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Misunderstanding of isomorphism and automorphism

  1. Aug 26, 2009 #1

    I was doing self studying abstract algebra from the online lecture notes posted by Robert Ash and I hit against the following theorem. I am posting it in the topology section because without a geometric/topological meaning to the concept I am never able to understand the topic and that is the reason during my undergraduate days 8 years ago I did not pass in algebra at all.

    Now for the theorem: Let E/F be a finite separable extension of degree n and let sigma be an embedding of F in C (where C is the algebraic closure of E). Then sigma extends to exactly n embeddings of E in C. In other words there are exactly n embeddings of tau in C such that restriction of tau to F coincides with sigma. In particular if sigma is the identity function on F then there are exactly n F-monomorphism of E into C.

    The above is reproduced in ditto from Thereom 3.5.2 of R.Ash lecture notes posted on his website.

    Now my question.

    Let me think of E as R^3 and F as R. Now surely R^3 is the closure of R. (The reason why I have taken R^3 is simply because I am successful in proving the theorem in 2D, using x^2 +1 = 0 and things like that. Now when I tried using x^3 +2 = 0, I understood that I would be getting 6 dimensions of the separable field. )

    My hunch would be automorphism could be thought of as vector rotations and hence I thought of coming out with this simple example. Surely having T(a) = a + 1 is an isomorphism but not an automorphism which fixes a.

    So I thought of taking the following isomorphisms of a vector .

    (x, y, z) ---> (x, y, z) (identity)
    (x, y, z) ---> (y, x, z)
    (x, y, z) ---> (z, y, x)
    (x, y, z) ---> (x, z, y)
    (x, y, z) ---> (z, x, y)
    (x, y, z) ---> (y, z, x)

    The above are some kind of transformations I thought which are possible.

    If the theorem above is true then at least 3 of the above isomorphisms should not be an automorphism. I am really not sure which 3 of them should not be and why not. I have tried a lot to think but have been unsuccessful.

    The theorem says that the transformation when restricted to F then it should fix F. Now I could think that since the extension of R^3 is over R hence x should be fixed but I only get 2 automorphisms. The identity and (x, y, z) ---> (x, z, y). Not sure why I am not getting the third one, because the theorem says that E/F is of degree n ( which is 3) and so it should be possible to get solutions of all the polynomials in R^3. So there is sth wrong in my understanding or the way that I am thinking of automorphisms as rotations and missing sth.

    I thought about whether a splitting field can have 3 dimensions and found that it could. Consider x^3 + 2 = 0 over F. Now it has got 3 roots. 2 complex one real but even the real does not fall in F. So the extension field is (modelling on R^3 ), x axis is F, y axis is (2)^1/3 , z complex root.

    Last edited: Aug 26, 2009
  2. jcsd
  3. Aug 27, 2009 #2


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    What? The algebraic closure of R (the field of real numbers) is C (the field of complex numbers).

    Since an automorphism on a field is simply an isomorphism from the field to itself, all of those are automorphisms on R3. But how are you making R3 a field? How are you defining multiplication?

  4. Aug 27, 2009 #3

    Sorry for the misunderstanding. Yes C is the closure of R. Since I was trying to think geometrically so I thought a similar thing is possible in R^3 vs R.

    So if E is not R^3 but say the field generated by the linear combination of the roots the solution of x^3 - 2 = 0. and the field F is Q then will this be true. One can visualize that each of the axes of R^3 will be somehow similar to the field generated by each of these roots. Surely the name suggests that it is a normal extension so this kind of a mental picture intusion is prob OK. Still willing to be proved wrong.

    Not sure about the point why if C is a field which is like R^2, then R^3 probably cannot be a field.
    Last edited: Aug 27, 2009
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