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I have read the next beautiful result about Galois group of polynomials.

Theorem: Let $f$ be an irreducible polynomial of prime degree $p\geqslant 5$ in $\mathbb{Q}[x]$. If $f$ has exactly two nonreal roots, then the Galois group $G_{f}=S_{p}$.

I want to know whether this theorem gives a necessary and sufficient condition? Can we replace the condition in it with some others to get the same result?

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Of course no. For example

$\textbf{Proposition}.$ There exists a polynomial $R\in \mathbb{Q}[x]$ of degree $p$, with only one real root, satisfying $gal(R)=S_p$.

$\textbf{Proof}$. We use the fact that the polynomials $Q$ with $gal(Q)=S_p$ are dense in $\mathbb{Q}[x]$.

Consider a monic polynomial $T=x^p+\sum_{i<p} a_ix^i$ with only one real root; randomly choose $(b_i)$ in a neighborhood of the $(a_i)$. Then, with probability $1$, $T_1=x^p+\sum_{i<p} b_ix^i$ has $S_p$ as Galois group. Moreover, $T_1$ has only one real root (by the continuity of the roots of a polynomial wrt. its coefficients, the non-real roots remain non-real).

Example. Let $T=x(x^2+1)^2=x^5+2x^3+x$ and

$T_1=x^5+(2022376873266539/1000000000000000)x^3+(1040578712917309/1000000000000000)x+1141656374703/100000000000000+(97561097757381/1000000000000000)x^2+(77370700889713/1000000000000000)x^4\approx$

$x^5+2.022376873x^3+1.040578713x+0.01141656375+0.09756109776x^2+0.07737070089x^4$.

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  • $\begingroup$ Do you have a reference for the stated fact? $\endgroup$ – Wojowu Mar 8 '19 at 9:46
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This is false already for $p=2$, just take any irreducible quadratic polynomial with two real roots.

If you dislike even primes, try some cubic polynomials - recall that for an irreducible cubic $f$, we have either $G_f=A_3$ or $S_3$, and the former happens iff the discriminant is a square.

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  • $\begingroup$ I have edited that $p\geqslant 5$. $\endgroup$ – user450201 Mar 8 '19 at 10:01

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