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What is the simplest radical expression of some root $a \in \overline{\mathbb{F}_5}$ of the polynomial $x^3+x+1 \in \mathbb{F}_5[x]$? I wonder if one can simplify the general formulas in this special case. I think $\sqrt[3]{\sqrt{2} - 3} - \sqrt[3]{\sqrt{2} + 3}$ is ok for the beginning, but this minus in the middle and the nested roots make it rather complicated.

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$\mathbb F_{5^3}$ is a cubic extension of $\mathbb F_5$, and the first thing to ask if it we can write it as $\mathbb F_5(\sqrt[3]a)$. Unfortunately, 5 is not congruent to 1 mod 3, so $x \mapsto x^3$ is a permutation of $F_5$, and the polynomials $X^3 - a$ are always reducible. In order to express $\alpha$ with radicals, you will need a cube root, so we have no choice but to go to $F_{5^2}$ first by taking a square root, then use a cube root to go to $F_{5^6}$, and then express $\alpha$ just like you did.

Now, rearranging your expression, we have $\alpha = \sqrt[3]{2 + \sqrt{2}} + \sqrt[3]{2 - \sqrt{2}}$. It is clear it is in $\mathbb F_{5^3}$ because it is invariant by the automorphism of $\mathbb F_{5^2}$ switching $\sqrt 2$ with $ - \sqrt 2$.

However it is unclear because we pick two different cube roots and someone might make a mistake and not take conjugated roots. It may be better to write $\sqrt[3]{2 - \sqrt{2}}$ = $\frac 1 {\sqrt[3]{2 + \sqrt{2}}}$ instead, so $\alpha = \sqrt[3]{2 + \sqrt{2}} + \frac 1 {\sqrt[3]{2 + \sqrt{2}}}$

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You don't offer a simplification, so why do you answer? I already know that $3$ is coprime to $4$ ;). – Martin Brandenburg Oct 5 '12 at 16:14
@ Martin : well I still explain why there is no simpler combination of radicals (unless you accept useless descriptions like x = 124th root of 1), and I give an alternate less ambiguous description, which is the best you can expect. I don't know what you want more, and of course yes 3 is also coprime to 7 =D – mercio Oct 6 '12 at 1:11

I think, this cannot be much simplified.

All elements ($0,\pm 1,\pm 2$) are cubic roots, so one $\sqrt[3]\ $ is not enough. The simplest you can get then (if you insist on root signs) is to introduce a $\sqrt\ $. But our field extension $\mathbb F_{125} \cong \mathbb F_5(a)$ has degree $3$, so doesn't contain any pure square root, so one additionally need a cubic root finally, and this minus trick with the conjugate, to get back of degree $3$..

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Yeah, but I hope for something like $\sqrt[3]{\sqrt{...}}$ without $\pm$. – Martin Brandenburg Oct 5 '12 at 16:14
Perhaps even two such simple radical expressions which generate the field. – Martin Brandenburg Oct 5 '12 at 18:47

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