Galois extension - minimum polynomial











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Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.



We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.




  1. Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.


  2. Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.



$$$$



Could you give me a hint how we could show these two points? I don't really have an idea.










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    Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.



    We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.




    1. Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.


    2. Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.



    $$$$



    Could you give me a hint how we could show these two points? I don't really have an idea.










    share|cite|improve this question


























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      Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.



      We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.




      1. Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.


      2. Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.



      $$$$



      Could you give me a hint how we could show these two points? I don't really have an idea.










      share|cite|improve this question















      Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.



      We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.




      1. Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.


      2. Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.



      $$$$



      Could you give me a hint how we could show these two points? I don't really have an idea.







      abstract-algebra field-theory galois-theory






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      edited Nov 29 at 14:46









      Bilo

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      asked Nov 19 at 23:09









      Mary Star

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          For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.



          For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$






          share|cite|improve this answer

















          • 1




            Do I have to write a full proof?
            – Bilo
            Nov 23 at 20:39










          • For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
            – Mary Star
            Nov 24 at 8:38






          • 1




            Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
            – Bilo
            Nov 24 at 14:57












          • Ah ok!! Could you explain to me also further the second point?
            – Mary Star
            Nov 24 at 16:41






          • 1




            Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
            – Bilo
            Nov 24 at 17:19













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          1 Answer
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          1 Answer
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          up vote
          2
          down vote



          accepted










          For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.



          For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$






          share|cite|improve this answer

















          • 1




            Do I have to write a full proof?
            – Bilo
            Nov 23 at 20:39










          • For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
            – Mary Star
            Nov 24 at 8:38






          • 1




            Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
            – Bilo
            Nov 24 at 14:57












          • Ah ok!! Could you explain to me also further the second point?
            – Mary Star
            Nov 24 at 16:41






          • 1




            Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
            – Bilo
            Nov 24 at 17:19

















          up vote
          2
          down vote



          accepted










          For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.



          For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$






          share|cite|improve this answer

















          • 1




            Do I have to write a full proof?
            – Bilo
            Nov 23 at 20:39










          • For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
            – Mary Star
            Nov 24 at 8:38






          • 1




            Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
            – Bilo
            Nov 24 at 14:57












          • Ah ok!! Could you explain to me also further the second point?
            – Mary Star
            Nov 24 at 16:41






          • 1




            Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
            – Bilo
            Nov 24 at 17:19















          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.



          For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$






          share|cite|improve this answer












          For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.



          For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 12:52









          Bilo

          1089




          1089








          • 1




            Do I have to write a full proof?
            – Bilo
            Nov 23 at 20:39










          • For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
            – Mary Star
            Nov 24 at 8:38






          • 1




            Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
            – Bilo
            Nov 24 at 14:57












          • Ah ok!! Could you explain to me also further the second point?
            – Mary Star
            Nov 24 at 16:41






          • 1




            Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
            – Bilo
            Nov 24 at 17:19
















          • 1




            Do I have to write a full proof?
            – Bilo
            Nov 23 at 20:39










          • For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
            – Mary Star
            Nov 24 at 8:38






          • 1




            Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
            – Bilo
            Nov 24 at 14:57












          • Ah ok!! Could you explain to me also further the second point?
            – Mary Star
            Nov 24 at 16:41






          • 1




            Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
            – Bilo
            Nov 24 at 17:19










          1




          1




          Do I have to write a full proof?
          – Bilo
          Nov 23 at 20:39




          Do I have to write a full proof?
          – Bilo
          Nov 23 at 20:39












          For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
          – Mary Star
          Nov 24 at 8:38




          For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
          – Mary Star
          Nov 24 at 8:38




          1




          1




          Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
          – Bilo
          Nov 24 at 14:57






          Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
          – Bilo
          Nov 24 at 14:57














          Ah ok!! Could you explain to me also further the second point?
          – Mary Star
          Nov 24 at 16:41




          Ah ok!! Could you explain to me also further the second point?
          – Mary Star
          Nov 24 at 16:41




          1




          1




          Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
          – Bilo
          Nov 24 at 17:19






          Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
          – Bilo
          Nov 24 at 17:19




















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