Rupture field and splitting field
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Is there a characterization of irreducible polynomials over $mathbb Q$ whose splitting field over $mathbb Q$ are isomorphic to a rupture field?
In other words, of polynomials $P in mathbb Q(X)$ that are irreducible over $mathbb Q$ and that split completely in $mathbb Q(X) /(P)$.
Equivalently, if $alpha$ is any root of $P$ then $mathbb Q(alpha)$ contains every root of $P$.
polynomials extension-field splitting-field
asked Dec 27 '18 at 15:43
RégisRégis
678310
$endgroup$
|
show 2 more comments
$begingroup$
Is there a characterization of irreducible polynomials over $mathbb Q$ whose splitting field over $mathbb Q$ are isomorphic to a rupture field?
In other words, of polynomials $P in mathbb Q(X)$ that are irreducible over $mathbb Q$ and that split completely in $mathbb Q(X) /(P)$.
Equivalently, if $alpha$ is any root of $P$ then $mathbb Q(alpha)$ contains every root of $P$.
polynomials extension-field splitting-field
asked Dec 27 '18 at 15:43
RégisRégis
678310
$endgroup$
2
$begingroup$
What is a rupture field?
$endgroup$
– lhf
Dec 27 '18 at 15:52
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It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $mathbb Q(X) / (P)$.
$endgroup$
– Régis
Dec 27 '18 at 15:54
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Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.)
$endgroup$
– quid♦
Dec 27 '18 at 16:11
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Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though.
$endgroup$
– Régis
Dec 27 '18 at 16:14
$begingroup$
I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though.
$endgroup$
– quid♦
Dec 27 '18 at 20:15
|
show 2 more comments
$begingroup$
Is there a characterization of irreducible polynomials over $mathbb Q$ whose splitting field over $mathbb Q$ are isomorphic to a rupture field?
In other words, of polynomials $P in mathbb Q(X)$ that are irreducible over $mathbb Q$ and that split completely in $mathbb Q(X) /(P)$.
Equivalently, if $alpha$ is any root of $P$ then $mathbb Q(alpha)$ contains every root of $P$.
polynomials extension-field splitting-field
asked Dec 27 '18 at 15:43
RégisRégis
678310
$endgroup$
Is there a characterization of irreducible polynomials over $mathbb Q$ whose splitting field over $mathbb Q$ are isomorphic to a rupture field?
In other words, of polynomials $P in mathbb Q(X)$ that are irreducible over $mathbb Q$ and that split completely in $mathbb Q(X) /(P)$.
Equivalently, if $alpha$ is any root of $P$ then $mathbb Q(alpha)$ contains every root of $P$.
polynomials extension-field splitting-field
polynomials extension-field splitting-field
asked Dec 27 '18 at 15:43
RégisRégis
678310
asked Dec 27 '18 at 15:43
RégisRégis
678310
edited Dec 27 '18 at 15:59
Régis
asked Dec 27 '18 at 15:43
RégisRégis
678310
asked Dec 27 '18 at 15:43
RégisRégis
678310
asked Dec 27 '18 at 15:43
RégisRégis
678310
678310
2
$begingroup$
What is a rupture field?
$endgroup$
– lhf
Dec 27 '18 at 15:52
$begingroup$
It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $mathbb Q(X) / (P)$.
$endgroup$
– Régis
Dec 27 '18 at 15:54
$begingroup$
Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.)
$endgroup$
– quid♦
Dec 27 '18 at 16:11
$begingroup$
Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though.
$endgroup$
– Régis
Dec 27 '18 at 16:14
$begingroup$
I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though.
$endgroup$
– quid♦
Dec 27 '18 at 20:15
|
show 2 more comments
2
$begingroup$
What is a rupture field?
$endgroup$
– lhf
Dec 27 '18 at 15:52
$begingroup$
It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $mathbb Q(X) / (P)$.
$endgroup$
– Régis
Dec 27 '18 at 15:54
$begingroup$
Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.)
$endgroup$
– quid♦
Dec 27 '18 at 16:11
$begingroup$
Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though.
$endgroup$
– Régis
Dec 27 '18 at 16:14
$begingroup$
I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though.
$endgroup$
– quid♦
Dec 27 '18 at 20:15
2
2
$begingroup$
What is a rupture field?
$endgroup$
– lhf
Dec 27 '18 at 15:52
$begingroup$
What is a rupture field?
$endgroup$
– lhf
Dec 27 '18 at 15:52
$begingroup$
It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $mathbb Q(X) / (P)$.
$endgroup$
– Régis
Dec 27 '18 at 15:54
$begingroup$
It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $mathbb Q(X) / (P)$.
$endgroup$
– Régis
Dec 27 '18 at 15:54
$begingroup$
Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.)
$endgroup$
– quid♦
Dec 27 '18 at 16:11
$begingroup$
Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.)
$endgroup$
– quid♦
Dec 27 '18 at 16:11
$begingroup$
Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though.
$endgroup$
– Régis
Dec 27 '18 at 16:14
$begingroup$
Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though.
$endgroup$
– Régis
Dec 27 '18 at 16:14
$begingroup$
I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though.
$endgroup$
– quid♦
Dec 27 '18 at 20:15
$begingroup$
I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though.
$endgroup$
– quid♦
Dec 27 '18 at 20:15
|
show 2 more comments
1 Answer
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$begingroup$
Here are the easy cases:
For quadratic polynomials, every rupture field is a splitting field.
For cubic polynomials, a rupture field is a splitting field iff the discriminant is a square.
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
$endgroup$
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
add a comment |
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$begingroup$
Here are the easy cases:
For quadratic polynomials, every rupture field is a splitting field.
For cubic polynomials, a rupture field is a splitting field iff the discriminant is a square.
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
$endgroup$
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
add a comment |
$begingroup$
Here are the easy cases:
For quadratic polynomials, every rupture field is a splitting field.
For cubic polynomials, a rupture field is a splitting field iff the discriminant is a square.
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
$endgroup$
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
add a comment |
$begingroup$
Here are the easy cases:
For quadratic polynomials, every rupture field is a splitting field.
For cubic polynomials, a rupture field is a splitting field iff the discriminant is a square.
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
$endgroup$
Here are the easy cases:
For quadratic polynomials, every rupture field is a splitting field.
For cubic polynomials, a rupture field is a splitting field iff the discriminant is a square.
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
answered Dec 27 '18 at 16:40
lhflhf
165k10171396
165k10171396
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
add a comment |
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
$begingroup$
See also math.stackexchange.com/questions/1767252/…
$endgroup$
– lhf
Dec 27 '18 at 16:42
add a comment |
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2
$begingroup$
What is a rupture field?
$endgroup$
– lhf
Dec 27 '18 at 15:52
$begingroup$
It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $mathbb Q(X) / (P)$.
$endgroup$
– Régis
Dec 27 '18 at 15:54
$begingroup$
Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.)
$endgroup$
– quid♦
Dec 27 '18 at 16:11
$begingroup$
Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though.
$endgroup$
– Régis
Dec 27 '18 at 16:14
$begingroup$
I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though.
$endgroup$
– quid♦
Dec 27 '18 at 20:15