10
$\begingroup$

Let $K$ be a field and $x$ be trnacendental over $K$. Compute $[K(x):K(\frac{x^5}{1+x})]$.

I've never came across questions like these. It's easy to see that this degree is at most $5$, since: $$x^5-\alpha(x+1)=0$$ ($\alpha$ being $\frac{x^5}{1+x}$). But how can I show the other direction? Why can't there be any $f\in K(\alpha)[X]$ with $\deg f<5$ and $f(x)=0$? Or can there be?

Help would be appriciated. Thank you!

$\endgroup$
3
  • 1
    $\begingroup$ Small nitpick: be careful using $x$ both as the transcendental element over $K$ and as the polynomial variable. It is a recipe for confusion. $\endgroup$
    – Arthur
    Jul 22, 2017 at 9:51
  • 1
    $\begingroup$ Oops, right. I changed the polynomial cariable to capital X. $\endgroup$
    – 35T41
    Jul 22, 2017 at 10:39
  • $\begingroup$ Well, $\alpha$ is certainly transcendental over $K$, so that $K[\alpha]$ is a polynomial ring, so Principal Ideal Domain. The polynomial $X^5-\alpha X-\alpha$ is Eisenstein over the PID $K[\alpha]$, thus irreducible. I think that suffices. $\endgroup$
    – Lubin
    Jul 22, 2017 at 13:47

2 Answers 2

4
$\begingroup$

We only need to show that the polynomial $x^5-\alpha x - \alpha$ is irreducible over $K(\alpha)[x]$. Note that the polynomial is primitive, thus it suffices to show it is irreducible over $K[\alpha][x]$.

We use the idea of Eisenstein criterion, let $\varphi:K[\alpha]\to K$ be the evaluation map sending $\alpha \mapsto 0$. Extend this homomorphism to get $\pi: K[\alpha][x]\to K[x]$. If $x^5-\alpha x -\alpha = fg$, with $f,g \in K[\alpha][x]$, $f,g$ non-unit. Then $$\pi(f)\pi(g) = x^5$$ Note that $\ker \pi = (\alpha)$, the ideal generated by $\alpha$.

Since $K[x]$ is a UFD, $f = x^r + \alpha f_0, g = x^{5-r} + \alpha g_0$, where $f_0,g_0\in K[\alpha][x]$. But this would imply that the constant term of $fg$ is a multiple of $\alpha^2$. Hence $x^5-\alpha x -\alpha$ is irreducible over $K[\alpha][x]$.

$\endgroup$
2
  • $\begingroup$ Thanks a nice extension of these terms that I know from the $\mathbb{Q}$ and $\mathbb{Z}$ world to these rather abstract fields. Thank you for that! $\endgroup$
    – 35T41
    Jul 22, 2017 at 13:48
  • $\begingroup$ Very nice, I didn't think of Eisenstein criterion, +1. Though, you could have just applied it directly since $(\alpha)$ is prime ideal (you basically wrote a proof of Eisenstein criterion). $\endgroup$
    – Ennar
    Jul 22, 2017 at 14:48
1
$\begingroup$

What you have to do is show that $f(X)=X^5-\alpha(X+1)$ is irreducible over $K(\alpha)$ and you are done.

This is because we would have $[K(\alpha)(x):K(x)][K(x):K(\alpha)] = [K(\alpha)(x):K(\alpha)] = 5$ and $K(x)\neq K(\alpha)$ would imply $[K(x):K(\alpha)]=5$ by $5$ being prime.

Now, $f$ is primitive polynomial in $K[\alpha][X]$, so by Gauss's lemma, it is irreducible over $K(\alpha)$ if and only if it is irreducible over $K[\alpha]$.

First of all, $f$ can't have a root in $K[\alpha]$ since it would mean that $$p(\alpha)^5-\alpha(p(\alpha)+1)=0$$ and that $\alpha$ is algebraic over $K$.

Thus, if $f$ weren't irreducible, it would have to allow factorization of the form

$$f(X) = (X^3 +a(\alpha)X^2+b(\alpha)X+c(\alpha))(X^2+d(\alpha)X+e(\alpha))$$ which would lead to system

\begin{align} a(\alpha)+d(\alpha)&=0\\ b(\alpha)+a(\alpha)d(\alpha)+e(\alpha)&= 0\\ c(\alpha)+b(\alpha)d(\alpha)+a(\alpha)e(\alpha) &= 0\\ c(\alpha)d(\alpha)+b(\alpha)e(\alpha)&= -\alpha\\ c(\alpha)e(\alpha)&=-\alpha \end{align}

and now you can use that $K[X]\cong K[\alpha]$ since $\alpha$ is transcedental and degree arguments to show that the above system is impossible to solve in $K[X]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .