Sum and Product of two transcendental numbers cannot be simultaneously algebraic 
If $\alpha$ and $\beta$ are real number and $\alpha$ and $\beta$ are transcendental over $\mathbb Q$, show that $\alpha \beta$ or $\alpha +\beta$ is also transcendental over $\mathbb Q$

Attempt: Our strategy should be that we will assume both $\alpha \beta$ and $\alpha +\beta$ are algebraic over $\mathbb Q$ and then arrive at a contradiction that $\alpha$ or $\beta$ is algebraic.
$\alpha$ and $\beta$ are transcendental over $\mathbb Q \implies ~\nexists f(x), g(x) \in \mathbb Q[x]$ such that $f(\alpha)=0, ~g(\beta)=0$. 
Lets assume both $\alpha \beta$ and $\alpha +\beta$ are algebraic over $\mathbb Q$. So, I must construct an equation which if, has roots  $\alpha \beta$ and $\alpha +\beta$, must have root $\alpha$ as well
I am not able to understand how the equation $~~x^2-( \alpha + \beta)x + \alpha \beta = (x-\alpha)(x-\beta)$ will help us in this regard. 
Thank you for your help.

$2.$ The splitting field for $x^4-x^2-2$ over $\mathbb Z_3$

Attempt: The roots of the given equation are $\pm i, \pm \sqrt 2$. Hence, splitting field is $\mathbb Z_3(i, \sqrt 2)$ . Am I correct?

$3$. Let $a$ be a complex  zero of $x^2+x+1$ over $\mathbb Q$. prove that $Q(\sqrt a)=Q(a).$

Attempt: Suppose $a = c+di~|~c,d \in \mathbb Q$, then : $c+di-c \in Q(a) \implies d^{-1}di \in \mathbb Q(a) \implies i \in \mathbb Q(a) \implies \alpha + \beta i \in \mathbb Q(a)~~\forall~~\alpha,\beta \in \mathbb Q \implies Q(a) = \mathbb C$
Since, $\sqrt a$ is also a complex number, $\implies Q(a) = Q(\sqrt a) = \mathbb C$.
Is my proof correct? 
Thank you for your help.
 A: Hints/comments/whatnot:


*

*If $K$ is the field of numbers (in $\Bbb{C}$) that are algebraic over $\Bbb{Q}$, then by your contrapositive assumption the polynomial
$$
(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta
$$
has coefficients in $K$. Hence its zeros are algebraic over $K$. Therefore...

*Your anwser is correct. However, your teacher may want you to observe that in $\Bbb{Z}_3$ we have $2=-1$. Hence ... BTW can you factor this polynomial in $\Bbb{Z}_3[x]$.

*I'm afraid you made a mistake at the starting gate. Why do you claim that the real and imaginary parts of $a$ would both be rational? What does the formula for the solutions of a quadratic tell you? (Also what Tomasz says. +1)

A: Those are really separate questions and should be asked separately.
I think the general fact you're supposed to use is that algebraic numbers are algebraically closed, i.e. a polynomial with algebraic coefficients has algebraic roots (so the result follows immediately from considering $(x-\alpha)(x-\beta)$).
For the second one, $\sqrt 2$ and $i$ are complex numbers. While $2$ and $-1$ certainly do have roots in an extension of ${\bf Z}_3$, no good will come out of calling them the same as you call their complex counterparts. Otherwise, I guess that is as good a description as any.
For the third, you're way off: ${\bf Q}(a)$ and ${\bf Q}(\sqrt a)$ are both countable, while ${\bf C}$ is certainly not.
