I have questions about the proof of:
Let $K \subset L$ be a field extension and $a,b \in L$ transcendental numbers over $K$.
How to show that $a+b$ and $ab$ are not both algebraic over $K$?
The proof is:
The formula is $[K(a,b):K]=[K(a,b):K(a+b,ab)] \cdot [K(a+b,ab):K]$.
It exists a polynomial $(x-a)(x-b)=x^2-(a+b)x+ab=0$ with roots $a,b$.
$\Rightarrow [K(a,b):K(a+b,ab)] \leq 2$
Suppose $a+b$ and $ab$ are both algebraic over $K$, then $[K(a+b,ab):K]<\infty$.
$\Rightarrow [K(a,b):K] <\infty$, and that is a contradiction to $a,b$ are transcendent over $K$.
My questions are: How to find the polynomial $x^2-(a+b)x+ab=0$ to see that $a,b$ are the roots? And why does it imply that $[K(a,b):K(a+b,ab)] \leq 2$?
$\textbf{Another approach of the proof:}$
The formula is $[K(a,b):K]=[K(a,b):K(a+b,ab)] \cdot [K(a+b,ab):K]$.
Let $m:=[K(a,b):K], p:=[K(a,b):K(a+b,ab)]$ and $q=[K(a+b,ab):K]$.
Since $a,b$ are transcendent over $K$, it follows that $m=\infty$.
So it has to be shown that $p$ is finite, as $\Rightarrow q=\infty$ and $a+b,ab$ are transcendent and not algebraic
It's $K(a,b)=K(a+b,ab)$
$L:=K(a,b)$ contains $a$ and $b \Rightarrow K(a+b,ab) \subset K(a,b)$
To show that $L \subset M:=K(a+b,ab)$, it has to be shown that $a,b \in M$
It's $\frac{a+b}{ab} \in M$ since $a+b,ab \in M$
$\frac{a+b}{ab}=\frac{a}{ab}+\frac{b}{ab}=\frac{1}{b}+\frac{1}{a}\in M$ and also $\frac{1}{b},\frac{1}{a} \in M$
$a=\frac{1}{b}(ab)\in M, b=\frac{1}{a}(ab)\in M$
$\Rightarrow K(a,b)=K(a+b,ab)$
$\Rightarrow p=1$
$\Rightarrow q=\infty$
$\Rightarrow a+b$ and $ab$ are not both algebraic over $K$
Is this right?
The comments provide an explanation of the proof provided by your text: here's an alternate approach.
If $a+b$ and $ab$ are algebraic, then $(a+b)^2-4ab=a^2-2ab+b^2=(a-b)^2$ is algebraic. Since the field of algebraic numbers are closed under roots, this implies that $a-b$ is algebraic.
But, this then shows that $(a+b)-(a-b)=2b$ is algebraic, a clear contradiction.