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Pe: 2011-02-20T16:31:16+02:00 In: In:

Izomorfism

Sa se arate ca inelele (

*** QuickLaTeX cannot compile formula:
\mathbf{Q (radical din 2)

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File ended while scanning use of \select@group.
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,+,*) si (

*** QuickLaTeX cannot compile formula:
\mathbf{Q (radical din 3)

*** Error message:
File ended while scanning use of \select@group.
Emergency stop.

,+,*) nu sunt izomorfe.

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1 raspuns

  1. sindex

    Fie { Q(\sqrt d ) = \left\{ {a + b\sqrt d |a,b \in \left. Q \right\}} \right. \cr (Q, + , \cdot ) - {\rm{inel comutativ}} \cr}
    Presupunem ca

        $${\rm{Q(}}\sqrt 2 ) \simeq Q(\sqrt 3 )$$

    Rezulta ca

        $$\exists {\rm{ o functie f : Q(}}\sqrt 2 ) \to Q(\sqrt 3 ){\rm{ - izomorfism de inele}}$$

    Asta inseamna ca :

        $$\left\{ \matrix{ {\rm{f(x + y) = f(x)}} \cdot {\rm{f(y)}} \hfill \cr {\rm{f(x}} \cdot {\rm{y) = f(x)}} \cdot {\rm{f(y)}} \hfill \cr} \right.{\rm{ }}{\rm{, }}\forall {\rm{ x}}{\rm{,y}} \in {\rm{Q(}}\sqrt 2 )$$

    si

        $$\left\{ \matrix{ f(0) = 0 \hfill \cr f(1) = 1 \hfill \cr} \right.$$

    Atunci :

        $$\left. {\left\{ \matrix{ f(2) = f(1) + f(1) = 1 + 1 = 2 \hfill \cr f(2) = f(\sqrt 2 \cdot \sqrt 2 ) = f(\sqrt 2 ) \cdot f(\sqrt 2 ) = {f^2}(\sqrt 2 ) \hfill \cr} \right.} \right| \Rightarrow {f^2}(\sqrt 2 ) = 2 \Rightarrow f(\sqrt 2 ) = \pm \sqrt 2 $$

    Presupunand ca e izomorfism,

        $$f(\sqrt 2 ) \in Q(\sqrt 3 )$$

    ceea ce este fals deci =>

        $$Q(\sqrt 2 ){\rm{ nu e izomorf cu Q(}}\sqrt 3 )$$

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