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Is it possible for a polynomial, $x^p -x-c$ where $p$ is prime, to be reducible in a field of characteristic $0$, yet have roots in that field?

I know for a fact that the general form is true, that a polynomial of degree 2 or 3 in $F[x]$ is irreducible if and only if it has no roots in $F$. Thus, I am looking at $p=5$ as the simplest counterexample.

user26857
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Harry Evans
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    You can't be irreducible and have a root unless you are of degree $1$. – Captain Lama Apr 13 '16 at 18:07
  • @CaptainLama, this is related to finding a counterexample to the problem posted here: http://math.stackexchange.com/questions/1740770/if-xp%E2%88%92x%E2%88%92c-is-irreducible-in-fx-then-it-has-no-root-in-the-field – Harry Evans Apr 13 '16 at 18:08
  • Irreducible polynomials have no roots (unless those of degree $1$)! – Crostul Apr 13 '16 at 18:09
  • @Crostul $(x^2 + 1)^2 \in \mathbb Q[x]$ has no roots in $\mathbb Q$, but may be reduced to $(x^2+1)(x^2+1)$. – Harry Evans Apr 13 '16 at 18:13
  • Yes, but this shows the converse : if $P$ has a root, it is reducible, but if it is reducible it doesn't necessarily have a root. – Captain Lama Apr 13 '16 at 18:15
  • @HarryEvans I was answering to your very first question: "is it possible that an irreducible polynomial [...] has a root"? The answer is no. – Crostul Apr 13 '16 at 18:15
  • And in the question you linked people have already explained you that. – Captain Lama Apr 13 '16 at 18:16
  • On the other hand, it is a priori possible that $x^p - x-c$ is reducible but has no roots, even though I don't have any specific example. – Captain Lama Apr 13 '16 at 18:19
  • @CaptainLama, I was actually leaning towards that. Is there a polynomial of the form $x^p-x-c \in\mathbb Q[x]$ that can be reduced, yet not have roots in $\mathbb Q$? – Harry Evans Apr 13 '16 at 18:23
  • I guess the correct method would be to compute the Galois group of $x^p - x - t$ over $\mathbb{Q}(t)$ and see if it contains a permutation without fixed points (given a standard embedding $G\to S_p$). Then conclude with the fact that $\mathbb{Q}$ is Hilbertian or something. – Captain Lama Apr 13 '16 at 18:34
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    I found one: $x^5-x+15=(x^2+x+3) (x^3-x^2-2 x+5) \in \mathbb Q$ but one root is not rational, while the other four roots are complex. – Harry Evans Apr 13 '16 at 18:49

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