Suppose that $f$ is a polynomial with coefficients in the function field $\mathbb Q(t)$ such that $f$ is irreducible over $\mathbb C(t)$. Is the Galois group of $f$ over $\mathbb C(t)$ isomorphic to its Galois group over $\mathbb Q(t)$?
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This is not true in general. For example, consider the polynomial $x^n - t$. The roots of this over $\mathbb{Q}(t)$ are $\zeta_n^i t^{1/n}$ where $\zeta_n$ is a primitive $n$-th root of unity. $\mathbb{C}(t)(t^{1/n})$ contains all of these roots and is thus the splitting field of $x^n - t$ over $\mathbb{C}(t)$. However, $\mathbb{Q}(t)(t^{1/n})$ only contains one of the roots, so the splitting field over $\mathbb{Q}(t)$ is strictly larger. Since the splitting fields have different degrees the Galois groups can't be the same.
