$a+b$ and $ab$ are not both algebraic over $K$ (proof)

Question

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 $K$?

I tried it with another approach of the proof.

My 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?

Answer 1

Your proof is not correct - and I think you might be getting lost in all the variables you defined (I certainly did!) - it's better to call things what they are if you can and to clearly segment the proof.

It starts off well; the equation

$$[K(a,b) : K] = [K(a,b) : K(a+b,ab)] \cdot [K(a+b,ab) : K]$$

is important and you correctly note that you need to show that $[K(a,b) : K(a+b,ab)]$ is finite in order to show that $[K(a+b,ab) : K]$ is infinite and hence that $a+b$ or $ab$ is transcendental.

However, you then try to show the stronger (and, in general, false!) statement that $$K(a+b,ab) = K(a,b)$$ by trying to show that $a$ and $b$ are in $K(a+b,ab)$. You do this by writing

$$\frac{a+b}{ab}=\frac{1}a+\frac{1}b \in K(a+b,ab)$$

which is correct, but then leap to saying that this means $\frac{1}a$ and $\frac{1}b$ are in $K(a+b,ab)$, which is not correct - it's like saying that, since $\sqrt{2}+ (-\sqrt{2})$ is in $\mathbb Q$, it must be that $\sqrt{2}$ and $-\sqrt{2}$ are as well. That is to say, a field is closed under addition, negation, multiplication and reciprocation - but this doesn't mean that whenever a sum of two things is in a field, both those summands were.

Instead, you have to note that if you know the sum and product of two numbers, then you can define a quadratic equation with those as roots by writing $$(x-a)(x-b)=x^2-(a+b)x+ab.$$ However, this is a polynomial in $K(a+b,ab)[x]$ and its roots are the desired values - so you can find that $[K(a,b):K(a+b,ab)]$ is at most $2$ because $a$ and $b$ are the roots of some quadratic polynomial.

Answer 2

Why not simply consider the tower of extensions: \begin{matrix} K(a,b)\\[-0.5ex] \big\vert &\mkern-36mu ^\text{quadratic extension}\\[-0.5ex] K(a+b,ab) \\[-0.5ex] \big\vert \\[-0.5ex] K \end{matrix} Indeed, $a,b$ are solutions of the quadratic equation $\;x^2-(a+b)x+ab=0$; so, if $a+b$ and $ab$ are both algebraic over $K$, $a$ and $b $ are algebraic over $K$ too, contrary to the hypothesis.