Can a degree $p^n$ field extension always be factored in a sequence of prime extensions?
$begingroup$
Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).
Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?
Using Galois theory, this problem translates into the following:
Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?
group-theory field-theory galois-theory
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
$endgroup$
add a comment |
$begingroup$
Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).
Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?
Using Galois theory, this problem translates into the following:
Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?
group-theory field-theory galois-theory
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
$endgroup$
$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05
1
$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24
1
$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34
$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20
add a comment |
$begingroup$
Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).
Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?
Using Galois theory, this problem translates into the following:
Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?
group-theory field-theory galois-theory
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
$endgroup$
Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).
Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?
Using Galois theory, this problem translates into the following:
Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?
group-theory field-theory galois-theory
group-theory field-theory galois-theory
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
edited Dec 7 '18 at 23:06
Asvin
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
asked Dec 7 '18 at 23:03
AsvinAsvin
3,38321331
3,38321331
$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05
1
$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24
1
$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34
$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20
add a comment |
$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05
1
$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24
1
$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34
$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20
$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05
$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05
1
1
$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24
$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24
1
1
$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34
$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34
$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20
$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20
add a comment |
1 Answer
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$begingroup$
No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
add a comment |
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1 Answer
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$begingroup$
No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
add a comment |
$begingroup$
No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
add a comment |
$begingroup$
No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
$endgroup$
No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
edited Dec 8 '18 at 6:43
Jyrki Lahtonen
109k13169372
109k13169372
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
answered Dec 7 '18 at 23:09
Eric WofseyEric Wofsey
184k13212338
184k13212338
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
add a comment |
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45
add a comment |
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$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05
1
$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24
1
$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34
$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20