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Can anyone explain this? It makes sense when I look at it... | Can anyone explain this? It makes sense when I look at it... | ||
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| + | If f(x) has roots <math>r_0, r_1, ... , r_n</math> then f(x+a) has roots <math>r_0 - a, r_1 - a, ... , r_n - a</math>. | ||
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| + | So, the splitting field of f(x) over F must contain <math>r_0, r_1, ... , r_n</math>. | ||
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| + | And the splitting field of f(x+a) over F must contain <math>r_0 - a, r_1 - a, ... , r_n - a</math>. But since F already contains a, this is the same as saying that the splitting field of f(x+a) over F must contain <math>r_0, r_1, ... , r_n</math>. | ||
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| + | So, the splitting field of f(x) over F is F adjoined any <math>r_i</math> not already in F. Same goes for f(x+a). | ||
Latest revision as of 17:49, 19 November 2008
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).
