Actually not that much! Given that we can easily jot down a polynomial with rational coefficients and known zeros. Adding my +1
– Jyrki Lahtonen
Jul 20 at 4:43
add a comment |Â
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Hint: Look at the degrees of the extensions and what are the irreducible polynomials?
– Michael Burr
Jul 17 at 11:02
I thought the degrees of the extensions depend on $n$ and less than $2^n$. I don't know the irreducible polynomials...
– w.sdka
Jul 17 at 11:05
I don't understand one thing: is $sqrt 2+2sqrt 3in Bbb Q[sqrt 2+sqrt 3]$?
– Mostafa Ayaz
Jul 17 at 11:18
$(sqrt2+sqrt3)^-1=sqrt3-sqrt2 in Q[sqrt2+sqrt3]$
– w.sdka
Jul 17 at 11:25
FWIW: See the comments to an old answer of mine for a Galois theoretic proof. Basically, an element generates a Galois extension iff the element is not fixed by any non-trivial automorphism.
– Jyrki Lahtonen
Jul 20 at 4:38