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I am reading Anderson and Feil - A First Course in Abstract Algebra.
I am currently focused on Ch. 47: Galois Groups... ...
I need some help with an aspect of the proof of Theorem 47.1 ...
Theorem 47.1 and its proof read as follows:
View attachment 6863
View attachment 6864
At the end of the above proof by Anderson and Feil, we read the following:
"... ... It then follows that $$| \text{Gal} ( F( \alpha ) | F ) | \le \text{deg}(f)$$.
The irreducibility of $$f$$ implies that $$\text{deg}(f) = | F( \alpha ) : F |$$ ... ... "
Can someone please explain exactly why the irreducibility of $$f$$ implies that $$\text{deg}(f) = | F( \alpha ) : F$$ | ... ... ?Peter
I am currently focused on Ch. 47: Galois Groups... ...
I need some help with an aspect of the proof of Theorem 47.1 ...
Theorem 47.1 and its proof read as follows:
View attachment 6863
View attachment 6864
At the end of the above proof by Anderson and Feil, we read the following:
"... ... It then follows that $$| \text{Gal} ( F( \alpha ) | F ) | \le \text{deg}(f)$$.
The irreducibility of $$f$$ implies that $$\text{deg}(f) = | F( \alpha ) : F |$$ ... ... "
Can someone please explain exactly why the irreducibility of $$f$$ implies that $$\text{deg}(f) = | F( \alpha ) : F$$ | ... ... ?Peter