# Cardinality Splitting field of $\varphi = X^5+2X^4+X^3+X^2+X+2 \in \mathbb F_3 [X]$

My task is, to find the splitting field of $\varphi = X^5+2X^4+X^3+X^2+X+2 \in \mathbb F_3 [X]$ and give the cardinality of it. I want to know, whether my solution is correct. Maybe there can be done some improvements.

I found out, that $\varphi$ factorizes into $(X^2+2X+2)(X^3+2X+1)$. Both factors are irreducible, so I take the first and define $\psi_0 := X^2+2X+2$.

No I construct the field $E_1:=\mathbb F_3 [X] / (\psi_0)$. $X^2+2X+2 \equiv 0 \mod \psi_0$ leads to $X^2 \equiv X+1 \mod \psi_0$, so the elements of $E_1$ are polynomials with degree < 2 and with $\alpha$ as root of $\psi_0$ I get $E_1=\{0,1,2,\alpha,\alpha+1,\alpha+2,2\alpha,2\alpha+1,2\alpha+2\}$.

Since the second factor $X^3+2X+1 =: \psi_1$ is irreducible in $E_1$, Ive got to extend to $E_2=E_1[X]/(\psi_0)$. Then I know, that this is the splitting field of $\varphi$, because both factors of $\varphi$ got roots now. I also know that the degree of the first extension is 2 and the degree of the second extension is 3, so all in all the splitting field has the cardinality $3^{2*3}=729$.

• Thank you. I edited it. – Myrkuls JayKay Jul 11 '17 at 14:47
• You also have to prove the irreducibility (by checking it has no zeroes, as the degrees are $\le 3$. – Henno Brandsma Jul 11 '17 at 14:50
• Ok, I added the elements to $E_1$. For sure it has to have $3^2=9$ elements, instead of 6. The proof of irreducibility was too much to write, so i let it out here. – Myrkuls JayKay Jul 11 '17 at 14:52
• $E_1$ has size $9$ and then $E_2$ has size $9^3$ (ground field size to the power of the degree), which agrees with your answer. – Henno Brandsma Jul 11 '17 at 14:52

So the final splitting field is just $F_3$ extended by $\alpha$ with $\alpha^2 = \alpha +1$ and $\beta$ with $\beta^3 = \beta + 2$ and an arbitrary element looks like $p_1(\alpha)\beta^2 + p_2(\alpha)\beta + p_3(\alpha)$ where the $p_i(\alpha)$ are linear functions of $\alpha$ with coefficients in $\{0,1,2\}$. This follows straight from the construction, and allows you to compute the size as indeed equal to $(3^2)^3 = 729$ (as there are $3^2$ choices for the $p_i(\alpha)$ and we have to choose $3$ of them.
• Why do I have to choose the same coefficients for $\beta^2$ and $\beta$? – Myrkuls JayKay Jul 11 '17 at 15:37