Mixing
Degree of the extension $mathbb{Q}(sqrt{3 + 2sqrt{2}})$.
$begingroup$
The following question is from Abstract Algebra by Dummit and Foote.
Determine the degree of the extension $mathbb{Q}(sqrt{3 +
2sqrt{2}})$ over $Bbb Q $.
This question has been answered here and the correct answer is $2$. But I got the answer as $4$ and couldn't find my mistake.
My attempt : I started out by constructing a polynomial $f (x)in Bbb{Q}[x]$ such that $sqrt{3 +
2sqrt{2}}$ is a root of $f(x)$.
$$x=sqrt{3 +2sqrt{2}};implies x^2=3 +2sqrt{2}.$$
Taking $3$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$
Hence $sqrt{3 +2sqrt{2}}$ is a root of $$f (x)=x^4-6x^2+1.$$
By Rational Root Theorem, the only possible rational roots are $pm1$ and neither of these satisfy $f(x)=0$. Hence $f(x)$ is a monic irreducible polynomial which has $sqrt{3 +2sqrt{2}}$ as a root. By the uniqueness property of minimal polynomial, $f(x)$ is the minimal polynomial for $sqrt{3 +2sqrt{2}}$. Since the degree of the extension is same as the degree of minimal polynomial, we have $[mathbb{Q}(sqrt{3 +2sqrt{2}}):Bbb Q]=4.$
So,where did I go wrong? Please help me find out my mistake.
Thank you.
abstract-algebra proof-verification field-theory extension-field
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
add a comment |
$begingroup$
The following question is from Abstract Algebra by Dummit and Foote.
Determine the degree of the extension $mathbb{Q}(sqrt{3 +
2sqrt{2}})$ over $Bbb Q $.
This question has been answered here and the correct answer is $2$. But I got the answer as $4$ and couldn't find my mistake.
My attempt : I started out by constructing a polynomial $f (x)in Bbb{Q}[x]$ such that $sqrt{3 +
2sqrt{2}}$ is a root of $f(x)$.
$$x=sqrt{3 +2sqrt{2}};implies x^2=3 +2sqrt{2}.$$
Taking $3$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$
Hence $sqrt{3 +2sqrt{2}}$ is a root of $$f (x)=x^4-6x^2+1.$$
By Rational Root Theorem, the only possible rational roots are $pm1$ and neither of these satisfy $f(x)=0$. Hence $f(x)$ is a monic irreducible polynomial which has $sqrt{3 +2sqrt{2}}$ as a root. By the uniqueness property of minimal polynomial, $f(x)$ is the minimal polynomial for $sqrt{3 +2sqrt{2}}$. Since the degree of the extension is same as the degree of minimal polynomial, we have $[mathbb{Q}(sqrt{3 +2sqrt{2}}):Bbb Q]=4.$
So,where did I go wrong? Please help me find out my mistake.
Thank you.
abstract-algebra proof-verification field-theory extension-field
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
2
$begingroup$
Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $mathbb{Q}$, for instance.
$endgroup$
– André 3000
Jan 23 at 13:42
1
$begingroup$
@André3000 Thanks a lot. That cleared everything.
$endgroup$
– Thomas Shelby
Jan 23 at 13:55
$begingroup$
Someone else did the same mistake here.
$endgroup$
– Thomas Shelby
Jan 24 at 1:53
add a comment |
$begingroup$
The following question is from Abstract Algebra by Dummit and Foote.
Determine the degree of the extension $mathbb{Q}(sqrt{3 +
2sqrt{2}})$ over $Bbb Q $.
This question has been answered here and the correct answer is $2$. But I got the answer as $4$ and couldn't find my mistake.
My attempt : I started out by constructing a polynomial $f (x)in Bbb{Q}[x]$ such that $sqrt{3 +
2sqrt{2}}$ is a root of $f(x)$.
$$x=sqrt{3 +2sqrt{2}};implies x^2=3 +2sqrt{2}.$$
Taking $3$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$
Hence $sqrt{3 +2sqrt{2}}$ is a root of $$f (x)=x^4-6x^2+1.$$
By Rational Root Theorem, the only possible rational roots are $pm1$ and neither of these satisfy $f(x)=0$. Hence $f(x)$ is a monic irreducible polynomial which has $sqrt{3 +2sqrt{2}}$ as a root. By the uniqueness property of minimal polynomial, $f(x)$ is the minimal polynomial for $sqrt{3 +2sqrt{2}}$. Since the degree of the extension is same as the degree of minimal polynomial, we have $[mathbb{Q}(sqrt{3 +2sqrt{2}}):Bbb Q]=4.$
So,where did I go wrong? Please help me find out my mistake.
Thank you.
abstract-algebra proof-verification field-theory extension-field
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
The following question is from Abstract Algebra by Dummit and Foote.
Determine the degree of the extension $mathbb{Q}(sqrt{3 +
2sqrt{2}})$ over $Bbb Q $.
This question has been answered here and the correct answer is $2$. But I got the answer as $4$ and couldn't find my mistake.
My attempt : I started out by constructing a polynomial $f (x)in Bbb{Q}[x]$ such that $sqrt{3 +
2sqrt{2}}$ is a root of $f(x)$.
$$x=sqrt{3 +2sqrt{2}};implies x^2=3 +2sqrt{2}.$$
Taking $3$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$
Hence $sqrt{3 +2sqrt{2}}$ is a root of $$f (x)=x^4-6x^2+1.$$
By Rational Root Theorem, the only possible rational roots are $pm1$ and neither of these satisfy $f(x)=0$. Hence $f(x)$ is a monic irreducible polynomial which has $sqrt{3 +2sqrt{2}}$ as a root. By the uniqueness property of minimal polynomial, $f(x)$ is the minimal polynomial for $sqrt{3 +2sqrt{2}}$. Since the degree of the extension is same as the degree of minimal polynomial, we have $[mathbb{Q}(sqrt{3 +2sqrt{2}}):Bbb Q]=4.$
So,where did I go wrong? Please help me find out my mistake.
Thank you.
abstract-algebra proof-verification field-theory extension-field
abstract-algebra proof-verification field-theory extension-field
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
asked Jan 23 at 13:17
Thomas ShelbyThomas Shelby
4,0342625
4,0342625
2
$begingroup$
Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $mathbb{Q}$, for instance.
$endgroup$
– André 3000
Jan 23 at 13:42
1
$begingroup$
@André3000 Thanks a lot. That cleared everything.
$endgroup$
– Thomas Shelby
Jan 23 at 13:55
$begingroup$
Someone else did the same mistake here.
$endgroup$
– Thomas Shelby
Jan 24 at 1:53
add a comment |
2
$begingroup$
Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $mathbb{Q}$, for instance.
$endgroup$
– André 3000
Jan 23 at 13:42
1
$begingroup$
@André3000 Thanks a lot. That cleared everything.
$endgroup$
– Thomas Shelby
Jan 23 at 13:55
$begingroup$
Someone else did the same mistake here.
$endgroup$
– Thomas Shelby
Jan 24 at 1:53
2
2
$begingroup$
Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $mathbb{Q}$, for instance.
$endgroup$
– André 3000
Jan 23 at 13:42
$begingroup$
Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $mathbb{Q}$, for instance.
$endgroup$
– André 3000
Jan 23 at 13:42
1
1
$begingroup$
@André3000 Thanks a lot. That cleared everything.
$endgroup$
– Thomas Shelby
Jan 23 at 13:55
$begingroup$
@André3000 Thanks a lot. That cleared everything.
$endgroup$
– Thomas Shelby
Jan 23 at 13:55
$begingroup$
Someone else did the same mistake here.
$endgroup$
– Thomas Shelby
Jan 24 at 1:53
$begingroup$
Someone else did the same mistake here.
$endgroup$
– Thomas Shelby
Jan 24 at 1:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Because $$3+2sqrt2=(1+sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
1
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
2
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
1
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
|
show 1 more comment
Your Answer
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1 Answer
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$begingroup$
Because $$3+2sqrt2=(1+sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
1
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
2
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
1
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
|
show 1 more comment
$begingroup$
Because $$3+2sqrt2=(1+sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
1
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
2
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
1
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
|
show 1 more comment
$begingroup$
Because $$3+2sqrt2=(1+sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Because $$3+2sqrt2=(1+sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 23 at 13:27
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 13:19
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
1
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
2
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
1
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
|
show 1 more comment
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
1
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
2
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
1
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
$begingroup$
I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"?
$endgroup$
– Thomas Shelby
Jan 23 at 13:25
1
1
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
$begingroup$
@Thomas Shelby I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:27
2
2
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
Okay. So lack of rational roots only means that they cannot be factored into linear factors in $Bbb Q $. The polynomial can still be factored as a product other irreducible polynomials. Am I right?
$endgroup$
– Thomas Shelby
Jan 23 at 13:37
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
$begingroup$
The polynomial, where $sqrt{3+2sqrt2}$ is a root must be irreducible above $mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $mathbb Q$, which says $mathbb Qleft(sqrt{3+2sqrt2}right):mathbb Q<4$.
$endgroup$
– Michael Rozenberg
Jan 23 at 13:40
1
1
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
$begingroup$
You are welcome, Thomas!
$endgroup$
– Michael Rozenberg
Jan 23 at 13:55
|
show 1 more comment
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$begingroup$
Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $mathbb{Q}$, for instance.
$endgroup$
– André 3000
Jan 23 at 13:42
1
$begingroup$
@André3000 Thanks a lot. That cleared everything.
$endgroup$
– Thomas Shelby
Jan 23 at 13:55
$begingroup$
Someone else did the same mistake here.
$endgroup$
– Thomas Shelby
Jan 24 at 1:53