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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.

Rgds

SM

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# Misunderstanding of isomorphism and automorphism

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