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?

• The splitting field contains all roots of the polynomial. If $a$ is a root, then check how the polynomial decomposes in ${\Bbb F}_5(a)$. – Wuestenfux Sep 24 '18 at 13:12
• So say a is root then $a^3+a+1=0$ so what can I do after that? – maths student Sep 24 '18 at 13:16
• Divide the polynomial into $x-a$. – Wuestenfux Sep 24 '18 at 13:19
• Can you please tell me how? – maths student Sep 24 '18 at 13:23
• @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. – maths student Sep 24 '18 at 13:27

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 $$f(X)=(X-a)(X-a^5)(X-a^{25})$$
• 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 $a^2+3a+4$ ? – lhf Sep 24 '18 at 16:45
• 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. – Lubin Sep 24 '18 at 19:57
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 $$x^3+x+1=0$$.