Irreducibility of polynomials over field of rational functions

Question

I was reading Dummit and Foote. On page 551 of the book, there is an example mentioned about irreducibile inseparable polynomials. I am facing difficulty in proving the irreducibility of polynomials.

How I can show this : - The polynomial $p(x) = x^{2^m} - t$ over $F = \mathbb{F}_2(t)$ is irreducible. For small $m$ like 2 or 4 one can enumerate cases and get done but what about a general m?

My first approach is a simple-minded cousin of this argument, possibly also seen many times on the site. Therefore I switched to CW. No new content here, really. — Jyrki Lahtonen, Jun 08 '19 at 06:10

Jyrki Lahtonen · Answer 1 · 2019-06-09T04:40:19.157

1

At least two simple ways of getting there:

If $u$ is a zero of your polynomial in an extension field $K$ of $F$ then $u^{2^m}=t$ and, by Freshman's dream, $$p(x)=(x-u)^{2^m}=x^{2^m}-u^{2^m}=x^{2^m}-t.$$ So, by uniqueness of polynomial factorization, any eventual factor of $p(x)$ in $F[x]$ must be of the form $(x-u)^i$ for some $i,0<i<2^m$. But the constant term $u^i$ is in $F$ only when $i=0$ or $i=2^m$, and we are done.
You can apply Eisenstein's criterion to the prime $t$, irreducible in $\Bbb{F}_2[t]$. All the coefficients of $p(x)$, save the leading $1$, are divisible by $t$, and the constant coefficient is not divisible by $t^2$. Therefore Eisenstein applies.

edited Jun 09 '19 at 04:40

answered Jun 08 '19 at 05:44

Jyrki Lahtonen

133,153

Thanks for your answer. However, in the first approach, $(x-u)^{2^m} = x^{2^m} + (- u)^{2^m} \neq x^{2^m} - u$ as $u$ is a formal variable not an element of $\mathbb{F}_{2^m}$ – user031197 Jun 08 '19 at 06:16
Not so, @ArihantJain. $u$ is a zero of $p(x)$, so $u^{2^m}=t$. – Jyrki Lahtonen Jun 08 '19 at 10:20
Or rather, I never claimed it to be $x^{2^m}-u$. I claim $(x-u)^{2^m}=x^{2^m}-t$. – Jyrki Lahtonen Jun 08 '19 at 19:46
@ArihantJain $(a+b)^{2^m} = a^{2^m}+b^{2^m}$ holds in any commutative ring where $2=0$ – reuns Jun 09 '19 at 04:49
@JyrkiLahtonen Oh! I am sorry. Yes that makes sense. Thank you for your answer, it was very helpful. – user031197 Jun 09 '19 at 08:08
Glad that we ended up on the same page, @ArihantJain. No worries. – Jyrki Lahtonen Jun 09 '19 at 10:09

Irreducibility of polynomials over field of rational functions

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