Kummer Theory Square Extension
$begingroup$
I have a question about an argument used in the proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
Let $K$ be a field which contains a n-th root $zeta_n$.
We consider $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. The autor deduces 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 kummer-theory
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
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add a comment |
$begingroup$
I have a question about an argument used in the proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
Let $K$ be a field which contains a n-th root $zeta_n$.
We consider $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. The autor deduces 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 kummer-theory
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
$endgroup$
add a comment |
$begingroup$
I have a question about an argument used in the proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
Let $K$ be a field which contains a n-th root $zeta_n$.
We consider $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. The autor deduces 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 kummer-theory
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
$endgroup$
I have a question about an argument used in the proof of Kummer theorem in Kedlaya's notes (see page 8)
www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
Let $K$ be a field which contains a n-th root $zeta_n$.
We consider $alpha in K* = K backslash {0}$ with the property that $alpha^{1/d} not in K$ for every proper divisor $d$ of $n$. The autor deduces 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 kummer-theory
field-theory kummer-theory
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
asked Dec 9 '18 at 13:28
KarlPeterKarlPeter
6131315
6131315
add a comment |
add a comment |
1 Answer
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Let $zeta$ be a primitive $n$-th root of unity. Over the Kummer extension
we have a factorisation
$$X^n-alpha=prod_{k=0}^{n-1}(X-zeta^kbeta).$$
If $f(X)$ is a non-trivial monic factor over $K$,
then $f(X)=X^r+cdotspmxibeta^r$
where $xi$ is some $n$-the root of unity. Therefore $beta^rin K$.
As $beta^n=alphain K$, then $beta^min K$ where $m=gcd(n,r)$
and $beta^m=alpha^{1/d}$ for $d=n/m$.
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $zeta$ be a primitive $n$-th root of unity. Over the Kummer extension
we have a factorisation
$$X^n-alpha=prod_{k=0}^{n-1}(X-zeta^kbeta).$$
If $f(X)$ is a non-trivial monic factor over $K$,
then $f(X)=X^r+cdotspmxibeta^r$
where $xi$ is some $n$-the root of unity. Therefore $beta^rin K$.
As $beta^n=alphain K$, then $beta^min K$ where $m=gcd(n,r)$
and $beta^m=alpha^{1/d}$ for $d=n/m$.
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
add a comment |
$begingroup$
Let $zeta$ be a primitive $n$-th root of unity. Over the Kummer extension
we have a factorisation
$$X^n-alpha=prod_{k=0}^{n-1}(X-zeta^kbeta).$$
If $f(X)$ is a non-trivial monic factor over $K$,
then $f(X)=X^r+cdotspmxibeta^r$
where $xi$ is some $n$-the root of unity. Therefore $beta^rin K$.
As $beta^n=alphain K$, then $beta^min K$ where $m=gcd(n,r)$
and $beta^m=alpha^{1/d}$ for $d=n/m$.
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
add a comment |
$begingroup$
Let $zeta$ be a primitive $n$-th root of unity. Over the Kummer extension
we have a factorisation
$$X^n-alpha=prod_{k=0}^{n-1}(X-zeta^kbeta).$$
If $f(X)$ is a non-trivial monic factor over $K$,
then $f(X)=X^r+cdotspmxibeta^r$
where $xi$ is some $n$-the root of unity. Therefore $beta^rin K$.
As $beta^n=alphain K$, then $beta^min K$ where $m=gcd(n,r)$
and $beta^m=alpha^{1/d}$ for $d=n/m$.
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
Let $zeta$ be a primitive $n$-th root of unity. Over the Kummer extension
we have a factorisation
$$X^n-alpha=prod_{k=0}^{n-1}(X-zeta^kbeta).$$
If $f(X)$ is a non-trivial monic factor over $K$,
then $f(X)=X^r+cdotspmxibeta^r$
where $xi$ is some $n$-the root of unity. Therefore $beta^rin K$.
As $beta^n=alphain K$, then $beta^min K$ where $m=gcd(n,r)$
and $beta^m=alpha^{1/d}$ for $d=n/m$.
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 9 '18 at 13:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
add a comment |
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
...and $beta^m in K$ in a consequence of Bezout?
$endgroup$
– KarlPeter
Dec 9 '18 at 14:09
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
$begingroup$
@KarlPeter Indeed
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 16:44
add a comment |
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