Results 1 to 3 of 3

Thread: Extension

  1. #1
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301

    Extension

    Theorem. Let $\displaystyle f(x) $ be an irreducible polynomial in $\displaystyle F[x] $. Then all the zeros of $\displaystyle f(x) $ in $\displaystyle \overline{F} $ have the same multiplicity.

    Proof. Suppose you have the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \to F(\beta) $. You extend this to an isomorphism $\displaystyle \tau: \overline{F} \to \overline{F} $. Then you form the following isomorphism $\displaystyle \tau_x: \overline{F}[x] \to \overline{F}[x] $ where $\displaystyle \tau_x $ leaves $\displaystyle f(x) $ fixed. We are assuming that $\displaystyle \alpha $ and $\displaystyle \beta $ are zeros of $\displaystyle f(x) $ in $\displaystyle \overline{F} $. So then $\displaystyle \tau_{x}((x-\alpha)^v) = (x-\beta)^v $ where $\displaystyle v $ is the multiplicity of $\displaystyle \alpha $ and $\displaystyle \beta $. Why does this imply that the multplicity of $\displaystyle \alpha \leq $ multiplicity of $\displaystyle \beta $?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Sampras View Post
    Theorem. Let $\displaystyle f(x) $ be an irreducible polynomial in $\displaystyle F[x] $. Then all the zeros of $\displaystyle f(x) $ in $\displaystyle \overline{F} $ have the same multiplicity.

    Proof. Suppose you have the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \to F(\beta) $. You extend this to an isomorphism $\displaystyle \tau: \overline{F} \to \overline{F} $. Then you form the following isomorphism $\displaystyle \tau_x: \overline{F}[x] \to \overline{F}[x] $ where $\displaystyle \tau_x $ leaves $\displaystyle f(x) $ fixed. We are assuming that $\displaystyle \alpha $ and $\displaystyle \beta $ are zeros of $\displaystyle f(x) $ in $\displaystyle \overline{F} $. So then $\displaystyle \tau_{x}((x-\alpha)^v) = (x-\beta)^v $ where $\displaystyle v $ is the multiplicity of $\displaystyle \alpha $ and $\displaystyle \beta $. Why does this imply that the multplicity of $\displaystyle \alpha \leq $ multiplicity of $\displaystyle \beta $?
    If you apply $\displaystyle \tau_x$ to a decomposition $\displaystyle f(x)=C\prod_\rho (X-\rho)^{m_\rho}$, you get another decomposition $\displaystyle f(x)=\tau_x(f(x))=C\prod_\rho (X-\tau(\rho))^{m_\rho}$, hence by unicity of the decomposition into irreducible factors in $\displaystyle \overline{F}[x]$, $\displaystyle m_{\tau(\rho)}=m_\rho$ for every root $\displaystyle \rho$, and in particular the multiplicities of $\displaystyle \alpha$ and $\displaystyle \beta$ coincide. Does that look correct?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by Laurent View Post
    If you apply $\displaystyle \tau_x$ to a decomposition $\displaystyle f(x)=C\prod_\rho (X-\rho)^{m_\rho}$, you get another decomposition $\displaystyle f(x)=\tau_x(f(x))=C\prod_\rho (X-\tau(\rho))^{m_\rho}$, hence by unicity of the decomposition into irreducible factors in $\displaystyle \overline{F}[x]$, $\displaystyle m_{\tau(\rho)}=m_\rho$ for every root $\displaystyle \rho$, and in particular the multiplicities of $\displaystyle \alpha$ and $\displaystyle \beta$ coincide. Does that look correct?
    It does look correct, but how would you do it using trichotomy? E.g $\displaystyle \tau_{x}(x-\alpha)^{v} = (x-\beta)^v \implies \text{multiplicity of} \ \alpha \leq \text{multiplicity of} \ \beta $ and $\displaystyle \tau_{x}(x-\beta)^{v} = (x-\alpha)^v \implies \text{multiplicity of} \ \beta \leq \text{multiplicity of} \ \alpha $. Would you use $\displaystyle \psi_{\alpha, \beta} $ and $\displaystyle \psi_{\beta, \alpha} $?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. extension
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 28th 2011, 07:07 AM
  2. Prove that an extension is a continuous extension
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Aug 12th 2011, 05:50 AM
  3. HNN extension
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Feb 23rd 2010, 06:46 AM
  4. Extension
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 2nd 2009, 06:52 AM
  5. Odd extension, even extension?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 25th 2009, 11:48 PM

Search Tags


/mathhelpforum @mathhelpforum