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$F$ - algebraically closed field. $k$ is a subfield of $F$. The dimension of $F$ over $k$ is a finite number $n$.

GENERAL QUESTION: What power can have then irreducible polynomials over $k$?

My attempts to solve this task:

1) Let's consider the irreducible polynomial $f$ (which has a solution $l$ in $F$).

2) Look at $k[l]$ - subring (and it is the same field), that generate all elements of $k$ and $l$ - one solution of $f$ in $F$.

3) I'll take a polynom $g$ and divide with remainder it into my polinom $f$.

$g=fq+r$

$g(l)=f(l)q(l)+r(l)=r(l)$

QUESTION: if i take information about this subring $k[l]$ I'll know about powers of my polynoms in $k$.

Can anybody help me with an idea, please? Thank you in advance!

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    $\begingroup$ Hint: Artin-Schreier Theorem. If $k\ne F$, the answer will be $1$ or $2$. $\endgroup$ – André Nicolas Dec 26 '13 at 2:10
  • $\begingroup$ Note that the title asks a much different Question than turns out to be asked in the body of the post, because the body posits an algebraically closed field $F$ of finite dimension over $k$. This is then not a "general question" but a very narrow situation. We speak of the degree of an irreducible polynomial, but the meaning of its power is not immediately clear (to me). $\endgroup$ – hardmath Dec 26 '13 at 2:17
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    $\begingroup$ @AndréNicolas +1 for making sense of the question., $\endgroup$ – Igor Rivin Dec 26 '13 at 2:17
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Your problem is easily settled using the Artin-Schreier Theorem. And nothing less will do. For an accessible proof, please see these notes by Conrad.

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  • $\begingroup$ can you tell in detail how you got degree polynomials, please? I don't see the logic. $\endgroup$ – user44647 Dec 26 '13 at 8:19
  • $\begingroup$ On page 2 of the linked notes, you will find the statement of the theorem. It says (with different symbols) that if $F$ is algebraically closed, and $F$ has finite dimension over $k$, then our fields have characteristic $0$, and $F=k(i)$, where $i^2=-1$. In particular, $F$ is of degree $2$ over $k$. Thus if $P$ is any polynomial irreducible over $k$, then the degree of $P$ divides $2$. Thus the degree of $P$ is $1$ or $2$. If $k\ne F$, then $2$ is possible, just use $x^2+1$. $\endgroup$ – André Nicolas Dec 26 '13 at 13:51

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