If $E/F$ is a finite extension and $E = F(alpha)$ for some $alphain E,$ is it true that $E/F$ separable?
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Let $E$ and $K$ be fields such that $Ksubseteq E.$
It is well-known that if the extension field $E/F$ is finite and separable, then $E/F$ is simple, that is, there exists $alpha in E$ such that
$$E = F(alpha).$$
I am interested in its converse, that is,
Question: If $E/F$ is a finite extension and $E = F(alpha)$ for some $alphain E,$ is it true that $E/F$ separable?
abstract-algebra field-theory extension-field
asked 20 hours ago
Idonknow
2,155746110
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up vote
0
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Let $E$ and $K$ be fields such that $Ksubseteq E.$
It is well-known that if the extension field $E/F$ is finite and separable, then $E/F$ is simple, that is, there exists $alpha in E$ such that
$$E = F(alpha).$$
I am interested in its converse, that is,
Question: If $E/F$ is a finite extension and $E = F(alpha)$ for some $alphain E,$ is it true that $E/F$ separable?
abstract-algebra field-theory extension-field
asked 20 hours ago
Idonknow
2,155746110
5
Certainly not. Take the following standard example: let $F = mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E = F[x]/(x^p - t) = F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
– André 3000
20 hours ago
1
@André3000, that comment could be an answer
– lhf
18 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $E$ and $K$ be fields such that $Ksubseteq E.$
It is well-known that if the extension field $E/F$ is finite and separable, then $E/F$ is simple, that is, there exists $alpha in E$ such that
$$E = F(alpha).$$
I am interested in its converse, that is,
Question: If $E/F$ is a finite extension and $E = F(alpha)$ for some $alphain E,$ is it true that $E/F$ separable?
abstract-algebra field-theory extension-field
asked 20 hours ago
Idonknow
2,155746110
Let $E$ and $K$ be fields such that $Ksubseteq E.$
It is well-known that if the extension field $E/F$ is finite and separable, then $E/F$ is simple, that is, there exists $alpha in E$ such that
$$E = F(alpha).$$
I am interested in its converse, that is,
Question: If $E/F$ is a finite extension and $E = F(alpha)$ for some $alphain E,$ is it true that $E/F$ separable?
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked 20 hours ago
Idonknow
2,155746110
asked 20 hours ago
Idonknow
2,155746110
asked 20 hours ago
Idonknow
2,155746110
asked 20 hours ago
Idonknow
2,155746110
asked 20 hours ago
Idonknow
2,155746110
2,155746110
5
Certainly not. Take the following standard example: let $F = mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E = F[x]/(x^p - t) = F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
– André 3000
20 hours ago
1
@André3000, that comment could be an answer
– lhf
18 hours ago
add a comment |
5
Certainly not. Take the following standard example: let $F = mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E = F[x]/(x^p - t) = F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
– André 3000
20 hours ago
1
@André3000, that comment could be an answer
– lhf
18 hours ago
5
5
Certainly not. Take the following standard example: let $F = mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E = F[x]/(x^p - t) = F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
– André 3000
20 hours ago
Certainly not. Take the following standard example: let $F = mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E = F[x]/(x^p - t) = F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
– André 3000
20 hours ago
1
1
@André3000, that comment could be an answer
– lhf
18 hours ago
@André3000, that comment could be an answer
– lhf
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Certainly not. Take the following standard example: let $F= mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E=F[x]/(x^p−t)=F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
In general, a simple finite extension $E = F(alpha)$ is separable iff the minimal polynomial $m$ of $alpha$ is squarefree. This holds iff $gcd(m, m') = 1$, where $m'$ its formal derivative. You can see how this fails in the example above, since $m' = p x^{p-1} = 0$.
answered 13 hours ago
André 3000
12k22041
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Certainly not. Take the following standard example: let $F= mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E=F[x]/(x^p−t)=F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
In general, a simple finite extension $E = F(alpha)$ is separable iff the minimal polynomial $m$ of $alpha$ is squarefree. This holds iff $gcd(m, m') = 1$, where $m'$ its formal derivative. You can see how this fails in the example above, since $m' = p x^{p-1} = 0$.
answered 13 hours ago
André 3000
12k22041
add a comment |
up vote
1
down vote
accepted
Certainly not. Take the following standard example: let $F= mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E=F[x]/(x^p−t)=F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
In general, a simple finite extension $E = F(alpha)$ is separable iff the minimal polynomial $m$ of $alpha$ is squarefree. This holds iff $gcd(m, m') = 1$, where $m'$ its formal derivative. You can see how this fails in the example above, since $m' = p x^{p-1} = 0$.
answered 13 hours ago
André 3000
12k22041
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Certainly not. Take the following standard example: let $F= mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E=F[x]/(x^p−t)=F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
In general, a simple finite extension $E = F(alpha)$ is separable iff the minimal polynomial $m$ of $alpha$ is squarefree. This holds iff $gcd(m, m') = 1$, where $m'$ its formal derivative. You can see how this fails in the example above, since $m' = p x^{p-1} = 0$.
answered 13 hours ago
André 3000
12k22041
Certainly not. Take the following standard example: let $F= mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E=F[x]/(x^p−t)=F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
In general, a simple finite extension $E = F(alpha)$ is separable iff the minimal polynomial $m$ of $alpha$ is squarefree. This holds iff $gcd(m, m') = 1$, where $m'$ its formal derivative. You can see how this fails in the example above, since $m' = p x^{p-1} = 0$.
answered 13 hours ago
André 3000
12k22041
answered 13 hours ago
André 3000
12k22041
answered 13 hours ago
André 3000
12k22041
answered 13 hours ago
André 3000
12k22041
12k22041
add a comment |
add a comment |
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5
Certainly not. Take the following standard example: let $F = mathbb{F}_p(t)$ be the rational function field over $mathbb{F}_p$ and let $E = F[x]/(x^p - t) = F[sqrt[p]{t}]$. The extension $E/F$ is simple but is purely inseparable.
– André 3000
20 hours ago
1
@André3000, that comment could be an answer
– lhf
18 hours ago