Galois Groups .... A&W Theorem 47.1 .... ....

[Math Amateur]

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

 

[Euge]

Hi Peter,

Recall that the minimum polynomial $m_\alpha$ of $\alpha$ is the unique monic irreducible polynomial having $\alpha$ as a root. Since $f$ is irreducible and $f(\alpha) = 0$, $f$ is a constant multiple of $m_\alpha$. Therefore, $\text{deg}(f) = \text{deg}(m_\alpha) = |F(\alpha) : F|$.

 

[Math Amateur]

Hi Peter,

Recall that the minimum polynomial $m_\alpha$ of $\alpha$ is the unique monic irreducible polynomial having $\alpha$ as a root. Since $f$ is irreducible and $f(\alpha) = 0$, $f$ is a constant multiple of $m_\alpha$. Therefore, $\text{deg}(f) = \text{deg}(m_\alpha) = |F(\alpha) : F|$.

Thanks for the help, Euge,

Peter