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Is it true that if $x_1, \dots, x_n \in \Bbb C$ are algebraically independent over $\mathbb Q$, then they are algebraically independent over $\overline{\mathbb Q}$?

I know how to prove it for $n=1$, by the contrapositive. If $P(x)=0$ and $P \neq 0$ has coefficients $a_0,...,a_n$ in $\overline{\mathbb Q}$, then $x$ is algebraic over $\mathbb Q(a_0,...,a_n)$ which is itself algebraic over $\mathbb Q$, therefore $x$ is algebraic over $\mathbb Q$, so it satisfies some relation $Q(x)=0$ where $Q \neq 0$ is a polynomial with rational coefficients.

For $n \geq 2$, I tried induction, but I was not sure how to proceed.

Thank you for your help!

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    The same question was asked [here](http://math.stackexchange.com/questions/1926799/equivalent-criterion-for-algebraic-independence-over-mathbbq?rq=1). The answer is given in the comments. – 2017-02-02
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    Do you know some Galois Theory? Denote $(X_1 , \dots , X_n) = \mathbf{X}$. Then, if $f(\mathbf{X}) \in \bar{\Bbb Q}[\mathbf{X}]$ is a nonzero polynomial vanishing at $x_1, \dots , x_n$, then you can find a suitable $F$ Galois extension of $\Bbb{Q}$ such that $f(\mathbf{X}) \in F[\mathbf{X}]$. Then $$\prod_{\sigma \in \mathrm{Gal} (F / \Bbb{Q})} \sigma f(\mathbf{X} ) \in \Bbb{Q}[\mathbf{X}]$$ is a nonzero polynomial vanishing at $x_1, \dots , x_n$. – 2017-02-02

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Induction seems to work here. Sketch of the proof:

  • You covered the base case $n=1$ in your question
  • To proceed, note that $\{x_1,\dots,x_n\}$ algebraically independent over $\overline{\mathbb{Q}}$ is equivalent with proving that $\{x_1\dots,x_{n-1}\}$ are algebraically independent over $\overline{\mathbb{Q}}$ and $x_n$ transcendental over $\overline{\mathbb{Q}}(x_1,\dots,x_{n-1})$. By the induction hypothesis we only have to prove this last assertion.

If $x_n$ were algebraic over $\overline{\mathbb{Q}}(x_1,\dots,x_{n-1})$, then in fact $x_n$ would be algebraic over $\mathbb{Q}(x_1,\dots,x_{n-1},c_0,\dots,c_m)$ where $c_0,\dots,c_m$ are the coefficients of the polynomial relation. Since $\mathbb{Q}(x_1,\dots,x_{n-1},c_0,\dots,c_m)$ is a finite (thus algebraic) extension of $\mathbb{Q}(x_1,\dots,x_{n-1})$, we see that $x_n$ would be algebraic over $\mathbb{Q}(x_1,\dots,x_{n-1})$, which is a contradiction.

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    Thank you for your answer! I have a question: why would $x_n$ be algebraic over $\mathbb{Q}(x_1,\dots,x_{n-1},c_0,\dots,c_m)$? Even if $c_i \in \overline{\mathbb Q}$, for me we only have that $x_n$ is algebraic over $\overline{\mathbb{Q}}(x_1,\dots,x_{n-1},c_0,\dots,c_m)$. – 2017-02-02