2
$\begingroup$

Suppose K is a field with characteristic p which is not a perfect field.

Then how do we prove that there does exist an irreducible polynomial which is not separable.

I have no idea how to proceed for this.what all i know is if K is a finite field then every irreducible polynomial is separable.

any hint/suggestion would be appreciated :)

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ If $K=F_p(t)$, then $x^p-t$ is irreducible and not separable. $\endgroup$ – Gerry Myerson Jul 29 '13 at 13:30
  • 1
    $\begingroup$ Since you say in your third paragraph that such a polynomial does not exist if $K$ is finite, I assume you left out a condition on $K$. $\endgroup$ – Hagen von Eitzen Jul 29 '13 at 13:34
  • $\begingroup$ @HagenvonEitzen : edited :) thanks :) $\endgroup$ – user87543 Jul 29 '13 at 13:38
  • 1
    $\begingroup$ Dear Praphulla, I'm not sure if this is what you meant to imply, but it's not true that for every infinite field $K$ of char. $p$, there is an inseparable irreducible polynomial in $K[x]$. In general a field is called perfect if every irred. polynomial in $K[x]$ is separable. As you note, all finite fields are perfect, but some infinite fields of char. $p$ are perfect too (e.g. all algebraically closed fields). The field $K = \mathbb F_p(t)$ is the simplest example of a non-perfect field. Regards, $\endgroup$ – Matt E Jul 29 '13 at 13:50
  • 3
    $\begingroup$ So you are trying to prove that, if $K\not=K^p$, then $K$ is not perfect? If so, then pick $t\in K-K^p$, and consider the polynomial $x^p-t$. If you are asking why we think of this, I am afraid that I cannot help then... $\endgroup$ – awllower Jul 29 '13 at 14:21
3
$\begingroup$

As mentioned by Gerry Myerson let $K=F_p(t)$ and the sub-field $k=F(t^p)$ and finaly let the polynomial $$P=x^p-t^p\in K[x]$$ so in $K[x]$ we have $P=(x-t)^p$ and since $t^i\not\in k,\ i=1,\ldots,p-1$ then $P$ is irreductible in $k[x]$ but it's not separable as $t$ is its root with multiplicity $p$ in $K$.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy