Condition for the complex conjugation to be in Gal(P/k)












0












$begingroup$


Let $Pin k[X]$ be a separable polynomial over a field k.



if P has a unique real root in $overline{k}$ then complex conjugation $epsilon$ verifies $epsilon in Gal(P/k)$.



To which extent is this a true statement regarding the degree $n$ of $P$?



I think $n$ needs to be odd for this to be true in order for $epsilon$ to stabilize any splitting field of P over k?



Thank you for your help.






abstract-algebra







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$endgroup$












  • $begingroup$
    You mean $k subset mathbb{R}$ so for any field $mathbb{C} supset L supset k$ the complex conjugaison $varepsilon$ is a field isomorphism $L to varepsilon(L)$ leaving $k$ fixed. Now if $L$ is the splitting field of $P in k[x]$ then so is $ varepsilon(L)$ and hence $L = varepsilon(L)$ and $varepsilon in Gal(L/k)$.
    $endgroup$
    – reuns
    Dec 27 '18 at 0:08












  • $begingroup$
    I mean the following: let's take $k=mathbb{Q}$ and $n=3$. Suppose P has a unique real root and 2 complex roots. we know that for a splitting field of P: $epsilon(L) = L$ because the two complex roots are conjugates. can we generalize this for any n?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 0:14








  • 1




    $begingroup$
    Yes. All you need is that $P(a) =0 Leftrightarrow 0 = varepsilon(P(a)) = P(varepsilon(a))$. So $P$ splits completely in $L$ iff it splits completely in $varepsilon(L)$, and $L$ is the smallest such field implies so is $varepsilon(L)$ and $L = varepsilon(L)$
    $endgroup$
    – reuns
    Dec 27 '18 at 0:20


















0












$begingroup$


Let $Pin k[X]$ be a separable polynomial over a field k.



if P has a unique real root in $overline{k}$ then complex conjugation $epsilon$ verifies $epsilon in Gal(P/k)$.



To which extent is this a true statement regarding the degree $n$ of $P$?



I think $n$ needs to be odd for this to be true in order for $epsilon$ to stabilize any splitting field of P over k?



Thank you for your help.






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    You mean $k subset mathbb{R}$ so for any field $mathbb{C} supset L supset k$ the complex conjugaison $varepsilon$ is a field isomorphism $L to varepsilon(L)$ leaving $k$ fixed. Now if $L$ is the splitting field of $P in k[x]$ then so is $ varepsilon(L)$ and hence $L = varepsilon(L)$ and $varepsilon in Gal(L/k)$.
    $endgroup$
    – reuns
    Dec 27 '18 at 0:08












  • $begingroup$
    I mean the following: let's take $k=mathbb{Q}$ and $n=3$. Suppose P has a unique real root and 2 complex roots. we know that for a splitting field of P: $epsilon(L) = L$ because the two complex roots are conjugates. can we generalize this for any n?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 0:14








  • 1




    $begingroup$
    Yes. All you need is that $P(a) =0 Leftrightarrow 0 = varepsilon(P(a)) = P(varepsilon(a))$. So $P$ splits completely in $L$ iff it splits completely in $varepsilon(L)$, and $L$ is the smallest such field implies so is $varepsilon(L)$ and $L = varepsilon(L)$
    $endgroup$
    – reuns
    Dec 27 '18 at 0:20
















0












0








0





$begingroup$


Let $Pin k[X]$ be a separable polynomial over a field k.



if P has a unique real root in $overline{k}$ then complex conjugation $epsilon$ verifies $epsilon in Gal(P/k)$.



To which extent is this a true statement regarding the degree $n$ of $P$?



I think $n$ needs to be odd for this to be true in order for $epsilon$ to stabilize any splitting field of P over k?



Thank you for your help.






abstract-algebra







share|cite|improve this question











$endgroup$




Let $Pin k[X]$ be a separable polynomial over a field k.



if P has a unique real root in $overline{k}$ then complex conjugation $epsilon$ verifies $epsilon in Gal(P/k)$.



To which extent is this a true statement regarding the degree $n$ of $P$?



I think $n$ needs to be odd for this to be true in order for $epsilon$ to stabilize any splitting field of P over k?



Thank you for your help.






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 0:14









Bernard

121k740116




121k740116










asked Dec 26 '18 at 23:53









PerelManPerelMan

544311




544311












  • $begingroup$
    You mean $k subset mathbb{R}$ so for any field $mathbb{C} supset L supset k$ the complex conjugaison $varepsilon$ is a field isomorphism $L to varepsilon(L)$ leaving $k$ fixed. Now if $L$ is the splitting field of $P in k[x]$ then so is $ varepsilon(L)$ and hence $L = varepsilon(L)$ and $varepsilon in Gal(L/k)$.
    $endgroup$
    – reuns
    Dec 27 '18 at 0:08












  • $begingroup$
    I mean the following: let's take $k=mathbb{Q}$ and $n=3$. Suppose P has a unique real root and 2 complex roots. we know that for a splitting field of P: $epsilon(L) = L$ because the two complex roots are conjugates. can we generalize this for any n?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 0:14








  • 1




    $begingroup$
    Yes. All you need is that $P(a) =0 Leftrightarrow 0 = varepsilon(P(a)) = P(varepsilon(a))$. So $P$ splits completely in $L$ iff it splits completely in $varepsilon(L)$, and $L$ is the smallest such field implies so is $varepsilon(L)$ and $L = varepsilon(L)$
    $endgroup$
    – reuns
    Dec 27 '18 at 0:20




















  • $begingroup$
    You mean $k subset mathbb{R}$ so for any field $mathbb{C} supset L supset k$ the complex conjugaison $varepsilon$ is a field isomorphism $L to varepsilon(L)$ leaving $k$ fixed. Now if $L$ is the splitting field of $P in k[x]$ then so is $ varepsilon(L)$ and hence $L = varepsilon(L)$ and $varepsilon in Gal(L/k)$.
    $endgroup$
    – reuns
    Dec 27 '18 at 0:08












  • $begingroup$
    I mean the following: let's take $k=mathbb{Q}$ and $n=3$. Suppose P has a unique real root and 2 complex roots. we know that for a splitting field of P: $epsilon(L) = L$ because the two complex roots are conjugates. can we generalize this for any n?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 0:14








  • 1




    $begingroup$
    Yes. All you need is that $P(a) =0 Leftrightarrow 0 = varepsilon(P(a)) = P(varepsilon(a))$. So $P$ splits completely in $L$ iff it splits completely in $varepsilon(L)$, and $L$ is the smallest such field implies so is $varepsilon(L)$ and $L = varepsilon(L)$
    $endgroup$
    – reuns
    Dec 27 '18 at 0:20


















$begingroup$
You mean $k subset mathbb{R}$ so for any field $mathbb{C} supset L supset k$ the complex conjugaison $varepsilon$ is a field isomorphism $L to varepsilon(L)$ leaving $k$ fixed. Now if $L$ is the splitting field of $P in k[x]$ then so is $ varepsilon(L)$ and hence $L = varepsilon(L)$ and $varepsilon in Gal(L/k)$.
$endgroup$
– reuns
Dec 27 '18 at 0:08






$begingroup$
You mean $k subset mathbb{R}$ so for any field $mathbb{C} supset L supset k$ the complex conjugaison $varepsilon$ is a field isomorphism $L to varepsilon(L)$ leaving $k$ fixed. Now if $L$ is the splitting field of $P in k[x]$ then so is $ varepsilon(L)$ and hence $L = varepsilon(L)$ and $varepsilon in Gal(L/k)$.
$endgroup$
– reuns
Dec 27 '18 at 0:08














$begingroup$
I mean the following: let's take $k=mathbb{Q}$ and $n=3$. Suppose P has a unique real root and 2 complex roots. we know that for a splitting field of P: $epsilon(L) = L$ because the two complex roots are conjugates. can we generalize this for any n?
$endgroup$
– PerelMan
Dec 27 '18 at 0:14






$begingroup$
I mean the following: let's take $k=mathbb{Q}$ and $n=3$. Suppose P has a unique real root and 2 complex roots. we know that for a splitting field of P: $epsilon(L) = L$ because the two complex roots are conjugates. can we generalize this for any n?
$endgroup$
– PerelMan
Dec 27 '18 at 0:14






1




1




$begingroup$
Yes. All you need is that $P(a) =0 Leftrightarrow 0 = varepsilon(P(a)) = P(varepsilon(a))$. So $P$ splits completely in $L$ iff it splits completely in $varepsilon(L)$, and $L$ is the smallest such field implies so is $varepsilon(L)$ and $L = varepsilon(L)$
$endgroup$
– reuns
Dec 27 '18 at 0:20






$begingroup$
Yes. All you need is that $P(a) =0 Leftrightarrow 0 = varepsilon(P(a)) = P(varepsilon(a))$. So $P$ splits completely in $L$ iff it splits completely in $varepsilon(L)$, and $L$ is the smallest such field implies so is $varepsilon(L)$ and $L = varepsilon(L)$
$endgroup$
– reuns
Dec 27 '18 at 0:20












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If $k$ is a subfield of $mathbb{R}$, then complex conjugation gives an element of the Galois group of any polynomial $Pin k[X]$ whatsoever. Indeed, if $Lsubseteqmathbb{C}$ is the splitting field of $P$ over $k$, then complex conjugation maps $L$ to itself (since it maps $k$ to itself and thus permutes the roots of $P$) and so is an automorphism of $L$.






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    $begingroup$

    If $k$ is a subfield of $mathbb{R}$, then complex conjugation gives an element of the Galois group of any polynomial $Pin k[X]$ whatsoever. Indeed, if $Lsubseteqmathbb{C}$ is the splitting field of $P$ over $k$, then complex conjugation maps $L$ to itself (since it maps $k$ to itself and thus permutes the roots of $P$) and so is an automorphism of $L$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $k$ is a subfield of $mathbb{R}$, then complex conjugation gives an element of the Galois group of any polynomial $Pin k[X]$ whatsoever. Indeed, if $Lsubseteqmathbb{C}$ is the splitting field of $P$ over $k$, then complex conjugation maps $L$ to itself (since it maps $k$ to itself and thus permutes the roots of $P$) and so is an automorphism of $L$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $k$ is a subfield of $mathbb{R}$, then complex conjugation gives an element of the Galois group of any polynomial $Pin k[X]$ whatsoever. Indeed, if $Lsubseteqmathbb{C}$ is the splitting field of $P$ over $k$, then complex conjugation maps $L$ to itself (since it maps $k$ to itself and thus permutes the roots of $P$) and so is an automorphism of $L$.






        share|cite|improve this answer









        $endgroup$



        If $k$ is a subfield of $mathbb{R}$, then complex conjugation gives an element of the Galois group of any polynomial $Pin k[X]$ whatsoever. Indeed, if $Lsubseteqmathbb{C}$ is the splitting field of $P$ over $k$, then complex conjugation maps $L$ to itself (since it maps $k$ to itself and thus permutes the roots of $P$) and so is an automorphism of $L$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 0:25









        Eric WofseyEric Wofsey

        186k14215342




        186k14215342






























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