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I need to compute the degree of the splitting field of the polynomial $X^{4}+X^{3}+X^{2}+X+1$ over the field $\mathbb{F}_{3}$. Quite honestly I don't really know where to begin, I know the polynomial is irreducible in this field. So I thought we could consider some element $\alpha \in E$ where E is some field extensions of $\mathbb{F}_{3}$ and try to find a relation between $\alpha$ and the other roots, but I am not 100% sure. Also I think the degree is $4$ but I am not sure why. Any hints would be apprecaited. Please beare in mind that I am only a few weeks into my galois theory course so it might take a while for me to follow.

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    $\begingroup$ The polynomial is a divisor of $X^5-1$ and so its roots are elements of order $5$ in some extension of $\mathbb F_3$. Now, $\mathbb F_{3^n}$ contains a fifth root of unity exactly when $3^n-1$ (the order of the multiplicative group of the field) is divisible by $5$. What is the smallest extension field that contains fifth roots of unity? We need to find the smallest $n$ such that $3^n-1$ is divisible by $5$. Just start from $n=1$ and work your way up.... $\endgroup$ – Dilip Sarwate Oct 23 '14 at 14:26
  • $\begingroup$ ok so the smallest n such that $5\mid3^{n}-1$ is 4 so this means $\mathbb{F}_{3^{n}}$ contains a fifth root of unity according to what you said. so the smallest extension field containing fifth roots of unity is $\mathbb{F}_{81}$ ? $\endgroup$ – ENAFMTH Oct 23 '14 at 14:36
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    $\begingroup$ Yes, $X^4+X^3+X^2+X+1$ splits in $\mathbb F_{81}$ (and extensions thereof). You don't need to work too hard at finding "relationships" between the roots of $X^4+X^3+X^2+X+1$. $$X^5-1 = \prod_{i=0}^4(X-\alpha^i) = (X-1)(X^4+X^3+X^2+X+1) \Rightarrow X^4+X^3+X^2+X+1 = \prod_{i=1}^4(X-\alpha^i).$$ $\endgroup$ – Dilip Sarwate Oct 23 '14 at 14:51
  • $\begingroup$ Ok, apologies for being so slow here but i think im starting to understand. So we have that $X^{5}-1=(X-1)(X^{5}+X^{4}+X^{3}+X^{2}+X+1)$ and as $X^{5}-1$ has the 5 roots of unity and so $X^{5}+X^{4}+X^{3}+X^{2}+X+1$ has 4 roots namely $\omega$ , $\omega^{2}$ , $-\omega$ and $-\omega^{2}$ where $\omega = e^{\dfrac{2i\pi}{5}}$ so we have splitting field $\mathbb{F}_{3}(\omega)$ so all i need to do is find the degree of the minimum polynomial for $\omega$ over $\mathbb{F}_{3}$? what would this polynomial be? $\endgroup$ – ENAFMTH Oct 23 '14 at 15:14
  • $\begingroup$ @user130289: What is this exponential function over $\mathbb{F}_3$? $\endgroup$ – Martin Brandenburg Oct 23 '14 at 15:24
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If $K$ is a finite field and $f \in K[X]$ is irreducible, then $K[X]/(f)$ is a splitting field of $f$. This follows from the fact that finite extensions of finite fields are always normal. If $\alpha$ is a root, the other roots are $\alpha^{p^n}$, $n \in \mathbb{N}$, where $p=\mathrm{char}(K)$. In particular, the degree is $\mathrm{deg}(f)$.

If $n$ is a natural number coprime to $p$, then it is a fact that the cyclotomic polynomial $\Phi_n$ is irreducible in $\mathbb{F}_p[x]$ if and only if $[p]$ generates $(\mathbb{Z}/n)^\times$.

Since $[3]$ generates $(\mathbb{Z}/5)^\times$, it follows that $\Phi_5=X^4+\dotsc+X+1$ is irreducible in $\mathbb{F}_3[X]$ and hence $\mathbb{F}_{3^4}$ is a splitting field. Explicitly, if $\alpha$ is a root of $\Phi_5$, then the other roots are $\alpha^2,\alpha^3,\alpha^4$.

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  • $\begingroup$ But surely if $\mathbb{F}_{3^{4}}$ is a splitting field (and all splitting fields are isomorphic to one another) and the cardinality of $\mathbb{F}_{3^{4}}$ is well $3^{4}-1$ how do we have degree 4? $\endgroup$ – ENAFMTH Oct 23 '14 at 18:22
  • $\begingroup$ sorry for the slow reply btw, thank you for your help i just need a bit more explanation as i am confused as to why the method above of trying to find the minimum polynomials degree over our constructed field would not work $\endgroup$ – ENAFMTH Oct 23 '14 at 18:23

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