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

1. Jul 4, 2017

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:

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

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2. Jul 5, 2017

Staff: Mentor

What is $|F(\alpha):F|$? Or better: how can it's value be determined given only $\alpha$ and $F\,$?

3. Jul 5, 2017

Math Amateur

Thanks fresh_42 ...

You write:

" ... What is $|F(\alpha):F|$? ... "

Well ... $|F(\alpha):F|$ is the degree of the vector space $F(\alpha)$ over $F$ ....

To answer your second question ... the degree of $F(\alpha)$ over $F$ is equal to the degree of the minimal polynomial of $\alpha$ over $F$ ... and that minimal polynomial is the unique monic irreducible polynomial in $F[x]$ of minimal degree having $\alpha$ as a root ... ...

Is that correct?

Peter

***EDIT***

Presumably, $f$ being an irreducible polynomial having $\alpha$ as a root means that $f$ is the minimum polynomial ...

BUT ...

... just thinking over how we know that there isn't an irreducible polynomial $g$ of lesser degree than $f$ having $\alpha$ as a root ... in which case $g$ would be the minimum polynomial, not $f$ ... how do we know that there is not such a polynomial $g$ ...

Last edited: Jul 5, 2017
4. Jul 5, 2017

Staff: Mentor

Yes, that's correct. $m_{\alpha,F}(x)$ is the polynomial which satisfies $m_{\alpha,F}(\alpha)=0$ and which has no factors $f(x)$ that also would satisfy $f(\alpha)=0$. You can see it by division: $f(x) = q(x)\cdot m_{\alpha,F}(x) +r(x)$ and if $f(\alpha)=0$, then $r(\alpha)=0$ with $\operatorname{deg} r(x) < \operatorname{deg} m_{\alpha,F}(x)$ which means $m_{\alpha,F}(x)$ wasn't minimal.

So what does the irreducibility of a polynomial $f(x)$ imply in the given situation?

5. Jul 5, 2017

Math Amateur

Thanks fresh_42 ...

You write:

"... ... So what does the irreducibility of a polynomial $f(x)$ imply in the given situation? ... ... "

From what you have said ... it means that $f$ is the minimal polynomial ... and so then ... $\text{deg}(f) = [ F( \alpha ) : F ]$ ... ...

Is that correct ... ?

Peter

6. Jul 5, 2017

Staff: Mentor

Yes, you can use the same argument. Since the remainder polynomial $r(x)$ would have a smaller degree than the minimal polynomial, $r(x)$ has to be the zero polynomial itself. But then $f(x)=q(x)\cdot m_{\alpha,F}(x)$ and irreducibility of $f(x)$ guarantees $q(x)=q_0 \in F$. Whether $f(x)$ is the minimal polynomial depends on whether it is monic (highest coefficient $= 1$) or not. It could well be a multiple of $m_{\alpha,F}(x)$ with a factor from $F$. But this isn't important for the degree.

Edit:
There might be one, but then $f(x)\,\vert \,g(x)$. And since $f(x)$ is irreducible, there cannot be any polynomial of smaller degree. It is always the division formula $g=q\cdot m + r$ which gives the argument.

7. Jul 6, 2017

mathwonk

Mr Amateur, has it occurred to you that you might be better off spending a bit more time and effort trying to answer your own questions? You seem to ask an awful lot of questions here over a relatively short amount of time. I'm just worried you are not learning as much as you would if you struggled a bit more with them yourself. Try it, you may surprise yourself. Often one only gets maybe half way or less to a solution, but even that is real progress toward understanding. Feel free to ignore the suggestion if it doesn't help, but many people find this is the way to reach the next level in learning math.