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If $R$ is a UFD and $f \in R[x]$, then we can't always use Eisenstein's Rule directly to show that $f$ is irreducible in Frac$(R)[x]$. However if we have $\phi$, an automorphism of Frac$(R)[x]$, then $f$ is irreducible if and only if $\phi(f)$ is irreducible. In order to apply Eisenstein to the resulting polynomial, you need $\phi(f) \in R[x]$, then you can deduce that your original $f$ was irreducible (we are using a 'shifted' version of the rule).

For example, for the cyclotomic polynomials $x^{p-1} + \dots + x + 1$, you cannot use Eisenstein directly, however if you consider $\phi : \mathbb{Q}[x] \rightarrow \mathbb{Q}[x]$ which sends $x$ to $x+1$ and does nothing on $\mathbb{Q}$, then using the binomial the resulting polynomial is Eisenstein at $p$.

What I am wondering is given an irreducible polynomial, can you always find an automorphism such that you can apply Eisenstein to show it's irreducible?
If not in general then what about the case where $R = \mathbb{Z}$ and $\phi$ is an automorphism of $\mathbb{Q}[x]$?

user26857
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James
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1 Answers1

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There is a family of examples I like very much.

Assume that $a,b\in\mathbb{Z}\setminus\{0\}$ are chosen in such a way that $a,b,ab$ are not squares. Then the polynomial $f_{a,b}=X^2-2(a+b)X^2+(a-b)^2$ is irreducible in $\mathbb{Q}[X]$, but you can show that for any prime $p$, and for all $u\in \mathbb{Z}$, $ f_{a,b}(X+u)$ is not $p$-Eisenstein, and is reducible modulo $p$.

Then the usual criteria for irreducibilty do not apply.

user26857
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GreginGre
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  • What about $f_{a,b}(wX + u)$ for $w,u \in \mathbb{Z}$ ? If you can prove this is irreducible then the first will be too. – James Oct 19 '19 at 11:13
  • it is not p-Eisenstein either for all $p\nmid w$ (which is necessary anayway to a $p$-Eisenstein) Same for irreducibility mod $p$. – GreginGre Oct 19 '19 at 17:19