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Galois number fields that have the imaginary unit.
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$newcommandQmathbb Q$Is it possible to find an irreducible polynomial $fin Q[x]$ of degree $4$ such that the following holds:
- All roots of $f$ are non-real
- The splitting field of $f$, $K_f$, contains "the" imaginary unit $mathrm i$
- The maximum real subfield of $K_f$, i.e. $K_fcap mathbb R$, is not Galois over $Q$
- $mathrmGal(K_f/Q)cong A_4$
I have the feeling that this cannot be done but I'm not sure how to prove this. Item 3. suggests that $K_f$ should not be some sort of cyclotomic extension but that is all I have figured. Can it be some radical extension? there is not much room to work with because I have only degree $4$.
In general, I would also like to have an intuition what the implication of having the imaginary unit is in a Galois number field. Will the Galois group have some specific property or will there be some characterizations for such fields?
Edit: I managed to find an example that satisfies 1,3 and 4. For instance $f=x^4 - x^3 - 3x + 4$.
field-theory galois-theory algebraic-number-theory splitting-field
asked Aug 9 at 5:40
quantum
43929
add a comment |Â
up vote
3
down vote
favorite
$newcommandQmathbb Q$Is it possible to find an irreducible polynomial $fin Q[x]$ of degree $4$ such that the following holds:
- All roots of $f$ are non-real
- The splitting field of $f$, $K_f$, contains "the" imaginary unit $mathrm i$
- The maximum real subfield of $K_f$, i.e. $K_fcap mathbb R$, is not Galois over $Q$
- $mathrmGal(K_f/Q)cong A_4$
I have the feeling that this cannot be done but I'm not sure how to prove this. Item 3. suggests that $K_f$ should not be some sort of cyclotomic extension but that is all I have figured. Can it be some radical extension? there is not much room to work with because I have only degree $4$.
In general, I would also like to have an intuition what the implication of having the imaginary unit is in a Galois number field. Will the Galois group have some specific property or will there be some characterizations for such fields?
Edit: I managed to find an example that satisfies 1,3 and 4. For instance $f=x^4 - x^3 - 3x + 4$.
field-theory galois-theory algebraic-number-theory splitting-field
asked Aug 9 at 5:40
quantum
43929
2
Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation.
– daruma
Aug 9 at 5:55
2
We have $i in K_f$ if and only if $mathbf Q(i) subset K_f$. For number fields $E$ and $F$ that are Galois over $mathbf Q$, Bauer's Theorem implies that $F subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = mathbf Q(i)$ we get that $mathbf Q(i) subset E$ if and only if the primes splitting completely in $E$ all satisfy $p equiv 1 bmod 4$. So that is one characterization of Galois number fields containing $i$.
– KCd
Aug 9 at 6:05
2
If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_fcap mathbb R$ and for this field to be galois over $mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$.
– ArithmeticGeometer
Aug 9 at 6:07
1
@KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question.
– quantum
Aug 9 at 7:09
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$newcommandQmathbb Q$Is it possible to find an irreducible polynomial $fin Q[x]$ of degree $4$ such that the following holds:
- All roots of $f$ are non-real
- The splitting field of $f$, $K_f$, contains "the" imaginary unit $mathrm i$
- The maximum real subfield of $K_f$, i.e. $K_fcap mathbb R$, is not Galois over $Q$
- $mathrmGal(K_f/Q)cong A_4$
I have the feeling that this cannot be done but I'm not sure how to prove this. Item 3. suggests that $K_f$ should not be some sort of cyclotomic extension but that is all I have figured. Can it be some radical extension? there is not much room to work with because I have only degree $4$.
In general, I would also like to have an intuition what the implication of having the imaginary unit is in a Galois number field. Will the Galois group have some specific property or will there be some characterizations for such fields?
Edit: I managed to find an example that satisfies 1,3 and 4. For instance $f=x^4 - x^3 - 3x + 4$.
field-theory galois-theory algebraic-number-theory splitting-field
asked Aug 9 at 5:40
quantum
43929
$newcommandQmathbb Q$Is it possible to find an irreducible polynomial $fin Q[x]$ of degree $4$ such that the following holds:
- All roots of $f$ are non-real
- The splitting field of $f$, $K_f$, contains "the" imaginary unit $mathrm i$
- The maximum real subfield of $K_f$, i.e. $K_fcap mathbb R$, is not Galois over $Q$
- $mathrmGal(K_f/Q)cong A_4$
I have the feeling that this cannot be done but I'm not sure how to prove this. Item 3. suggests that $K_f$ should not be some sort of cyclotomic extension but that is all I have figured. Can it be some radical extension? there is not much room to work with because I have only degree $4$.
In general, I would also like to have an intuition what the implication of having the imaginary unit is in a Galois number field. Will the Galois group have some specific property or will there be some characterizations for such fields?
Edit: I managed to find an example that satisfies 1,3 and 4. For instance $f=x^4 - x^3 - 3x + 4$.
field-theory galois-theory algebraic-number-theory splitting-field
asked Aug 9 at 5:40
quantum
43929
edited Aug 20 at 18:08
asked Aug 9 at 5:40
quantum
43929
asked Aug 9 at 5:40
quantum
43929
asked Aug 9 at 5:40
quantum
43929
43929
2
Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation.
– daruma
Aug 9 at 5:55
2
We have $i in K_f$ if and only if $mathbf Q(i) subset K_f$. For number fields $E$ and $F$ that are Galois over $mathbf Q$, Bauer's Theorem implies that $F subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = mathbf Q(i)$ we get that $mathbf Q(i) subset E$ if and only if the primes splitting completely in $E$ all satisfy $p equiv 1 bmod 4$. So that is one characterization of Galois number fields containing $i$.
– KCd
Aug 9 at 6:05
2
If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_fcap mathbb R$ and for this field to be galois over $mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$.
– ArithmeticGeometer
Aug 9 at 6:07
1
@KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question.
– quantum
Aug 9 at 7:09
add a comment |Â
2
Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation.
– daruma
Aug 9 at 5:55
2
We have $i in K_f$ if and only if $mathbf Q(i) subset K_f$. For number fields $E$ and $F$ that are Galois over $mathbf Q$, Bauer's Theorem implies that $F subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = mathbf Q(i)$ we get that $mathbf Q(i) subset E$ if and only if the primes splitting completely in $E$ all satisfy $p equiv 1 bmod 4$. So that is one characterization of Galois number fields containing $i$.
– KCd
Aug 9 at 6:05
2
If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_fcap mathbb R$ and for this field to be galois over $mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$.
– ArithmeticGeometer
Aug 9 at 6:07
1
@KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question.
– quantum
Aug 9 at 7:09
2
2
Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation.
– daruma
Aug 9 at 5:55
Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation.
– daruma
Aug 9 at 5:55
2
2
We have $i in K_f$ if and only if $mathbf Q(i) subset K_f$. For number fields $E$ and $F$ that are Galois over $mathbf Q$, Bauer's Theorem implies that $F subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = mathbf Q(i)$ we get that $mathbf Q(i) subset E$ if and only if the primes splitting completely in $E$ all satisfy $p equiv 1 bmod 4$. So that is one characterization of Galois number fields containing $i$.
– KCd
Aug 9 at 6:05
We have $i in K_f$ if and only if $mathbf Q(i) subset K_f$. For number fields $E$ and $F$ that are Galois over $mathbf Q$, Bauer's Theorem implies that $F subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = mathbf Q(i)$ we get that $mathbf Q(i) subset E$ if and only if the primes splitting completely in $E$ all satisfy $p equiv 1 bmod 4$. So that is one characterization of Galois number fields containing $i$.
– KCd
Aug 9 at 6:05
2
2
If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_fcap mathbb R$ and for this field to be galois over $mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$.
– ArithmeticGeometer
Aug 9 at 6:07
If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_fcap mathbb R$ and for this field to be galois over $mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$.
– ArithmeticGeometer
Aug 9 at 6:07
1
1
@KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question.
– quantum
Aug 9 at 7:09
@KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question.
– quantum
Aug 9 at 7:09
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
Items 2 and 4 are incompatible. Assume that
$$
BbbQsubset BbbQ(i)subset K_f,
$$
where $K_f$ is Galois over $BbbQ$, $G=Gal(K_f/BbbQ)simeq A_4$. By Galois correspondence $BbbQ(i)$ is the fixed field of a subgroup $Hle G$. Because $[BbbQ(i):BbbQ]=2$ it follows that $|H|=|A_4|/2=6$.
But $A_4$ has no subgroups of order six.
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
1
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Items 2 and 4 are incompatible. Assume that
$$
BbbQsubset BbbQ(i)subset K_f,
$$
where $K_f$ is Galois over $BbbQ$, $G=Gal(K_f/BbbQ)simeq A_4$. By Galois correspondence $BbbQ(i)$ is the fixed field of a subgroup $Hle G$. Because $[BbbQ(i):BbbQ]=2$ it follows that $|H|=|A_4|/2=6$.
But $A_4$ has no subgroups of order six.
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
1
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
 |Â
show 1 more comment
up vote
7
down vote
accepted
Items 2 and 4 are incompatible. Assume that
$$
BbbQsubset BbbQ(i)subset K_f,
$$
where $K_f$ is Galois over $BbbQ$, $G=Gal(K_f/BbbQ)simeq A_4$. By Galois correspondence $BbbQ(i)$ is the fixed field of a subgroup $Hle G$. Because $[BbbQ(i):BbbQ]=2$ it follows that $|H|=|A_4|/2=6$.
But $A_4$ has no subgroups of order six.
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
1
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
 |Â
show 1 more comment
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Items 2 and 4 are incompatible. Assume that
$$
BbbQsubset BbbQ(i)subset K_f,
$$
where $K_f$ is Galois over $BbbQ$, $G=Gal(K_f/BbbQ)simeq A_4$. By Galois correspondence $BbbQ(i)$ is the fixed field of a subgroup $Hle G$. Because $[BbbQ(i):BbbQ]=2$ it follows that $|H|=|A_4|/2=6$.
But $A_4$ has no subgroups of order six.
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
Items 2 and 4 are incompatible. Assume that
$$
BbbQsubset BbbQ(i)subset K_f,
$$
where $K_f$ is Galois over $BbbQ$, $G=Gal(K_f/BbbQ)simeq A_4$. By Galois correspondence $BbbQ(i)$ is the fixed field of a subgroup $Hle G$. Because $[BbbQ(i):BbbQ]=2$ it follows that $|H|=|A_4|/2=6$.
But $A_4$ has no subgroups of order six.
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
edited Aug 9 at 6:40
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
answered Aug 9 at 6:33
Jyrki Lahtonen
105k12161358
105k12161358
1
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
 |Â
show 1 more comment
1
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
1
1
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $sqrt2,sqrt3,sqrt-2notin K_f$.
– Jyrki Lahtonen
Aug 9 at 6:38
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
Wonderful argument. Thank you.
– quantum
Aug 9 at 7:06
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible.
– quantum
Aug 10 at 9:47
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
@quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question.
– Jyrki Lahtonen
Aug 10 at 10:14
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though!
– quantum
Aug 10 at 10:46
 |Â
show 1 more comment
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2
Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation.
– daruma
Aug 9 at 5:55
2
We have $i in K_f$ if and only if $mathbf Q(i) subset K_f$. For number fields $E$ and $F$ that are Galois over $mathbf Q$, Bauer's Theorem implies that $F subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = mathbf Q(i)$ we get that $mathbf Q(i) subset E$ if and only if the primes splitting completely in $E$ all satisfy $p equiv 1 bmod 4$. So that is one characterization of Galois number fields containing $i$.
– KCd
Aug 9 at 6:05
2
If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_fcap mathbb R$ and for this field to be galois over $mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$.
– ArithmeticGeometer
Aug 9 at 6:07
1
@KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question.
– quantum
Aug 9 at 7:09