degree of field extension with a trick or not ? [closed]












1












$begingroup$


Given $[Bbb Q(a):Bbb Q]$ and $b=icdot a$, with $i$ imaginary number, I want to know $[Bbb Q(a,b):Bbb Q]$, I can use $$[Bbb Q(a,b):Bbb Q]=[Bbb Q(a,b):Bbb Q(a)]cdot [Bbb Q(a):Bbb Q]$$



Is it possible to conclude the degrees or do I have to do it with the minimal polynomial etc. ?






abstract-algebra extension-field







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closed as unclear what you're asking by jgon, Eric Wofsey, metamorphy, Lee David Chung Lin, Shailesh Dec 22 '18 at 8:53


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    I don't understand your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 3:56






  • 2




    $begingroup$
    You’ll need to know $[Bbb Q(a,b):Bbb Q(a)]$ of course. This number could equal $1$.
    $endgroup$
    – Lubin
    Dec 11 '18 at 4:08






  • 1




    $begingroup$
    @jgon: I want to know [Q(a,b):Q] with b=i*a
    $endgroup$
    – kratos88
    Dec 11 '18 at 8:13






  • 1




    $begingroup$
    In that case Lubin's comment answers your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 13:07
















1












$begingroup$


Given $[Bbb Q(a):Bbb Q]$ and $b=icdot a$, with $i$ imaginary number, I want to know $[Bbb Q(a,b):Bbb Q]$, I can use $$[Bbb Q(a,b):Bbb Q]=[Bbb Q(a,b):Bbb Q(a)]cdot [Bbb Q(a):Bbb Q]$$



Is it possible to conclude the degrees or do I have to do it with the minimal polynomial etc. ?






abstract-algebra extension-field







share|cite|improve this question









$endgroup$



closed as unclear what you're asking by jgon, Eric Wofsey, metamorphy, Lee David Chung Lin, Shailesh Dec 22 '18 at 8:53


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    I don't understand your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 3:56






  • 2




    $begingroup$
    You’ll need to know $[Bbb Q(a,b):Bbb Q(a)]$ of course. This number could equal $1$.
    $endgroup$
    – Lubin
    Dec 11 '18 at 4:08






  • 1




    $begingroup$
    @jgon: I want to know [Q(a,b):Q] with b=i*a
    $endgroup$
    – kratos88
    Dec 11 '18 at 8:13






  • 1




    $begingroup$
    In that case Lubin's comment answers your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 13:07














1












1








1


1



$begingroup$


Given $[Bbb Q(a):Bbb Q]$ and $b=icdot a$, with $i$ imaginary number, I want to know $[Bbb Q(a,b):Bbb Q]$, I can use $$[Bbb Q(a,b):Bbb Q]=[Bbb Q(a,b):Bbb Q(a)]cdot [Bbb Q(a):Bbb Q]$$



Is it possible to conclude the degrees or do I have to do it with the minimal polynomial etc. ?






abstract-algebra extension-field







share|cite|improve this question









$endgroup$




Given $[Bbb Q(a):Bbb Q]$ and $b=icdot a$, with $i$ imaginary number, I want to know $[Bbb Q(a,b):Bbb Q]$, I can use $$[Bbb Q(a,b):Bbb Q]=[Bbb Q(a,b):Bbb Q(a)]cdot [Bbb Q(a):Bbb Q]$$



Is it possible to conclude the degrees or do I have to do it with the minimal polynomial etc. ?






abstract-algebra extension-field




abstract-algebra extension-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 22:42









kratos88kratos88

566




566




closed as unclear what you're asking by jgon, Eric Wofsey, metamorphy, Lee David Chung Lin, Shailesh Dec 22 '18 at 8:53


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by jgon, Eric Wofsey, metamorphy, Lee David Chung Lin, Shailesh Dec 22 '18 at 8:53


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    $begingroup$
    I don't understand your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 3:56






  • 2




    $begingroup$
    You’ll need to know $[Bbb Q(a,b):Bbb Q(a)]$ of course. This number could equal $1$.
    $endgroup$
    – Lubin
    Dec 11 '18 at 4:08






  • 1




    $begingroup$
    @jgon: I want to know [Q(a,b):Q] with b=i*a
    $endgroup$
    – kratos88
    Dec 11 '18 at 8:13






  • 1




    $begingroup$
    In that case Lubin's comment answers your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 13:07














  • 1




    $begingroup$
    I don't understand your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 3:56






  • 2




    $begingroup$
    You’ll need to know $[Bbb Q(a,b):Bbb Q(a)]$ of course. This number could equal $1$.
    $endgroup$
    – Lubin
    Dec 11 '18 at 4:08






  • 1




    $begingroup$
    @jgon: I want to know [Q(a,b):Q] with b=i*a
    $endgroup$
    – kratos88
    Dec 11 '18 at 8:13






  • 1




    $begingroup$
    In that case Lubin's comment answers your question.
    $endgroup$
    – jgon
    Dec 11 '18 at 13:07








1




1




$begingroup$
I don't understand your question.
$endgroup$
– jgon
Dec 11 '18 at 3:56




$begingroup$
I don't understand your question.
$endgroup$
– jgon
Dec 11 '18 at 3:56




2




2




$begingroup$
You’ll need to know $[Bbb Q(a,b):Bbb Q(a)]$ of course. This number could equal $1$.
$endgroup$
– Lubin
Dec 11 '18 at 4:08




$begingroup$
You’ll need to know $[Bbb Q(a,b):Bbb Q(a)]$ of course. This number could equal $1$.
$endgroup$
– Lubin
Dec 11 '18 at 4:08




1




1




$begingroup$
@jgon: I want to know [Q(a,b):Q] with b=i*a
$endgroup$
– kratos88
Dec 11 '18 at 8:13




$begingroup$
@jgon: I want to know [Q(a,b):Q] with b=i*a
$endgroup$
– kratos88
Dec 11 '18 at 8:13




1




1




$begingroup$
In that case Lubin's comment answers your question.
$endgroup$
– jgon
Dec 11 '18 at 13:07




$begingroup$
In that case Lubin's comment answers your question.
$endgroup$
– jgon
Dec 11 '18 at 13:07










1 Answer
1






active

oldest

votes


















0












$begingroup$

We first recall that



$Bbb Q(a) subset Bbb Q(a, b), tag 0$



thus if $a$ is transcendental over $Bbb Q$, $[Bbb Q(a):Bbb Q]$ is not finite, and so neither is $[Bbb Q(a, b):Bbb Q]$; we thus turn to the case



$[Bbb Q(a):Bbb Q] < infty; tag 1$



note that, with $b = ia$,



$i = dfrac{b}{a} in Bbb Q(a, b); tag 1$



hence,



$Bbb Q(a, i) subset Bbb Q(a, b); tag 2$



also, since



$b = ia; tag 3$



$Bbb Q(a, b) subset Bbb Q(a, i); tag 4$



taken together, (2) and (4) imply



$Bbb Q(a, b) = Bbb Q(a, i); tag 5$



thus,



$[Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)]; tag 6$



now, we have



$x^2 + 1 in Bbb Q(a)[x]; tag 6$



if $x^2 + 1$ irreducible in $Bbb Q(a)[x]$, then



$[Bbb Q(a, i):Bbb Q(a)] = [Bbb Q(a)(i):Bbb Q] = 2; tag 7$



if reducible,



$x^2 + 1 = (x + alpha)(x + beta), ; alpha, beta in Bbb Q(a); tag 8$



$(x + alpha)(x + beta) = x^2 + (alpha + beta)x + alpha beta; tag 9$



$beta = -alpha, ; -alpha^2 = 1 Longrightarrow alpha = pm i in Bbb Q(a) Longrightarrow [Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)] = 1, tag{10}$



and we therefore arrive at the conclusion that



$[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = [Bbb Q(a):Bbb Q] tag{11}$



if $x^2 + 1$ is reducible over $Bbb Q(a)$; if $x^2 + 1$ is irreducible, then



$[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = 2[Bbb Q(a):Bbb Q]. tag{12}$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    We first recall that



    $Bbb Q(a) subset Bbb Q(a, b), tag 0$



    thus if $a$ is transcendental over $Bbb Q$, $[Bbb Q(a):Bbb Q]$ is not finite, and so neither is $[Bbb Q(a, b):Bbb Q]$; we thus turn to the case



    $[Bbb Q(a):Bbb Q] < infty; tag 1$



    note that, with $b = ia$,



    $i = dfrac{b}{a} in Bbb Q(a, b); tag 1$



    hence,



    $Bbb Q(a, i) subset Bbb Q(a, b); tag 2$



    also, since



    $b = ia; tag 3$



    $Bbb Q(a, b) subset Bbb Q(a, i); tag 4$



    taken together, (2) and (4) imply



    $Bbb Q(a, b) = Bbb Q(a, i); tag 5$



    thus,



    $[Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)]; tag 6$



    now, we have



    $x^2 + 1 in Bbb Q(a)[x]; tag 6$



    if $x^2 + 1$ irreducible in $Bbb Q(a)[x]$, then



    $[Bbb Q(a, i):Bbb Q(a)] = [Bbb Q(a)(i):Bbb Q] = 2; tag 7$



    if reducible,



    $x^2 + 1 = (x + alpha)(x + beta), ; alpha, beta in Bbb Q(a); tag 8$



    $(x + alpha)(x + beta) = x^2 + (alpha + beta)x + alpha beta; tag 9$



    $beta = -alpha, ; -alpha^2 = 1 Longrightarrow alpha = pm i in Bbb Q(a) Longrightarrow [Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)] = 1, tag{10}$



    and we therefore arrive at the conclusion that



    $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = [Bbb Q(a):Bbb Q] tag{11}$



    if $x^2 + 1$ is reducible over $Bbb Q(a)$; if $x^2 + 1$ is irreducible, then



    $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = 2[Bbb Q(a):Bbb Q]. tag{12}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We first recall that



      $Bbb Q(a) subset Bbb Q(a, b), tag 0$



      thus if $a$ is transcendental over $Bbb Q$, $[Bbb Q(a):Bbb Q]$ is not finite, and so neither is $[Bbb Q(a, b):Bbb Q]$; we thus turn to the case



      $[Bbb Q(a):Bbb Q] < infty; tag 1$



      note that, with $b = ia$,



      $i = dfrac{b}{a} in Bbb Q(a, b); tag 1$



      hence,



      $Bbb Q(a, i) subset Bbb Q(a, b); tag 2$



      also, since



      $b = ia; tag 3$



      $Bbb Q(a, b) subset Bbb Q(a, i); tag 4$



      taken together, (2) and (4) imply



      $Bbb Q(a, b) = Bbb Q(a, i); tag 5$



      thus,



      $[Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)]; tag 6$



      now, we have



      $x^2 + 1 in Bbb Q(a)[x]; tag 6$



      if $x^2 + 1$ irreducible in $Bbb Q(a)[x]$, then



      $[Bbb Q(a, i):Bbb Q(a)] = [Bbb Q(a)(i):Bbb Q] = 2; tag 7$



      if reducible,



      $x^2 + 1 = (x + alpha)(x + beta), ; alpha, beta in Bbb Q(a); tag 8$



      $(x + alpha)(x + beta) = x^2 + (alpha + beta)x + alpha beta; tag 9$



      $beta = -alpha, ; -alpha^2 = 1 Longrightarrow alpha = pm i in Bbb Q(a) Longrightarrow [Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)] = 1, tag{10}$



      and we therefore arrive at the conclusion that



      $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = [Bbb Q(a):Bbb Q] tag{11}$



      if $x^2 + 1$ is reducible over $Bbb Q(a)$; if $x^2 + 1$ is irreducible, then



      $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = 2[Bbb Q(a):Bbb Q]. tag{12}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We first recall that



        $Bbb Q(a) subset Bbb Q(a, b), tag 0$



        thus if $a$ is transcendental over $Bbb Q$, $[Bbb Q(a):Bbb Q]$ is not finite, and so neither is $[Bbb Q(a, b):Bbb Q]$; we thus turn to the case



        $[Bbb Q(a):Bbb Q] < infty; tag 1$



        note that, with $b = ia$,



        $i = dfrac{b}{a} in Bbb Q(a, b); tag 1$



        hence,



        $Bbb Q(a, i) subset Bbb Q(a, b); tag 2$



        also, since



        $b = ia; tag 3$



        $Bbb Q(a, b) subset Bbb Q(a, i); tag 4$



        taken together, (2) and (4) imply



        $Bbb Q(a, b) = Bbb Q(a, i); tag 5$



        thus,



        $[Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)]; tag 6$



        now, we have



        $x^2 + 1 in Bbb Q(a)[x]; tag 6$



        if $x^2 + 1$ irreducible in $Bbb Q(a)[x]$, then



        $[Bbb Q(a, i):Bbb Q(a)] = [Bbb Q(a)(i):Bbb Q] = 2; tag 7$



        if reducible,



        $x^2 + 1 = (x + alpha)(x + beta), ; alpha, beta in Bbb Q(a); tag 8$



        $(x + alpha)(x + beta) = x^2 + (alpha + beta)x + alpha beta; tag 9$



        $beta = -alpha, ; -alpha^2 = 1 Longrightarrow alpha = pm i in Bbb Q(a) Longrightarrow [Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)] = 1, tag{10}$



        and we therefore arrive at the conclusion that



        $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = [Bbb Q(a):Bbb Q] tag{11}$



        if $x^2 + 1$ is reducible over $Bbb Q(a)$; if $x^2 + 1$ is irreducible, then



        $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = 2[Bbb Q(a):Bbb Q]. tag{12}$






        share|cite|improve this answer









        $endgroup$



        We first recall that



        $Bbb Q(a) subset Bbb Q(a, b), tag 0$



        thus if $a$ is transcendental over $Bbb Q$, $[Bbb Q(a):Bbb Q]$ is not finite, and so neither is $[Bbb Q(a, b):Bbb Q]$; we thus turn to the case



        $[Bbb Q(a):Bbb Q] < infty; tag 1$



        note that, with $b = ia$,



        $i = dfrac{b}{a} in Bbb Q(a, b); tag 1$



        hence,



        $Bbb Q(a, i) subset Bbb Q(a, b); tag 2$



        also, since



        $b = ia; tag 3$



        $Bbb Q(a, b) subset Bbb Q(a, i); tag 4$



        taken together, (2) and (4) imply



        $Bbb Q(a, b) = Bbb Q(a, i); tag 5$



        thus,



        $[Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)]; tag 6$



        now, we have



        $x^2 + 1 in Bbb Q(a)[x]; tag 6$



        if $x^2 + 1$ irreducible in $Bbb Q(a)[x]$, then



        $[Bbb Q(a, i):Bbb Q(a)] = [Bbb Q(a)(i):Bbb Q] = 2; tag 7$



        if reducible,



        $x^2 + 1 = (x + alpha)(x + beta), ; alpha, beta in Bbb Q(a); tag 8$



        $(x + alpha)(x + beta) = x^2 + (alpha + beta)x + alpha beta; tag 9$



        $beta = -alpha, ; -alpha^2 = 1 Longrightarrow alpha = pm i in Bbb Q(a) Longrightarrow [Bbb Q(a, b):Bbb Q(a)] = [Bbb Q(a, i):Bbb Q(a)] = 1, tag{10}$



        and we therefore arrive at the conclusion that



        $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = [Bbb Q(a):Bbb Q] tag{11}$



        if $x^2 + 1$ is reducible over $Bbb Q(a)$; if $x^2 + 1$ is irreducible, then



        $[Bbb Q(a, b):Bbb Q] = [Bbb Q(a, b):Bbb Q(a)[Bbb Q(a):Bbb Q] = 2[Bbb Q(a):Bbb Q]. tag{12}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 5:20









        Robert LewisRobert Lewis

        45.7k23065




        45.7k23065















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