How to prove that the sum and product of two algebraic numbers is algebraic? Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$

Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)? 

The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension.

Consider fields $ E \supseteq F $, and elements $ \alpha, \beta \in E $ algebraic over $ F $. We want to show $ \alpha + \beta $, $ \alpha \beta $ are algebraic over $ F $ too. If even one of $ \alpha, \beta $ are $ 0 $, the result is trivial, so let's take both $ \alpha, \beta $ to be non-zero.
We have $ \alpha ^m + a_{m-1} \alpha ^{m-1} + \ldots + a_0 = 0 $ ( each $ a_i \in F $ ), and $ \beta ^n + b_{n-1} \beta ^{n-1} + \ldots + b_0 = 0 $ ( each $ b_j \in F $ ).
(The first equation lets us express all powers of $ \alpha $ as $F$-combinations of $ 1, \alpha, \ldots, \alpha ^{m-1} $. Similarly for $ \beta $)
Let $$ Z := \, [ \, \alpha ^0 \beta ^0, \alpha ^0 \beta ^1, \ldots, \alpha ^0 \beta ^{n-1} ; \alpha ^1 \beta ^0, \ldots, \alpha ^1 \beta ^{n-1} ; \ldots ; \alpha ^{m-1} \beta ^{0}, \ldots, \alpha ^{m-1} \beta ^{n-1} \, ]^{T} \in E^{mn} $$
Now notice we can express $ (\alpha + \beta)Z $ as $ M_1 Z $ with $ M_1 \in F^{mn \times mn} $. So $ ( ( \alpha + \beta ) I - M_1 ) Z = 0 $, and as $ Z \neq 0 $ we have $ \det( (\alpha + \beta)I - M_1 ) = 0 $. Hence $ \alpha + \beta $ is a root of the polynomial $ P(t) := \det( tI - M_1 ) \in F[t] $, and is therefore algebraic over $ F $. Similarly we can show $ \alpha \beta $ is algebraic over $ F $ (Write $ \alpha \beta Z = M_2 Z $ and proceed as above).
