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I'm trying to show that the polynomial $f(x) = \frac{x^5}{32}-3x-2$ is irreducible over $\mathbb Q$.

Obviously $f$ doesn't have a root over $\mathbb Q$ so I tried to use Gauss lemma for $32f$ and that led to nothing, also trying to use $x+a$ for several integers $a$ instead of $x$ in $f$ wasn't useful.

It seems like it would be the best to show that $f(2x)$ is irreducible (which is easily obtained by Eisenstein criteria), but it seems like subtituting $x$ with $2x$ will hurt the irreducibility (for instance consider $x\in \mathbb Z [x]$ is irreducible but $2x = (2)(x)$).

So what else can I try?

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If $f(x)$ is reducible - say $f(x) = g(x)h(x)$, then $f(2x) = g(2x)h(2x)$ is reducible. So here it is fine to work with $f(2x)$.

Your counterexample doesn't work, as Gauss' Lemma - that irreducibility over $\mathbb Q$ is equivalent to irreducibility over $\mathbb Z$ - only works for monic polynomials. $2x$ is indeed irreducible over $\mathbb Q$.

And $f(2x) = x^5 - 6x - 2 $ is irreducible by Eisenstein's criterion.


In general, $f(x)$ is irreducible, if and only if $f(ax+b)$ is irreducible by the same argument as above:$$f(x) = g(x)h(x) \implies f(ax + b) = g(ax + b)h(ax + b)\\f(ax + b) = g(ax + b)h(ax + b)\implies f(x) = f\left(\frac1a(ax+b)-\frac ba\right) = g(x)h(x)$$

This will, however, not work over $\mathbb Z$ if $a \ne \pm1,\ b\not\in\mathbb Z$, as the resulting polynomial will no longer be monic and/or in $\mathbb Z[X]$.

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  • $\begingroup$ are there any guildlines for when its okay or not to use those subtitutions? $\endgroup$ – Snufsan Jul 22 '14 at 11:59
  • $\begingroup$ I guess when you can write something similar to the first line - $f(x)$ is reducible $\implies$ $f(2x)$ reducible. Working over $\mathbb Q$ this would normally be fine... and you'd only really want to do this if you were working over $\mathbb Q$ anyway. $\endgroup$ – Mathmo123 Jul 22 '14 at 12:02
  • $\begingroup$ so over $\mathbb Q$ any linear subtitute $x\rightarrow ax+b$ will work? $\endgroup$ – Snufsan Jul 22 '14 at 12:04

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