3 minor detail

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. 2 spelling correction 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. 1 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.