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What is splitting field of polynomial $X^3+X+\bar 1$ in $\mathbb{F_5}$.
Attempt: To show this we first want to check irreducibility which is clear in case of finite field since there are only 5 elements so put values and check it has no root in $\mathbb{F_5}$. After that how to proceed?

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  • $\begingroup$ The splitting field contains all roots of the polynomial. If $a$ is a root, then check how the polynomial decomposes in ${\Bbb F}_5(a)$. $\endgroup$ – Wuestenfux Sep 24 '18 at 13:12
  • $\begingroup$ So say a is root then $a^3+a+1=0$ so what can I do after that? $\endgroup$ – maths student Sep 24 '18 at 13:16
  • $\begingroup$ Divide the polynomial into $x-a$. $\endgroup$ – Wuestenfux Sep 24 '18 at 13:19
  • $\begingroup$ Can you please tell me how? $\endgroup$ – maths student Sep 24 '18 at 13:23
  • $\begingroup$ @Wuestenfux I know if we quotient out it with irreducible then we get field with 125 elements but then how to show polynomial completely splits over there. $\endgroup$ – maths student Sep 24 '18 at 13:27
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Your polynomial is an irreducible cubic over the finite field $\Bbb F_5$. Adjoining one root gives you a field with $5^3$ elements. It’s not hard to see that any extension of finite fields is (automatically) normal. So you may say that the splitting field is $\Bbb F_{125}$.

Knowing a little more about finite fields, you can determine that, with $f(X)=X^3+X+1$ and $a$ your constructed root of $f$ in $\Bbb F_{125}$, you get $a^5=4a^2+a+1$ and $a^{25}=a^2+3a+4$, and $f(X)=(X-a)(X-a^5)(X-a^{25})$

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  • $\begingroup$ Is it obvious that $f(X)=(X-a)(X-a^5)(X-a^{25})$ ? Or even that the other roots are $4a^2+a+1$ and $a^2+3a+4$ ? $\endgroup$ – lhf Sep 24 '18 at 16:45
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    $\begingroup$ Well, @lhf, we know that if $a$ is a root of $f(X)\in\Bbb F_q[X]$, then $a^q$ is also a root. Because $z\mapsto z^q$ is an $\Bbb F_q$-automorphism of any algebraic extension of $\Bbb F_q$. In this case, $q=5$ and our field is cubic, so $a$, $a^5$, and $a^{25}$ are the only conjugates of $a$. It’s certainly not obvious that the conjugates of $a$ have the form they turned out to have: that required a computation. $\endgroup$ – Lubin Sep 24 '18 at 19:57
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My answer to this question with software is in the following form.

First, I find a positive integer $n$ such that in the decomposition of $x^n-1$ in modulo $5$ we have the factor $x^3+x+1$. In your question we get $n=62$. In fact, we get $$ (x^{62}-1) \pmod{5}= \left( {x}^{3}+2\,x+4 \right) \left( {x}^{3}+2\,x+1 \right) \left( {x}^{3}+{x}^{2}+3\,x+4 \right) \left( {x}^{3}+4\,{x}^{2}+x+1 \right) \left( x+1 \right) \left( {x}^{3}+3\,{x}^{2}+4 \right) \left( {x}^{ 3}+{x}^{2}+3\,x+1 \right) \left( {x}^{3}+2\,{x}^{2}+x+4 \right) \left( {x}^{3}+{x}^{2}+1 \right) \left( {x}^{3}+3\,{x}^{2}+4\,x+1 \right) \left( {x}^{3}+x+4 \right) \left( {x}^{3}+4\,{x}^{2}+3\,x+4 \right) \left( {x}^{3}+3\,{x}^{2}+x+1 \right) \left( {x}^{3}+x+1 \right) \left( {x}^{3}+{x}^{2}+4\,x+1 \right) \left( {x}^{3}+4\,{x} ^{2}+4 \right) \left( {x}^{3}+{x}^{2}+x+4 \right) \left( x+4 \right) \left( {x}^{3}+4\,{x}^{2}+3\,x+1 \right) \left( {x}^{3}+2\, {x}^{2}+1 \right) \left( {x}^{3}+2\,{x}^{2}+4\,x+4 \right) \left( {x }^{3}+4\,{x}^{2}+4\,x+4 \right) $$ Then, we obtain a positive integer $k$ such that $62\mid 5^k-1$ which is $k=3$. Then, we construct $\mathbb{F}_{5^3}$ with any irreducible polynomial such as $\bf f$.

Finally, we choose elements of the field $\mathbb{F}_{5^3}$ that their order is $62$ and test which of these elements satisfy $x^3+x+1=0$.

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