Kummer Theory Proof
$begingroup$
I have a question about a step in proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
We start with a $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. Then he concludes that $x^n- alpha$ is irreducible over $K$.
Why? The fact that $alpha^{1/d} not in K$ states just that $x^n- alpha$ doesn't splits in linear factors in $K$ but naively it could be split in non linear factors, right? Or where is the error in my reasonings?
field-theory extension-field kummer-theory
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
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add a comment |
$begingroup$
I have a question about a step in proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
We start with a $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. Then he concludes that $x^n- alpha$ is irreducible over $K$.
Why? The fact that $alpha^{1/d} not in K$ states just that $x^n- alpha$ doesn't splits in linear factors in $K$ but naively it could be split in non linear factors, right? Or where is the error in my reasonings?
field-theory extension-field kummer-theory
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
$endgroup$
add a comment |
$begingroup$
I have a question about a step in proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
We start with a $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. Then he concludes that $x^n- alpha$ is irreducible over $K$.
Why? The fact that $alpha^{1/d} not in K$ states just that $x^n- alpha$ doesn't splits in linear factors in $K$ but naively it could be split in non linear factors, right? Or where is the error in my reasonings?
field-theory extension-field kummer-theory
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
$endgroup$
I have a question about a step in proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
We start with a $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. Then he concludes that $x^n- alpha$ is irreducible over $K$.
Why? The fact that $alpha^{1/d} not in K$ states just that $x^n- alpha$ doesn't splits in linear factors in $K$ but naively it could be split in non linear factors, right? Or where is the error in my reasonings?
field-theory extension-field kummer-theory
field-theory extension-field kummer-theory
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:09
KarlPeterKarlPeter
6131315
6131315
add a comment |
add a comment |
1 Answer
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Indeed not. For instance take $K=Bbb Q$, $n=4$ and $alpha=-4$. Then $alpha
$ is not a square in $K$, but
$$x^n-alpha=x^4+4=(x^2+2x+2)(x^2-2x+2)$$
is reducible over $K$.
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
Indeed not. For instance take $K=Bbb Q$, $n=4$ and $alpha=-4$. Then $alpha
$ is not a square in $K$, but
$$x^n-alpha=x^4+4=(x^2+2x+2)(x^2-2x+2)$$
is reducible over $K$.
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
add a comment |
$begingroup$
Indeed not. For instance take $K=Bbb Q$, $n=4$ and $alpha=-4$. Then $alpha
$ is not a square in $K$, but
$$x^n-alpha=x^4+4=(x^2+2x+2)(x^2-2x+2)$$
is reducible over $K$.
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
add a comment |
$begingroup$
Indeed not. For instance take $K=Bbb Q$, $n=4$ and $alpha=-4$. Then $alpha
$ is not a square in $K$, but
$$x^n-alpha=x^4+4=(x^2+2x+2)(x^2-2x+2)$$
is reducible over $K$.
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
Indeed not. For instance take $K=Bbb Q$, $n=4$ and $alpha=-4$. Then $alpha
$ is not a square in $K$, but
$$x^n-alpha=x^4+4=(x^2+2x+2)(x^2-2x+2)$$
is reducible over $K$.
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 9 '18 at 13:12
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
add a comment |
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
$begingroup$
Hi. Thank you for this answer which obviosly solves the problem for given conditions as above. But I think that I forgot to mention a substantial assumption: In the script he assumes that $K$ contains a primitive n-th root $zeta_n$. Here I started a new thread since in this case the problem has fundamentally another color: math.stackexchange.com/questions/3032382/…. Do you see how to cope with this one? Here your example obviously doesn't work.
$endgroup$
– KarlPeter
Dec 9 '18 at 13:28
add a comment |
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