Let $K$ be a field, say it is of type $n\in\mathbb{N}$ if all its irreducible polynomials are of degree $\leq n$ and if there is an irreducible polynomial of degree $n$. Say $K$ is of infinite type if it has irreducible polynomials of all degrees.

There are easy examples of fields of type $1$ ($\mathbb{C}$ and more generally algebraically closed fields), $2$ ($\mathbb{R}$), and of infinite type ($\mathbb{Q}$, finite fields, etc.)

But I couldn't come up with a field of type $n >2$, so I was wondering if there was any.

There isn't any perfect field of (finite) type $n>2$, since if $K$ is perfect and of type $n$, then the primitive element theorem proves that any algebraic extension of $K$ is of degree $\leq n$, in particular the algebraic closure $L$ of $K$ has $[L:K]< \infty$, and (see this ) this implies $K$ is of type at most $2$.

So in conclusion my question is : is it known whether there are (imperfect) fields of type $n>2$, if so what are some examples, or what is a proof that there are none ?

EDIT: while looking at stuff about the Artin-Schreier theorem mentioned in the comments, I found the following lemma : if $K$ is a field of characteristic $p>0$ and if $a\in K\setminus K^p$, then for all $m\geq 1$, $X^{p^m}-a$ is irreducible over $K$, which proves that a field that isn't perfect is of infinite type, which concludes the question

  • $\begingroup$ Try using Artin Schreier's theorem $\endgroup$ – user379195 May 9 '17 at 14:30
  • $\begingroup$ But how do you show that $[L:K]<\infty$ when $K$ is not perfect ? $\endgroup$ – Max May 9 '17 at 14:31
  • $\begingroup$ Why don't you assume the characteristic to be 0 at first ? $\endgroup$ – user379195 May 9 '17 at 14:32
  • $\begingroup$ Well I have already dealt with the case where the characteristic is $0$. A field of characteristic $0$ is perfect $\endgroup$ – Max May 9 '17 at 14:34
  • $\begingroup$ Try working with the perfect closure $\endgroup$ – user379195 May 9 '17 at 14:35

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