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## Produce an irreducible polynomial that can’t be proved irreducible by using Eisenstein [closed]

give An example of an irreducible polynomial that cannot prove it by using the Eisenstein criterion even with the use of all linear change variable($x-c=y$).

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mathoverflow.net/faq#whatnot – Martin Brandenburg Nov 8 2011 at 11:49
It would be nice if you didn't formulate this as a command ("give an example...") and explained why you are asking (idle curiosity, homework,...). Your question isn't at the intended level of MO, but I'll make a comment which I think is: if $K$ is a number field in which (1) the ring of integers has the form ${\mathbf Z}[\alpha]$ and (2) no prime number is totally ramified, then the minimal polynomial of $\alpha$ over ${\mathbf Q}$ has the feature you seek. Many cyclotomic extensions of ${\mathbf Q}$ fits these properties. – KConrad Nov 8 2011 at 11:59
Newton polygon scenarios systematically give examples just-slightly-more-complicated than Eisenstein-criterion examples. E.g., $x^{n+1}+2x+4$: the slopes are $1/n$ $n$ times and a single $1$. Thus, this has at least an irreducible degree-$n$ factor. Excluding a rational root is easy (not $\pm 1,\pm 2\pm 4$), so it's irreducible. – paul garrett Nov 8 2011 at 16:40

## closed as too localized by Felipe Voloch, Mark Sapir, Martin Brandenburg, Bruce Westbury, Torsten EkedahlNov 8 2011 at 12:39

$x^2+8$ is an example.

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It is possible to produce a polynomial that cannot (provably) be proved to be irreducible considering the valuations of the roots, or even any polynomial function in the roots (which can be much more general than a linear substitution), for every possible valuation over the base field.

Let $L/K$ be an unramifed extension of numeber fields (for instance the Hilbert class field of a $K$ with non-trivial class group), generated by $\alpha$ say. Then the minimal polynomial for $\alpha$ over $K$ will do.

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