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Let $f\in \mathbb Z[X]$ be an irreducible polynomial. Suppose, the discriminant of $f$ is a perfect square.

Can the galois group of $f$ over $\mathbb Q$ be $S_d$, where $d$ denotes the degree of $f$ ?

Additional question : What can we conclude if the discrimiant is the negative of a perfect square, lets say $-81$ ?

Here

http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf

it is suggested to try to prove that the galois group is $A_n$, is the discriminant happens to be a perfect square. It is not mentioned directly, but I assume this is because it cannot be $S_n$ in this case.

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Consider the discriminant of $f$ defined as $\Delta= \prod_{i < j} (\alpha_i - \alpha_j)^2$, where $\alpha_i$ are some ordering of the roots of $f$. This is invariant under the full $S_d$ action so lies in $\mathbb{Q}$.

If this is a square, then that means $\delta = \prod_{i < j} (\alpha_i - \alpha_j) \in \mathbb{Q}$, so is fixed by the Galois action. However the largest subgroup that fixes this is $A_d$, hence the Galois group is contained in $A_d$ since otherwise we would have an odd permutation which would ma $\delta$ to $-\delta$ contradicting the fact that it is rational and therefore fixed.

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  • $\begingroup$ Can this argument also be used for the negative of a perfect square ? $\endgroup$ – Peter Feb 8 '16 at 22:30
  • $\begingroup$ @Peter the full result is that the Galois group is contained in $A_d$ if and only if the discriminant is a perfect square. So the only thing this tells you is that it is not contained in $A_d$ (but will be over $\mathbb{Q}(i)$ ). $\endgroup$ – Matt B Feb 8 '16 at 22:33

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