# 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

## 1 Answer

You have, I think, succesfully applied the "algorithm" to find the splitting field (split into irreducibles and apply successive group extensions as needed, this was the existence proof in my old Galois theory notes). As I said in the comments, a full proof would require to show the irreducibility in full (but that's a lot of work and can be assisted by a computer package like Sage).

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
• @MyrkulsJayKay sorry, I'll edit – Henno Brandsma Jul 11 '17 at 15:38