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How can I prove (if it's true) that $X^n - t$ is irreducible over $k(t)$, the field of fractions of $k$ ?

Arnaud
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2 Answers2

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You can use Eisenstein's criterion. By Gauss' lemma, $X^n - t$ is irreducible over $k(t)$ iff it's irreducible over $k[t]$. Now, consider $\mathfrak{p} = (t)$. It is prime, and $t \not \in \mathfrak{p}^2$, so $X^n - t$ is irreducible over $k[t]$ by Eisenstein's criterion.

xyzzyz
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Viewing $X^n-t$ as a polynomial over $k[t]$, this is a primitive polynomial, because the $\gcd$ of the (non-zero) coefficients $1,-t$ is $1$. Therefore by a lemma of Gauss, it will be irreducible over $k(t)$ if and only if it is irreducible over $k[t]$. But over $k[t]$ the polynomial satisfies the Schönemann-Eisenstein criterion for the irreducible element $t$, hence it is irreducible.

Marc van Leeuwen
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  • Why is it necessary to check if the polynomial is primitive ? I thought that it was only necessary if we want to show that $P$ is irreducible over $k[t]$, knowing that it is over $k(t)$. – Arnaud Mar 31 '13 at 13:39
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    @Arnaud: That is right, but then I said "if and only if". The point is, if the polynomial hadn't been primitive, then it would stand no chance of being irreducible over $k[t]$, and it would be necessary to make it primitive before proceeding. – Marc van Leeuwen Mar 31 '13 at 14:40
  • @Arnaud: And actually there is a special case where a polynomial is irreducible over $k[t]$ while not being irreducible over $k(t)$, namely if it is a constant (w.r.t. $X$) polynomial (therefore invertible in $k(t)$) where the constant is irreducible as element of $k[t]$. That's not the case here of course, it just shows you cannot blindly deduce irreducibility from the other side in either direction. – Marc van Leeuwen Mar 31 '13 at 14:47