Mixing
What is the minimal polynomial of $sqrt{3+4i}+sqrt{3-4i}$ over $mathbb{Q}$?
$begingroup$
My attempt:
I know that $P=mathbb{Q(sqrt{3+4i}+sqrt{3-4i})} subseteq mathbb{Q(sqrt{3+4i},sqrt{3-4i})}=K$. But I don't know whether $K subseteq P$. Assuming $K=P$. I can see that K is Galois over $mathbb{Q}$(as the minimal polynomial $x^4-6x^2+25$ which splits into distinct roots in K) and that would be the minimal polynomial of $sqrt{3+4i}+sqrt{3-4i}$.
Thanks for any help!
abstract-algebra field-theory galois-theory minimal-polynomials
asked Jan 11 at 18:33
InfinityInfinity
326112
$endgroup$
add a comment |
$begingroup$
My attempt:
I know that $P=mathbb{Q(sqrt{3+4i}+sqrt{3-4i})} subseteq mathbb{Q(sqrt{3+4i},sqrt{3-4i})}=K$. But I don't know whether $K subseteq P$. Assuming $K=P$. I can see that K is Galois over $mathbb{Q}$(as the minimal polynomial $x^4-6x^2+25$ which splits into distinct roots in K) and that would be the minimal polynomial of $sqrt{3+4i}+sqrt{3-4i}$.
Thanks for any help!
abstract-algebra field-theory galois-theory minimal-polynomials
asked Jan 11 at 18:33
InfinityInfinity
326112
$endgroup$
5
$begingroup$
By $sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $sqrt{3-4i}$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:36
$begingroup$
@LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$.
$endgroup$
– 6005
Jan 11 at 19:18
$begingroup$
Hmm, actually maybe it does matter. When adjoining $z = sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it?
$endgroup$
– 6005
Jan 11 at 19:25
$begingroup$
And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$.
$endgroup$
– 6005
Jan 11 at 19:27
1
$begingroup$
Cool, $mathbb{3+4i}$ looks fancy.
$endgroup$
– Dietrich Burde
Jan 11 at 19:39
add a comment |
$begingroup$
My attempt:
I know that $P=mathbb{Q(sqrt{3+4i}+sqrt{3-4i})} subseteq mathbb{Q(sqrt{3+4i},sqrt{3-4i})}=K$. But I don't know whether $K subseteq P$. Assuming $K=P$. I can see that K is Galois over $mathbb{Q}$(as the minimal polynomial $x^4-6x^2+25$ which splits into distinct roots in K) and that would be the minimal polynomial of $sqrt{3+4i}+sqrt{3-4i}$.
Thanks for any help!
abstract-algebra field-theory galois-theory minimal-polynomials
asked Jan 11 at 18:33
InfinityInfinity
326112
$endgroup$
My attempt:
I know that $P=mathbb{Q(sqrt{3+4i}+sqrt{3-4i})} subseteq mathbb{Q(sqrt{3+4i},sqrt{3-4i})}=K$. But I don't know whether $K subseteq P$. Assuming $K=P$. I can see that K is Galois over $mathbb{Q}$(as the minimal polynomial $x^4-6x^2+25$ which splits into distinct roots in K) and that would be the minimal polynomial of $sqrt{3+4i}+sqrt{3-4i}$.
Thanks for any help!
abstract-algebra field-theory galois-theory minimal-polynomials
abstract-algebra field-theory galois-theory minimal-polynomials
asked Jan 11 at 18:33
InfinityInfinity
326112
asked Jan 11 at 18:33
InfinityInfinity
326112
asked Jan 11 at 18:33
InfinityInfinity
326112
asked Jan 11 at 18:33
InfinityInfinity
326112
asked Jan 11 at 18:33
InfinityInfinity
326112
326112
5
$begingroup$
By $sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $sqrt{3-4i}$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:36
$begingroup$
@LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$.
$endgroup$
– 6005
Jan 11 at 19:18
$begingroup$
Hmm, actually maybe it does matter. When adjoining $z = sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it?
$endgroup$
– 6005
Jan 11 at 19:25
$begingroup$
And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$.
$endgroup$
– 6005
Jan 11 at 19:27
1
$begingroup$
Cool, $mathbb{3+4i}$ looks fancy.
$endgroup$
– Dietrich Burde
Jan 11 at 19:39
add a comment |
5
$begingroup$
By $sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $sqrt{3-4i}$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:36
$begingroup$
@LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$.
$endgroup$
– 6005
Jan 11 at 19:18
$begingroup$
Hmm, actually maybe it does matter. When adjoining $z = sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it?
$endgroup$
– 6005
Jan 11 at 19:25
$begingroup$
And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$.
$endgroup$
– 6005
Jan 11 at 19:27
1
$begingroup$
Cool, $mathbb{3+4i}$ looks fancy.
$endgroup$
– Dietrich Burde
Jan 11 at 19:39
5
5
$begingroup$
By $sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $sqrt{3-4i}$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:36
$begingroup$
By $sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $sqrt{3-4i}$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:36
$begingroup$
@LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$.
$endgroup$
– 6005
Jan 11 at 19:18
$begingroup$
@LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$.
$endgroup$
– 6005
Jan 11 at 19:18
$begingroup$
Hmm, actually maybe it does matter. When adjoining $z = sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it?
$endgroup$
– 6005
Jan 11 at 19:25
$begingroup$
Hmm, actually maybe it does matter. When adjoining $z = sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it?
$endgroup$
– 6005
Jan 11 at 19:25
$begingroup$
And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$.
$endgroup$
– 6005
Jan 11 at 19:27
$begingroup$
And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$.
$endgroup$
– 6005
Jan 11 at 19:27
1
1
$begingroup$
Cool, $mathbb{3+4i}$ looks fancy.
$endgroup$
– Dietrich Burde
Jan 11 at 19:39
$begingroup$
Cool, $mathbb{3+4i}$ looks fancy.
$endgroup$
– Dietrich Burde
Jan 11 at 19:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As the minimal polynomial can be solved, depending on which roots you denote $;sqrt{3+4i}$ and $sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $mathbf Q$. Hence it can't generate an extension of degree $4$.
On the other hand, it is easy to see that $;K=mathbf Q(sqrt{3+4i})=mathbf Q(sqrt{3-4i})$, since
$$sqrt{3+4i},sqrt{3-4i}=pm 5.$$
answered Jan 11 at 19:07
BernardBernard
120k740116
$endgroup$
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
add a comment |
$begingroup$
Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $mathbb{Q}$.
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
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votes
$begingroup$
As the minimal polynomial can be solved, depending on which roots you denote $;sqrt{3+4i}$ and $sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $mathbf Q$. Hence it can't generate an extension of degree $4$.
On the other hand, it is easy to see that $;K=mathbf Q(sqrt{3+4i})=mathbf Q(sqrt{3-4i})$, since
$$sqrt{3+4i},sqrt{3-4i}=pm 5.$$
answered Jan 11 at 19:07
BernardBernard
120k740116
$endgroup$
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
add a comment |
$begingroup$
As the minimal polynomial can be solved, depending on which roots you denote $;sqrt{3+4i}$ and $sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $mathbf Q$. Hence it can't generate an extension of degree $4$.
On the other hand, it is easy to see that $;K=mathbf Q(sqrt{3+4i})=mathbf Q(sqrt{3-4i})$, since
$$sqrt{3+4i},sqrt{3-4i}=pm 5.$$
answered Jan 11 at 19:07
BernardBernard
120k740116
$endgroup$
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
add a comment |
$begingroup$
As the minimal polynomial can be solved, depending on which roots you denote $;sqrt{3+4i}$ and $sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $mathbf Q$. Hence it can't generate an extension of degree $4$.
On the other hand, it is easy to see that $;K=mathbf Q(sqrt{3+4i})=mathbf Q(sqrt{3-4i})$, since
$$sqrt{3+4i},sqrt{3-4i}=pm 5.$$
answered Jan 11 at 19:07
BernardBernard
120k740116
$endgroup$
As the minimal polynomial can be solved, depending on which roots you denote $;sqrt{3+4i}$ and $sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $mathbf Q$. Hence it can't generate an extension of degree $4$.
On the other hand, it is easy to see that $;K=mathbf Q(sqrt{3+4i})=mathbf Q(sqrt{3-4i})$, since
$$sqrt{3+4i},sqrt{3-4i}=pm 5.$$
answered Jan 11 at 19:07
BernardBernard
120k740116
edited Jan 11 at 19:35
answered Jan 11 at 19:07
BernardBernard
120k740116
answered Jan 11 at 19:07
BernardBernard
120k740116
answered Jan 11 at 19:07
BernardBernard
120k740116
120k740116
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
add a comment |
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
$begingroup$
Oh! yes. (albeit for me, it's a $mathbf Q$). Thanks for pointing it!
$endgroup$
– Bernard
Jan 11 at 19:37
add a comment |
$begingroup$
Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $mathbb{Q}$.
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $mathbb{Q}$.
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $mathbb{Q}$.
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
$endgroup$
Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $mathbb{Q}$.
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
edited Jan 11 at 20:17
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
answered Jan 11 at 20:05
robjohn♦robjohn
267k27308632
267k27308632
add a comment |
add a comment |
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5
$begingroup$
By $sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $sqrt{3-4i}$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:36
$begingroup$
@LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$.
$endgroup$
– 6005
Jan 11 at 19:18
$begingroup$
Hmm, actually maybe it does matter. When adjoining $z = sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it?
$endgroup$
– 6005
Jan 11 at 19:25
$begingroup$
And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$.
$endgroup$
– 6005
Jan 11 at 19:27
1
$begingroup$
Cool, $mathbb{3+4i}$ looks fancy.
$endgroup$
– Dietrich Burde
Jan 11 at 19:39