Suppose γ is a kth root of unity that satisfies a quadratic equation z 2 − m z − n = 0 with m , n ∈ Z . Then k = 3 , 4 or 6.Let k ∈ Z with k &gt; 2 and suppose γ is a kth root of unity that satisfies a quadratic equation z 2 − m z − n = 0 with m , n ∈ Z . Then k = 3 , 4 or 6.My knowledge on algebra and number theory is poor, I just need this lemma to complete the computation of the analytic automorphism groups of complex tori.Can you help me prove this lemma with the knowledge of field theory and Euler phi function as small as possible? Actually I forget almost everything on field theory and have never learned number theory.

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Suppose       γ   is a kth root of unity that satisfies a quadratic equation       z    2    −  m  z  −  n  =  0 with   m  ,  n  ∈      Z  . Then   k  =  3  ,  4 or 6.Let   k  ∈      Z   with   k  &amp;gt;  2 and suppose   γ is a kth root of unity that satisfies a quadratic equation       z    2    −  m  z  −  n  =  0 with   m  ,  n  ∈      Z  . Then   k  =  3  ,  4 or 6.My knowledge on algebra and number theory is poor, I just need this lemma to complete the computation of the analytic automorphism groups of complex tori.Can you help me prove this lemma with the knowledge of field theory and Euler phi function as small as possible? Actually I forget almost everything on field theory and have never learned number theory.

merlatas497 · Accepted Answer

Step 1If   γ is a kth root of unity,   k  ≥  1, and root of a quadratic polynomial       X    2    +  a  X  +  b  ∈      Z    [  X  ] then       X    2    +  a  X  +  b divides       X    k    −  1 or vice versa. As   k  ≥  2, we may restrict to the first case. It follows that   a  ∈  {  −  2  ,  −  1  ,  0  ,  1  ,  2  }. Then since       |        γ    2    ±  1      |    ≤  2 and       |    a  γ      |    =      |    a      |  , it follows that   a  ∈  {  −  2  ,  −  1  ,  0  ,  1  ,  2  }. Then   ℜ  γ  =  −      a    2    ∈  {  −  1  ,  −      1    2    ,  0  ,      1    2    ,  1  }, which implies that   γ is a primitive second or third or fourth or sixth or first root of unity.For each of these cases we can indeed name a quadratic polynomial as required:  γ is a primitive first root of unity, i.e.   γ  =  1: Then       γ    2    −  1  =  0.  γ is a primitive second root of unity, i.e.   γ  =  −  1: Then       γ    2    −  1  =  0.  γ is a primitive third root of unity, i.e.   γ  =            −      1      ±      i              3              2  : Then       γ    2    +  γ  +  1  =  0.  γ is a primitive fourth root of unity, i.e.   γ  =  ±  i: Then       γ    2    +  1  =  0.  γ is a primitive sixth root of unity, i.e.   γ  =            1      ±      i              3              2  : Then       γ    2    −  γ  +  1  =  0.Step 2Of course, a number like   γ  =  1 is also a kth root of unity for   k  =  42 or   k  =  666 (though not a primitive such root - that word is missing from the problem statement).

Filloltarninsv9p · Accepted Answer

Step 1Let       u    =          e              i        ψ              =    cos    ⁡    ψ    +    i    sin    ⁡    ψ   be a k th root of unity     ⟹    (      e          i      ψ            )    k    =  1  =      e          i      ψ      k        =  1So, we have       (    cos    ⁡    ψ    +    i    sin    ⁡    ψ          )      2        −    m    (    cos    ⁡    ψ    +    i    sin    ⁡    ψ    )    +    n    =    0             ⟹        cos    ⁡    2    ψ    +    i    sin    ⁡    2    ψ    −    m    (    cos    ⁡    ψ    +    i    sin    ⁡    ψ    )    +    n    =    0                        (    1    )  Equating the imaginary parts,       sin    ⁡    2    ψ    −    m    sin    ⁡    ψ    =    0        ⟺        sin    ⁡    ψ    (    2    cos    ⁡    ψ    −    m    )    =    0  If       sin    ⁡    ψ    =    0    ,    cos    ⁡    ψ    =    ±    1    ,    u    =    ±    1        ⟹        k    =    2   but given that   k  &amp;gt;  2Step 2(1) becomes  1  −  m  (  ±  1  )  +  n  =  0    ⟺    ∓  m  =  n  +  1Step 2Else       cos    ⁡    ψ    =          m      2            ⟹        −    1    ≤          m      2        ≤    1        ⟺        −    2    ≤    m    ≤    2  As   m  ∈      Z    ,  m  =  0  ,  ±  1  ,  ±  2              (  1  )Now check for values of m each of which gives exactly one value of   sin  ⁡  ψ

Suppose gamma is a kth root of unity that satisfies a quadratic equation z^2-mz-n=0 with m, n in Z. Then k=3,4 or 6.

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