First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: Q ( 2 3 , γ ), where γ 3 = 1 and γ 2 = 1 − γ. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.I know that my Galois group will consist of the following automorphisms: σ i = identity σ 1 : 2 3 ↦ 2 3 γ and γ is fixed. σ 2 : 1 3 ↦ 2 3 γ 2 and γ is fixed. σ 3 : γ ↦ γ 2 and 2 3 is fixed. σ 4 : 2 3 ↦ 2 3 γ and γ ↦ γ 2 . σ 5 : 1 3 ↦ 2 3 γ 2 and γ ↦ γ 2 .If I didn't miss something or make some error.My specific question is, when constructing my automorphisms, why exactly is it that I can't send 2 3 to something like 4 3 or γ or some other random thing? Why are these automorphisms so strictly defined?

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First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity:       Q    (      2    3    ,  γ  ), where       γ    3    =  1 and       γ    2    =  1  −  γ. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.I know that my Galois group will consist of the following automorphisms:      σ    i   = identity      σ    1    :      2    3    ↦      2    3    γ and   γ is fixed.      σ    2    :      1    3    ↦      2    3        γ    2   and   γ is fixed.      σ    3    :  γ  ↦      γ    2   and       2    3   is fixed.      σ    4    :      2    3    ↦      2    3    γ and   γ  ↦      γ    2  .      σ    5    :      1    3    ↦      2    3        γ    2   and   γ  ↦      γ    2  .If I didn&#039;t miss something or make some error.My specific question is, when constructing my automorphisms, why exactly is it that I can&#039;t send       2    3   to something like       4    3   or   γ or some other random thing? Why are these automorphisms so strictly defined?

Isaiah Haynes · Accepted Answer

Explanation:By the Fundamental Theorem of Galois Theorem, if the Galois group were abelian, then all subfields would be normal extensions. But this cannot be, as       Q    (      2    3    )      /        Q   is not a normal extension.

sooxicyiy · Accepted Answer

Step 1let   j  =      1    2    (  −  1  +  i      3    ). Clearly       2    3      ,      2    3      j   and             2      3                          j                    2             are roots of   f  (  x  )  =      x    3    −  2. We claim   E  =      Q    (      2    3    ,  j  ) is splitting field f over       Q   because  E  =      Q    (      2    3      ,      2    3      j    ,      j    2        2    3    )  ⊂      Q    (      2    3    ,  j  )and  j  =      1    2        2    3    (      2    3        )    2    j    ⟹        Q    (      2    3    ,  j  )  ⊂  E  =      Q    (      2    3      ,      2    3      j    ,      j    2        2    3    )Step 2The polynomial f is not irreducible over Q as a result minimal polynomial of       2    3     is f and   [      Q    (      2    3      )  :      Q    ]  =  3 on the other hand  [      Q    (      2    3      ,  j    )  :      Q    ]  =  [      Q    (      2    3    ,    j  )  :      Q    (      2    3    )  ]  [      Q    (      2    3      )  :      Q    ]  =  2  ×  3  =  6in the other words  [  E  :      Q    ]  =  6so the Galois group is order 6. letA: The metamorphism which fixes j and maps   j  →      j    2  B: The metamorphism which fixes       2    3   and maps       2    3    →      2    3      jThe Galois group is thus:      ⟨    A    ,    B              |                                        A                          2                          ,                      B                    3              ,      A  B  =                                  B                          −          1                      A    ⟩    ≅                    D                    6

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