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Deleted the old question, because tho whole question kind of changed. I am facing following problem:

Given extension of finite fields $L/K$ of degree $3$, prove that every polynomial of degree $3$ with coefficients in $K$ does have root in $L$.

I know now the proof using some knowledge about splitting fields of irreducible polynomials. However, I am trying now to do somewhat more elementary and quite nonelegant proof and I would be very glad if someone would have look at this. I don't know whether a) what I have so far is correct b) even if it is, whether it leads anywhere...

edit: thanks to both of you guys, I finally got the exact number of irreducible polynomials in $K[x]$ right, which should be $(q^3 - q)/3$. Still working on how to use this number to finish the original problem.

note: the extension is finite, hence algebraic, so every element of $F$ is root of some polynomial in $K[x]$. So in particular, I have $q^3 - q$ "new" roots.Above mentioned $(q^3 - q)/3$ irreducible monic polynomials can have at most $q^3 - q$ roots in any extension. This "new" roots are exactly the roots of those polynomials, if I am not mistaken, but I don't know how to prove that.

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  • $\begingroup$ As Hurkyl pointed out there may be repetitions among $a,b,c$ in $(x+a)(x+b)(x+c)$. In those cases you shouldn't divide by six. $\endgroup$ – Jyrki Lahtonen Sep 22 '14 at 4:18
  • $\begingroup$ And for the main question. You don't need to settle for an upper bound when the exact number is easy to derive. A little bit of searching on our site points at here, here, here, here for example :-) $\endgroup$ – Jyrki Lahtonen Sep 22 '14 at 4:29
  • $\begingroup$ But +1 for the question and effort shown. This does lead to the correct formula after correcting for the undercounting above. $\endgroup$ – Jyrki Lahtonen Sep 22 '14 at 4:32
  • $\begingroup$ thanks you very much, you are being incredibly helpful, I see what I was doing wrong. Now I got exactly $(q^3 - q)/3$ irreducible polynomials in $K[x]$. Still not sure, how to use this to finish the original problem. $\endgroup$ – rizzu Sep 22 '14 at 5:11
  • $\begingroup$ I made one more edit. What I don't see is why each element of $L$ that is not in $K$ has a minimal polynomial of degree 3. If I could, it seems pretty straightforward after that. $\endgroup$ – rizzu Sep 22 '14 at 5:21
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Each element of $L$ that isn't in $K$ has a minimal polynomial of degree $3$. At most three of them can share the same minimal polynomial.

You may wish to count more accurately: e.g. you're only counting $x^3$ as one sixth of a polynomial.

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  • $\begingroup$ I am not sure I'm following you. When I was counting, lets say polynomials of degree 3 that split into linear factors $(x-a)(x-b)(x-c)$ that gives $q^3$ possibilities for $a,b,c$ but I divided that by $6$, so that I would count every split just once (iow not counting linear splits that differ only in order of factors more than once). But probably you were saying something else, and I just misunderstood. $\endgroup$ – rizzu Sep 22 '14 at 1:59
  • $\begingroup$ @rizzu: Only one possibility, not six, gives $x^3$: namely, $(x-0)(x-0)(x-0)$. $\endgroup$ – Hurkyl Sep 22 '14 at 11:31

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