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This question is previously asked here, but there is no complete solution of it.

I understand that the root $\alpha$ exist in the algebraic closure of $\mathbb{F}_p(t)[x]$, and it is the only root because $f(X)=(X-\alpha)^p$, but how do we proceed to show that $\alpha\not\in\mathbb{F}_p(t)[x]$?

One solution I see writes $f(X)=g(X)h(X)$, and argue that since $g(X)=(X-\alpha)^i$ and $i<p$, then $\alpha\in \mathbb{F}_p(t)[x]$. How does this follow?

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2 Answers 2

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Eisenstein's criterion and Gauss's Lemma are useful here. By Gauss' Lemma, $X^p-t$ is irreducible in $\mathbb F_p(t)[X]$ if and only if it is irreducible in $\mathbb F_p[t][X]$. But the ideal $(t) \trianglelefteq \mathbb F_p[t]$ is prime ; this is proved by the fact that if the product of two polynomials in $t$ (with coefficients in $\mathbb F_p$) is divisible by $t$, then one of the polynomial factors has to be divisible by $t$ (hint : plug in $t=0$!). Knowing that $t$ is prime in $\mathbb F_p[t]$ allows us to apply Eisenstein's criterion.

Hope that helps,

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Because it only has one root, and none of them are in $\Bbb F_p(t)$. Recall for any root $\zeta$ we have $\zeta^p=t$, but then since $\Bbb F_p(t)[x]$ is a UFD, it means that $(x-\zeta)^p=x^p-t$ has just the one root. So if it is reducible, it reduces all the way, and in fact there is an element of $\Bbb F_p(t)$ such that $\left(\displaystyle{q(t)\over r(t)}\right)^p=t$.

But clearly this is impossible since then $r(t)^p = q(t)^p\cdot t$ and taking derivatives on both sides we get

$$0=pr(t)^{p-1}r'(t)=pq(t)^{p-1}\cdot q'(t)\cdot t + q(t)^p = q(t)^p$$

which would imply that $q(t) = 0$ which implies $t=0$.

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    $\begingroup$ How to make it clear that if $f$ is reducible, then $\zeta$ must be in $\Bbb F_p(t)$? $\endgroup$ Commented Apr 27, 2016 at 21:50
  • $\begingroup$ Use the Euclidean algorithm in an algebraic extension. $\endgroup$ Commented Apr 27, 2016 at 21:59

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