Prove that if $y,z \in\Bbb Q$ then $y^z \in\Bbb A$

Question

Question :

Prove that if $y,z \in\Bbb Q$ then $y^z \in\Bbb A$.

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.

Answer 1

Since $z$ is rational, write $z$ as $p/q$. Well, $y^p$ is still rational so that's good. The question now is, is the $q$-th root of $y^p$ the solution to some polynomial over $\mathbb{Q}$ (and therefore $\mathbb{Z}$). But $X^q - y^p$ is in $\mathbb{Q}[X]$ (and $\mathbb{Z}[X]$ after clearing denominators).

Answer 2

If $a, q \in \Bbb{N}$, then $a^{1/q}$ is algebraic. Take $p(a^{1/q}) = 0$ where $p(X) = X^{q} - a$.

Since algebraic numbers form a field we have that $(\frac{a}{b})^{1/q} = a^{1/q} (b^{1/q})^{-1}$ is also algebraic for $b \neq 0$.

Now $(\frac{a}{b})^{p/q}$ is algebraic for both positive and negative $p \in \Bbb{Z}$ since it's just a finite product of algebraic numbers.

Notice that all such $p/q$ covers $\Bbb{Q}$ since we let $p$ both negative and positive, or zero. Now we're only left to prove that the first part for negative $a$.

For instance, for $q = 2$ we can define $i^2 = -1$ and $\Bbb{Q}(i) \subset A$ (THINK OF COMPLEX NUMBERS!).

Then each $(-a)^{1/2} \in A$ since $i$ is algebraic and $(-a)^{1/2} = i a^{1/2}$. For the other $1/q$ powers. Define $i_q$ such that $(i_q)^q = -1$. Then $\Bbb{Q}(i_q)$, the rationals extended with $i_q$ is a subset of $A$.