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An easy way to ensure that a polynomial $g$ of degree m $m$ over $\mathbf{Z}$ has Galois group $S_m$ is to take primes $p_1$, $p_2$ and $p_3$ with $f$ g$ irreducible modulo $p_1$, a linear times an irreducible modulo $p_2$ and a bunch of distinct linears times an irreducible quadratic modulo $p_3$. Then the Galois group must be doubly transitive and have a transposition, so it's $S_m$.

Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by close rationals won't create any real roots. So replace the $x^k$ coefficient of $f$ by a sufficient close rational $a_k/b_k$ where $a_k$ and $b_k$ are congruent modulo $p_1 p_2 p_3$ respectively to the $x^k$ coefficient of $g$ and to $1$. Then the new polynomial has rational coefficients, no real roots and Galois group $S_{2n}$. You can easily convert it to one with these properties and integer coefficients should you wish.
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An easy way to ensure that a polynomial $g$ of degree m over $\mathbf{Z}$ has Galois group $S_m$ is to take primes $p_1$, $p_2$ and $p_3$ with $f$ irreducible modulo $p_1$, a linear times an irreducible modulo $p_2$ and a bunch of distinct linears times an irreducble irreducible quadratic modulo $p_3$. Then the Galois group must be doubly transitive and have a transposition, so it's $S_m$.

Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by close rationals won't create any real roots. So replace the $x^k$ coefficient of $f$ by a sufficient close rational $a_k/b_k$ where $a_k$ and $b_k$ are congruent modulo $p_1 p_2 p_3$ respectively to the $x^k$ coefficient of $g$ and to $1$. Then the new polynomial has rational coefficients, no real roots and Galois group $S_{2n}$. You can easily convert it to one with these properties and integer coefficients should you wish.
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An easy way to ensure that a polynomial $g$ of degree m over $\mathbf{Z}$ has Galois group $S_m$ is to take primes $p_1$, $p_2$ and $p_3$ with $f$ irreducible modulo $p_1$, a linear times an irreducible modulo $p_2$ and a bunch of distinct linears times an irreducble quadratic modulo $p_3$. Then the Galois group must be doubly transitive and have a transposition, so it's $S_m$.

Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by close rationals won't create any real roots. So replace the $x^k$ coefficient of $f$ by a sufficient close rational $a_k/b_k$ where $a_k$ and $b_k$ are congruent modulo $p_1 p_2 p_3$ respectively to the $x^k$ coefficient of $g$ and to $1$. Then the new polynomial has rational coefficients, no real roots and Galois group $S_{2n}$. You can easily convert it to one with these properties and integer coefficients should you wish.