Question : 

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> Prove that if $y,z \in\Bbb Q$ then $y^z \in\Bbb A$.
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My attempt:

Definition 2.7.8 states that a number $s$ is an *algebraic number* when there exists some $p \in\Bbb Z[x]$ such that $p(s)=0$. Let us denote the set $\Bbb A = \{x \in\Bbb C: x \text{ is algebraic}\}$.

The set of all polynomials in $x$ with coefficients from $\Bbb Z$ is denoted $\Bbb Z[x]$. Thus each element $p \in\Bbb Z[x]$ has the form

$p(x) = a\_nx^n+a\_{n-1}x^{n-1}+\cdots+a\_1x+a\_0$

where $n$ is a non-negative integer, $a\_i \in\Bbb Z$ for each $i \in \{0,1,...,n\}$ and $a\_n \neq 0$ except when $n=0$. 

Moreover, by proposition 2.7.9, all rational numbers are algebraic.

So, from $y, z \in\Bbb Q$, $y$ and $z$ must belong in a rational number set, and by the proposition, they must be algebraic. 

Suppose we let $y \in\Bbb Q$, then $y = \frac{a}{b}$ for some $a,b \in \Bbb Z , b \neq 0$. Then the polynomial $p(x) =bx-a$ belongs to $\Bbb Z[x]$ and $p(y) = p(\frac{a}{b}) = 0$. Hence $y \in\Bbb A $.

Now suppose we let $z \in\Bbb Q$. Then $ z = \frac{c}{d}$ for some $ c,d \in\Bbb Z , d \neq 0$. The polynomial $p(x) = dx-c$ belongs to $\Bbb Z[x]$ and $p(z) = p(\frac{c}{d}) = 0$. Hence $ z \in\Bbb A$.

My question is what is with the $y^z \in\Bbb A$? That means $y$ to the $z$ power is algebraic. How do I tackle this problem? 

There is a proposition 2.7.11 in my book that states if $ a \in\Bbb A$ and $q \in\Bbb Q$, then $qa \in\Bbb A$. That would mean that if a is algebraic, and q is rational, then qa is algebraic. However it is just qa, not $q^a$.

There is also another example about whether or not it's possible to have $a^b \in\Bbb Q$ if a and b are irrational, but I don't think that's going to work because if a and b are irrational, it breaks proposition 2.7.9, and a and b are not algebraic.