Closure of Algebraic Closure

I assume that you know some basic facts on field theory. I will prove that a multiplication of two algebraic numbers is again an algebraic number.

Let $\alpha$, $\beta$ be algebraic numbers over $\mathbb{Q}$. If you can show $[\mathbb{Q}(\alpha \beta):\mathbb{Q}]$ is finite, then you get $\alpha\beta$ is algebraic over $\mathbb{Q}$. Observe that $\mathbb{Q}(\alpha\beta)\subseteq \mathbb{Q}(\alpha,\beta)$, so we will prove that $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ is finite instead.

Let $n$ and $m$ be a degree of an minimal polynomial of $\alpha$ and $\beta$ respectively. Then $\{1,\alpha,\cdots, \alpha^{n-1}\}$ and $\{1,\beta,\cdots,\beta^{m-1}\}$ are bases of $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ over $\mathbb{Q}$ respectively. Now, you can see that the set $$\{\alpha^k\beta^l \mid 0\le k<n,\, 0\le l<m\}$$ generates $\mathbb{Q}(\alpha,\beta)$; that is, $\mathbb{Q}(\alpha,\beta)$ is generated by a finite set.

The OP points out that my proof only shows the set of all algebraic numbers is closed under multiplication, and we do not know the set of all algebraic numbers is the algebraic closure of $\mathbb{Q}$. It may depend on the definition of algebraic closure, but the question is meaningful if we define algebraic closure of $F$ as the least algebraically closed field containing $F$.

In fact, the field of algebraic numbers is algebraic closure of $\mathbb{Q}$. (The similar thing holds even if we replace $\mathbb{Q}$ to any general field.) Let me describe its proof: Let $\alpha$ be a solution of an equation $$\beta_n x^n+\cdots \beta_1 x+\beta_0=0,$$ where $\beta_0$, $\beta_1$, $\cdots$, $\beta_n$ are algebraic numbers. From this, we can conclude that $[\mathbb{Q}(\alpha,\beta_0,\cdots ,\beta_n):\mathbb{Q}(\beta_0,\cdots,\beta_n)]$ is finite (in fact, it is less than or equal to $n$.) By applying the previous argument repeatedly, we can also see that $[\mathbb{Q}(\beta_0,\cdots,\beta_n):\mathbb{Q}]$ is also finite. Therefore, $[\mathbb{Q}(\alpha,\beta_0,\cdots,\beta_n):\mathbb{Q}]$ is finite.

Especially, we know that $\mathbb{Q}(\alpha)$ is a subfield of $\mathbb{Q}(\alpha,\beta_0,\cdots,\beta_n)$, so $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is also finite. This proves $\alpha$ is algebraic.

Therefore, we can see that the field of algebraic numbers is algebraically closed.

What can be inferred from this? Observe that every algebraically closed field $K$ containing $\mathbb{Q}$ also contains a copy of the field of algebraic numbers. (Just collect all solutions of equation with rational coefficients in $F$.)

Therefore, we can conclude that the field of all algebraic numbers is the least algebraically closed field containing $\mathbb{Q}$!. Hence the field of algebraic numbers is the algebraic closure of $\mathbb{Q}$.

this wouldn't apply to any field

Not so! The following is actually true:

Theorem Let $F$ be any field. Then there exists a field $K$ containing $F$ such that:

1. $K$ is algebraically closed,
2. Every element of $K$ is algebraic over $F$.

Moreover, $K$ is unique up to isomorphism; it is called the algebraic closure of $F$.

But this is besides the point. What you need to show is the following:

Claim Let $S$ be the set of complex numbers which are algebraic over $\mathbb{Q}$. Then $S$ is closed under multiplication.

Here's one strategy you can try:

1. Take two elements $s_1, s_2 \in S$. We aim to show that $s_1 s_2 \in S$.

2. Find a polynomial $p \in \mathbb{Q}[x]$ which has both $s_1$ and $s_2$ as roots.

3. Let $K$ be the subfield of $\mathbb{C}$ generated by $\mathbb{Q}$ and the roots of $p$; prove that $K$ is finite-dimensional as a $\mathbb{Q}$-vector space.

4. Prove that every subfield of $\mathbb{C}$ which is finite-dimensional as a $\mathbb{Q}$-vector space is a subset of $S$.

5. Since $s_1 s_2 \in K$, conclude that $s_1 s_2 \in S$.