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Well, my question is essentially:

Let $R$ be a Factorial Ring (UFD, basically) and let $p$ be a prime element in $R$. Let $d$ be an integer larger than 2, and let

$f(t) = t^d + c_{d-1} t^{d-1} + ... + c_0$ be a polynomial belonging to $R[t]$. Let $n \ge 1$ be an integer.

Show that $g(t) = f(t) + \displaystyle \frac{p}{p^{nd}}$ is irreducible in $K[t]$, where $K$ is the quotient field of $R$.

Okay, well, I know that it is enough, from Gauss' Lemma, to show that (after multiplying through by $p^{nd}$) $g$ has no non-trivial factorizations in $R$ itself. EDIT: My apologies, I forgot that this holds specifically for primitive polynomials, and no such specification was given. Nevertheless, perhaps, I suppose the gcd of the coefficients could be factored out?

I also tried thinking of $g$ as essentially just a translation of the graph of $f$ slightly upward, and that $g$ converges to $f$ as $n \to \infty$, but I don't know exactly what to make of that, because, well, as $n$ gets larger, it starts to inch towards the roots of $f$, and would no longer be irreducible, but perhaps I've confused myself.

(As an auxilliary question, there was a previous exercise which asked us to prove that if $R = \mathbb{Z},$ then, if $f$ had $m$ real roots, then, "for sufficiently large $n$", $g$ also has $m$ EDIT: real roots. This should be fairly apparent, I should think, from the graph? But if not, then how would one rigorously establish this?)

Anyway, thanks a lot.

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As $p$ is a unit in $K$, the irreducibility of $g(t)$ is equivalent to the irreduciblity of $h(t):=p^{nd}g(t)$. Write $$h(t)=(p^nt)^{d}+c_{d-1}p^n (p^nt)^{d-1}+\cdots + c_1p^{(d-1)n}(p^nt)+p(1+p^{nd-1}c_0).$$ Now apply Eisenstein criterion to the polynomial $$ H(T)=T^d+c_{d-1}p^n T^{d-1}+\cdots + c_1p^{(d-1)n}T+p(1+p^{nd-1}c_0)$$ and Gauss lemma to conclude that $H(T)\in K[T]$ is irreducible. As $h(t)$ is obtained from $H(T)$ by a change of variables, it is also irreducible.

Edit Forgot the auxillary question. It is false: consider $f(t)=t^2$ and $p$ a prime number.

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  • $\begingroup$ Lovely. How did I not think of that!? $\endgroup$ – AlpArslan Sep 10 '13 at 14:55
  • $\begingroup$ @MWarsi: the second question has a positive answer for simple real roots. $\endgroup$ – Cantlog Sep 10 '13 at 15:53

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