Results 1 to 3 of 3

Math Help - Extension

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

    Extension

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

    Proof. Suppose you have the conjugation isomorphism  \psi_{\alpha, \beta}: F(\alpha) \to F(\beta) . You extend this to an isomorphism  \tau: \overline{F} \to \overline{F} . Then you form the following isomorphism  \tau_x: \overline{F}[x] \to \overline{F}[x] where  \tau_x leaves  f(x) fixed. We are assuming that  \alpha and  \beta are zeros of  f(x) in  \overline{F} . So then  \tau_{x}((x-\alpha)^v) = (x-\beta)^v where  v is the multiplicity of  \alpha and  \beta . Why does this imply that the multplicity of  \alpha  \leq multiplicity of  \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  f(x) be an irreducible polynomial in  F[x] . Then all the zeros of  f(x) in  \overline{F} have the same multiplicity.

    Proof. Suppose you have the conjugation isomorphism  \psi_{\alpha, \beta}: F(\alpha) \to F(\beta) . You extend this to an isomorphism  \tau: \overline{F} \to \overline{F} . Then you form the following isomorphism  \tau_x: \overline{F}[x] \to \overline{F}[x] where  \tau_x leaves  f(x) fixed. We are assuming that  \alpha and  \beta are zeros of  f(x) in  \overline{F} . So then  \tau_{x}((x-\alpha)^v) = (x-\beta)^v where  v is the multiplicity of  \alpha and  \beta . Why does this imply that the multplicity of  \alpha  \leq multiplicity of  \beta ?
    If you apply \tau_x to a decomposition f(x)=C\prod_\rho (X-\rho)^{m_\rho}, you get another decomposition f(x)=\tau_x(f(x))=C\prod_\rho (X-\tau(\rho))^{m_\rho}, hence by unicity of the decomposition into irreducible factors in \overline{F}[x], m_{\tau(\rho)}=m_\rho for every root \rho, and in particular the multiplicities of \alpha and \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 \tau_x to a decomposition f(x)=C\prod_\rho (X-\rho)^{m_\rho}, you get another decomposition f(x)=\tau_x(f(x))=C\prod_\rho (X-\tau(\rho))^{m_\rho}, hence by unicity of the decomposition into irreducible factors in \overline{F}[x], m_{\tau(\rho)}=m_\rho for every root \rho, and in particular the multiplicities of \alpha and \beta coincide. Does that look correct?
    It does look correct, but how would you do it using trichotomy? E.g  \tau_{x}(x-\alpha)^{v} = (x-\beta)^v \implies \text{multiplicity of} \ \alpha \leq \text{multiplicity of} \ \beta and  \tau_{x}(x-\beta)^{v} = (x-\alpha)^v \implies \text{multiplicity of} \ \beta \leq \text{multiplicity of} \ \alpha . Would you use  \psi_{\alpha, \beta} and  \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: December 28th 2011, 07:07 AM
  2. Prove that an extension is a continuous extension
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: August 12th 2011, 05:50 AM
  3. HNN extension
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 23rd 2010, 06:46 AM
  4. Extension
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 2nd 2009, 06:52 AM
  5. Odd extension, even extension?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 25th 2009, 11:48 PM

Search Tags


/mathhelpforum @mathhelpforum