# I Splitting Fields: Anderson and Feil, Theorem 45.6 ...

1. Jun 23, 2017

### Math Amateur

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 45: The Splitting Field ... ...

I need some help with an aspect of the proof of Theorem 45.6 ...

Theorem 45.6 and its proof read as follows:

At the start of the proof of Theorem 45.6 we read the following:

"Suppose that $K$ is a normal extension of $F$, a field with characteristic zero. Then by Theorem 45.5, $K = F( \alpha )$, where $\alpha$ is algebraic over $F$. ... .. "

Can someone please explain exactly how $K = F( \alpha )$ follows in the above statement ... ?

The quote mentions Anderson and Feil's Theorem 45.5 and also mentions that K is a normal extension so I am providing the statement of Theorem 45.5 and Anderson and Feil's definition of a normal extension as follows ... ...

Hope someone can help ...

Peter

#### Attached Files:

File size:
77.4 KB
Views:
17
File size:
17.3 KB
Views:
18
• ###### A&F - Defintion of a Normal Extension ... ....png
File size:
75.9 KB
Views:
19
2. Jun 23, 2017

### Staff: Mentor

Gather what you have: $F \subseteq K$ a finite extension. This means $\dim_FK =: n < \infty$.
Now what does that mean for $\{1,a,a^2,a^3,\ldots ,a^n\}\,$, if we take an arbitrary element $a \in K\,$?
Can you deduce from that, that the conditions for Theorem 45.5 are met?
And what does $F \subseteq K$ simple algebraic extension mean?

3. Jun 24, 2017

### Math Amateur

Thanks for the advice to "Gather what you have: ... "...

I was myopically focused on the first sentence of the proof where we are given

(i) $K$ is a normal extension of $F$
(ii) $F$ is a field with characteristic zero

and was wondering about how to get from this to the initial conditions of Theorem 45.5 that

(i) $F$ is a field with characteristic zero
(ii) $K$ was a finite algebraic extension of $F$

BUT ... in my tunnel vision I neglected that $K$ was given as a finite extension ... sorry for such a simple oversight ...

... HOWEVER ... still concerned that we need to establish that $K$ is algebraic over $F$ ... that is all elements of $K$ are algebraic over $F$ ... before applying Theorem 45..5 ...

Peter

4. Jun 24, 2017

### Staff: Mentor

That's why I wrote - with $n$ being the dimesion of $K$ over $F$
How many elements has this set? Can they be linear independent, and if not, what does that mean?

5. Jun 24, 2017

### Math Amateur

You write:

"Now what does that mean for $\{1,a,a^2,a^3,\ldots ,a^n \},$, if we take an arbitrary element $a \in K\,$?"

It means that $\{1,a,a^2,a^3,\ldots ,a^{n - 1} \} ,$ is a basis for the vector space $K$ over $F$ ... and that $\{1,a,a^2,a^3,\ldots ,a^n \},$ is a linearly dependent set ... ... is that correct?

You write:

"Can you deduce from that, that the conditions for Theorem 45.5 are met?"

... regarding conditions for Theorem 45.5 to be met ... see my previous post ...

You write:

"And what does $F \subseteq K$ simple algebraic extension mean?""

Means that $K$ is of the form $K = F( \alpha )$ ...

Peter

6. Jun 24, 2017

### Math Amateur

As mentioned in the previous post the set is linearly dependent ... so this means that any element of the set can be expressed as a linear combination of the others ...

Peter

7. Jun 24, 2017

### Staff: Mentor

I don't know. I don't care either. This would exclude e.g. all $a \in F \subseteq K$ without reason.

It means the set $\{1,a,a^2,a^3,\ldots ,a^{n}\}$ is linear dependent - in any case. Linear dependency means, we have a non trivial expression of zero: $0 = c_0\cdot 1 + c_1\cdot a + c_2\cdot a^2 + \ldots + c_n\cdot a^n$ with not all $F \ni c_i=0$. If we define $p(x)=c_0\cdot 1 + c_1\cdot x + c_2\cdot x^2 + \ldots + c_n\cdot x^n$ we get a polynomial in $F[x]$ with $p(a)=0$ and $a$ is algebraic.

$K=F(\alpha)$ for a certain $\alpha \in K$ is the result of theorem 45.5.

8. Jun 24, 2017

### Math Amateur

Thanks for all your help fresh_42 ...

Gradually building a basic understanding of Galois theory, thanks to your support ...

Peter