Norm of an element mod I

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Let $E|F$ be a finite field extension, $Rsubseteq R'subseteq E$ subrings
with $R'cap F=R$. Suppose we have ideals $Isubseteq R$, $I'subseteq R'$
with $I'=IR'$ and we know that $R'$ is free over $R$ of rank $[E:F]$.
Let $fin R'$ such that $f-1in I'$ or equivalent $fequiv1$ mod $I'$.



The norm $N_F(f)$ of $f$ is the determinant of the transformation matrix "multiplication with f" on a $R$-base of $R'$.



Why is the transformations matrix coefficient-wise equivalent to the unit matrix modulo $I$? Or equivalent



Why is $N_F(f)equiv1$ mod $I$ ?




algebraic-number-theory




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    Let $E|F$ be a finite field extension, $Rsubseteq R'subseteq E$ subrings
    with $R'cap F=R$. Suppose we have ideals $Isubseteq R$, $I'subseteq R'$
    with $I'=IR'$ and we know that $R'$ is free over $R$ of rank $[E:F]$.
    Let $fin R'$ such that $f-1in I'$ or equivalent $fequiv1$ mod $I'$.



    The norm $N_F(f)$ of $f$ is the determinant of the transformation matrix "multiplication with f" on a $R$-base of $R'$.



    Why is the transformations matrix coefficient-wise equivalent to the unit matrix modulo $I$? Or equivalent



    Why is $N_F(f)equiv1$ mod $I$ ?




    algebraic-number-theory




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $E|F$ be a finite field extension, $Rsubseteq R'subseteq E$ subrings
      with $R'cap F=R$. Suppose we have ideals $Isubseteq R$, $I'subseteq R'$
      with $I'=IR'$ and we know that $R'$ is free over $R$ of rank $[E:F]$.
      Let $fin R'$ such that $f-1in I'$ or equivalent $fequiv1$ mod $I'$.



      The norm $N_F(f)$ of $f$ is the determinant of the transformation matrix "multiplication with f" on a $R$-base of $R'$.



      Why is the transformations matrix coefficient-wise equivalent to the unit matrix modulo $I$? Or equivalent



      Why is $N_F(f)equiv1$ mod $I$ ?




      algebraic-number-theory




      share|cite|improve this question











      Let $E|F$ be a finite field extension, $Rsubseteq R'subseteq E$ subrings
      with $R'cap F=R$. Suppose we have ideals $Isubseteq R$, $I'subseteq R'$
      with $I'=IR'$ and we know that $R'$ is free over $R$ of rank $[E:F]$.
      Let $fin R'$ such that $f-1in I'$ or equivalent $fequiv1$ mod $I'$.



      The norm $N_F(f)$ of $f$ is the determinant of the transformation matrix "multiplication with f" on a $R$-base of $R'$.



      Why is the transformations matrix coefficient-wise equivalent to the unit matrix modulo $I$? Or equivalent



      Why is $N_F(f)equiv1$ mod $I$ ?





      algebraic-number-theory





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 20 at 11:47









      Boki

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