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Is a well known result that, for every $n \in \mathbb{N}$, there exist an irreducible polynomial $p \in \mathbb{Q}[x]$ such that the Galois Group of its splitting field is $S_n$.

Now my question:

Given a polynomial $g(x) \in \mathbb{Q}[x]$ of degree $n$ is true that there exist a rational number $q$ such that the Galois Group of $g(x)+q$ is $S_n$?

Edit: A more complicated question:

Let's fix $g(x) \in \mathbb{Q}[x]$ of degree $n$ and define $GAL(n)$ as the set of groups that can be realized as Galois Group of a polynomial of degree $n$.

Consider the map from $\mathbb{Q}$ to $GAL(n)$ that sends $q \in\mathbb{Q}$ to $Gal (g(x)+q)$.

Is this map surjective?

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    $\begingroup$ The answer to the first question (and hence also the second) is no, unless you assume something more about the polynomial (take $g(x) = x^n$ to see this). $\endgroup$ – Tobias Kildetoft Jun 26 '15 at 14:43
  • $\begingroup$ Do you think that could be more interesting if such polynomial is squarefree? $\endgroup$ – Sabino Di Trani Jun 26 '15 at 14:47
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    $\begingroup$ I am not sure what sort of condition on the polynomial will work (if any will). I think for degree $5$ the polynomial $x^5 + x$ works (but I might recall incorrectly. $\endgroup$ – Tobias Kildetoft Jun 26 '15 at 14:49
  • $\begingroup$ Maybe you should revert to the original here and ask the new version as a new question. It seems a shame that the answer now doesn't answer the question. $\endgroup$ – Tobias Kildetoft Jun 26 '15 at 15:03
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No. Take $g(x)=x^4$, and consider the polynomial $x^4+q$, where $q$ is a rational number.

If $q$ is negative, then the roots of $x^4+q$ in $\mathbb{C}$ are $\sqrt[4]{-q}, i\sqrt[4]{-q}, -\sqrt[4]{-q}$ and $-i\sqrt[4]{-q}$, where $\sqrt[4]{-q}$ is the positive real fourth root of $-q$. Then the decomposition field of $x^4+q$ is $$ \mathbb{Q}( \sqrt[4]{-q}, i\sqrt[4]{-q}) = \mathbb{Q}( \sqrt[4]{-q}, i)$$ which has degree at most $8$ over $\mathbb{Q}$. Thus the Galois group has order at most $8$, and cannot be $S_4$.

Now, if $q$ is positive, then take $\zeta = e^{i\pi/4}$, a primitive eight root of unity. Then the roots of $x^4+q$ are $\zeta \sqrt[4]{q}, \zeta^3 \sqrt[4]{q}, \zeta^5 \sqrt[4]{q}$ and $\zeta^7 \sqrt[4]{q}$. The field of decomposition is then $$ \mathbb{Q}(\zeta \sqrt[4]{q}, i),$$ which again is of degree at most $8$.

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  • $\begingroup$ I've just seen @TobiasKildetoft's comment above. This answer is just a computation of a special case of his remark. $\endgroup$ – Pierre-Guy Plamondon Jun 26 '15 at 14:48
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    $\begingroup$ Taking $g(x) = x^4+2$ instead, you get the same result, but this time $g(x)$ is both squarefree and irreducible. $\endgroup$ – Pierre-Guy Plamondon Jun 26 '15 at 14:55
  • $\begingroup$ Yes, you're right. I think the real interesting condition is $g'(x) \notin (x)$ $\endgroup$ – Sabino Di Trani Jun 26 '15 at 14:59
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    $\begingroup$ @JosephCurwen, your suggestion is good, but it’s specific to the point $x=0$ on the line, while the question is invariant under translations of the line. I think that when you smear your suggestion over the whole line, it becomes the hypothesis that $g$ has all roots simple. [ Whoops — not quite right, but I’ll leave what I said anyway. ] $\endgroup$ – Lubin Jun 26 '15 at 22:09

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