Why this polynomial is irreducible? [closed]

Let $K=\mathbb{Z}_p(t)$. How to prove $f(x)=x^p-t$ is irreducible in $K[x]$?

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closed as off-topic by user26857, JKnecht, Edward Jiang, Jon Mark Perry, Daniel W. FarlowApr 28 at 1:50

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Did you mean to write $\mathbb Z_p[t]$? If not, could you define the construct you did indicate? – Lubin Feb 5 '13 at 19:28
@Lubin $K=\mathbb F_p(t)$ is the field of rational functions in one variable over $\mathbb F_p$, the finite field with $p$ elements and $K[x]$ is the ring of polynomials in one variable over $K$. What's unclear? – JSchlather Feb 5 '13 at 19:44

$\mathbb Z_p$ is a UFD ring, therefore so is $A = \mathbb Z_p[t]$. The polynomial $f(x) = x^p - t$ is monic so, because of Gauß' lemma, $f$ is reducible in $A$ iff it is reducible in $\mathrm{Frac}\, A = K[t]$.

But now, we can apply the generalisation of Eisenstein's criterion. We have a prime ideal $\mathfrak p =(x) \in \mathbb{Z}_t[x]$ which contains every coefficient of $f$ except the dominant one, and the constant coefficient $t$ doesn't belong to $\mathfrak p^2$. Therefore, $f$ is irreducible in $A$.

Remark. If $p$ is a prime, on any field $F$, $f = x^p -a$ is irreducible in $F[x]$ unless it has a root in $F$. This is proved in Cox' Galois Theory, section 4.2.D.

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Very clear, thank you! – hxhxhx88 Feb 6 '13 at 1:18

Hint: Use Eisenstein's criterion by recalling that $t$ is prime in $\mathbb Z_p[t]$.

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I understand now, thank you! – hxhxhx88 Feb 6 '13 at 1:19