Prove every finite extension is a simple extension can someone help me understand the proof of this theorem : 
Where $F(c)$ is the extension field of $F$ with $c$, 
Prove every finite extension of $F$ is a simple extension $F(c)$.
I do not understand the end of the proof, which I included below from Pinter : let $p(x)$ be the minimum polynomial of $b$ over $F(c)$. If the degree of $p(x)$ is $1$, then $p(x)=x-b$, so $b\in F(C)$
If the minimum polynomial is $x-b$, doesn't that mean $b\in F \subset F(c)$ ? There is a theorem that states the degree of $F(c)$ over $F$ is the degree of the minimum polynomial of $c$ over $F$, so I think in this case $F(c)=F$ if the degree of $p(x)=1$. 
So, Pinter then proves by contradiction that the degree of $p(x)=1$. I think this means $b$ was in $F$ all along, so of coarse $F(a,b)=F(c)$ because $F(a,b)=F(a)$ if $b \in F$. Just set $c=a$. 
I am confused, because don't we need $b$ to not be in $F$ for this proof to be meaningful? 
Proof: 
Every field extension is a simple extension, we show $F(a,b) = F(c)$ for some $c$. 
Given $F(a,b)$, let $A(x)$ be the minimum polynomial of a over $F$, and let $B(x)$ be the minimum polynomial of $b$ over $F$. Let $K$ be the extension that holds all of the roots $a_1,...,a_n$ of $A(x)$ and $b_1,...b_n$ of $B(x)$. Let $a_1=a$ and $b_1=b$.
let $t$ be any non zero element of $F$ such that $t \neq \frac{a_i-a}{b-b_j}$ for every $i\neq 1$ and $j \neq 1$.
Cross multiplying and setting $c=a+tb$, it follows that $c\neq ai+tb_j$, that is, $c-tb_j\neq a_i$ for all $i\neq 1, j\neq 1$. 
Define $h(x)=A(c-tx);$ then, $h(b)=A(c-tb)=0$, while for every $j\neq 1, h(b_j)=A(c-tb_j)\neq 0$. Thus, $b$ is the only common root of $h(x)$ and $B(x)$. 
So it remains only to prove that $b\in F(c)$. let $p(x)$ be the minimum polynomial of $b$ over $F(c)$. If the degree of $p(x)$ is $1$, then $p(x)=x-b$, so $b\in F(C)$, and we are done. let us suppose $p(x)\geq 2$ and get a contradiction. 
Observe $h(x), B(x)$ must both be multiples of $p(x)$ because both have $b$ as a root, and $p(x)$ is the minimum polynomial of $b$. But if $h(x)$ and $p(x)$ have a common factor of degree $\geq 2$, they must have two or more roots in common, contrary to the fact that b is their only common root. Our proof is complete. 
 A: I figured it out. 
We have $F(a,b) : b_1,...,b_n$ are factors of min poly $B(x)$  and $a_1,...,a_m$ are factors of $A(x)$. 
We define $c$ such that $c=a+tb$ with $t$ in $F$, not $F(a,b)$, in such a way to construct a polynomial, $H(x)$, which has $b$ as a root: $H(b)=0$, but $H(x\neq b)\neq 0$.
This way, we can consider the minimum polynomial $p(x)$ of $b$ over $F(c)$. this polynomial must be of degree $1$, which means $p(x)=x-b$. This means $b \in F(c)$ because $x-b$ is in $F(c)$. If the degree is more than $1$, $B$ and $H$ would have to be multiples of $p(x)$, since all polynomials with $b$ as a factor are multiples of the minimum polynomial. If the minimum polynomial has $2$ factors, then $B$ and $H$ would have two common factors. $H$ was designed so that this is not the case. Therefore, $b \in F(c)$. 
A: The first question is about this statement:

Let $p(x)$ be the minimum polynomial of $b$ over $F(c)$. If the degree of $p(x)$ is $1$, then $p(x)$ is $x − b$, so $b \in F(c)$, and we are done.

Because if the degree of minimum polynomial $p(x)$ is $1$, it must be in the form of $(x - b)$, so as the statement $(x - b)$ itself must contain $b$, it is only valid if $b$ is already an element of $F(c)$.
( Also note that in the text before, it is wrongly stated that $t \in F$ which may not be the case as $a$ and $b$ are not necessarily elements of $F$. The correct statement should be $t \in F(a, b)$. )
And then the next question is about this statement:

But if $h(x)$ and $B(x)$ have a common factor of degree $\geq 2$, they must have two or more roots in common, contrary to the fact that $b$ is their only common root.

The statement is based on Theorem 1 above:

If $F$ has characteristic $0$, irreducible polynomials over $F$ can never have multiple roots.

Note that the word "multiple" tends to mislead, nevertheless what it means is that all roots of irreducible polynomials over $F$ must all be distinct roots.
If the degree of $p(x)$ is $\geq 2$, based on Theorem 1, it must have $2$ or more distinct roots, but it cannot be the case as $b$ has been showed it is the only common root of $h(x)$ and $B(x)$ (because for all roots of $B(x)$ only $b$ that is a root of $h(x)$). Hence it's a contradiction, or in other words $p(x)$ cannot have degree more than $1$.
As it has been proved above that for the case of degree $1$, $b \in F(c)$, this completes the proof.
