Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, JKnecht, Edward Jiang, Jon Mark Perry, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

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
up vote 2 down vote accepted

$\mathbb Z_p$ 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.

share|cite|improve this answer
Very clear, thank you! – hxhxhx88 Feb 6 '13 at 1:18

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

share|cite|improve this answer
I understand now, thank you! – hxhxhx88 Feb 6 '13 at 1:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.