# Arbitrary number of algebraically independent elements of prime field: when is that possible?

If we consider the prime field $\mathbb{Q}$, and the field extension $\mathbb{Q}\subseteq\mathbb{C}$, then for any natural $n$ we can find $a_1,...,a_n\in\mathbb{C}$ algebraically independent over $\mathbb{Q}$. To prove this, we use induction on $n$, where the case $n=1$ is trivially true. Now if there are $a_1,...,a_{n-1}$ algebraically independent, we can (if I'm not mistaken) consider an algebraic closure of $\mathbb{Q}(a_1,...,a_{n-1})$, which would still be in $\mathbb{C}$, but also still countable, since $\mathbb{Q}$ and $\mathbb{Q}(a_1,...,a_{n-1})$ are. Hence there must be an $a_n$ as desired.

Is this proof correct so far?

I'm also interested in a more general statement about fields in arbitrary characteristic, something like:

Let $F$ be a prime field. Then for any natural $n$ we can find a field $K$ containing $F$ and elements $a_1,...,a_n\in K$ algebraically independent over $F$.

Unfortunately, I'm not very used to fields, and finite fields especially. I don't know if this statement is true anyway. Could you tell me? If it should be true, is there a way (in finite characteristic) to choose one $K$ for all $n$, and even maybe use a similar way of proving the statement?

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As for your second question: for any field $K$ the set $K(x_1,\ldots ,x_n)$ of (formal) rational functions in $n$ variables is a field containing $K$, and the elements $x_1,\ldots ,x_n$ are algebraically independent over $K$.
One can generalize the last statement: let $X$ be a set. Then the set $K(X)$ of (formal) rational functions in the variables $x\in X$ is a field containing $K$, and the elements $x\in X$ are algebraically independent over $K$.
Note that in a specific rational function only finitely many $x\in X$ actually appear.
Hello @Hagen! Seems like I was being a bit dumb there ;) I was thinking of the existence of some field actually containing "numbers" (in a sense), and didn't even think of just constructing it formally. Thank you very much! So for $\mathbb{F}_p$, I could use $\mathbb{F}_p(x_1,x_2,...)$ as a field satisfying the above proposition. –  InvisiblePanda Dec 6 '12 at 12:55