Revision as of 17:49, 19 November 2008 by Ricec (Talk)

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Can anyone explain this? It makes sense when I look at it...


If f(x) has roots $ r_0, r_1, ... , r_n $ then f(x+a) has roots $ r_0 - a, r_1 - a, ... , r_n - a $.

So, the splitting field of f(x) over F must contain $ r_0, r_1, ... , r_n $.

And the splitting field of f(x+a) over F must contain $ r_0 - a, r_1 - a, ... , r_n - a $. But since F already contains a, this is the same as saying that the splitting field of f(x+a) over F must contain $ r_0, r_1, ... , r_n $.

So, the splitting field of f(x) over F is F adjoined any $ r_i $ not already in F. Same goes for f(x+a).

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