I am working on the following problem:
Find the splitting field of the polynomial $f(x)=x^5+2x^4+5x^2+x+4$ over $F_{11}$.
What I have done: So far I found that $-2$ is the only root of $f(x)$ in $F_{11}$. I have also factored $f(x)$ as $f(x)=(x+2)(x^2+5x+1)(x^2-5x+2)$ over $F_{11}$.
Apparently the splitting field of $f(x)$ over $F_{11}$ is $F_{11^4}$. Can someone help me understand how one can conclude that.
On
We are considering the field $F_{11}(\alpha,\beta)$ were $\alpha^2 + 5\alpha +1=0$ and $\beta^2 - 5\beta +2=0$. We are going to show that $$F_{11}(\alpha,\beta)=F_{11}(\alpha+\beta)$$
In fact summing the equations we get $$(\alpha + \beta)(\alpha - \beta + 5)=1$$ which means that these elements are inverse one another, thus adding $(\alpha + \beta)$ to the base field we add its inverse $(\alpha - \beta + 5)$ as well, and also their sum and their difference and thus both $\alpha$ and $\beta$. We proved the claim. Now notice that $(\alpha + \beta) = \frac{\sqrt{21}+\sqrt{17}}{2}$ simply solving the equations of degree two. It has minimum polynomial $x^4 -19x^2+1$. It can be factorized in $F_{11}$ as $x^4 -19x^2+1=(x^2+5)(x^2-2)$ thus $\alpha + \beta$ is a solution of one of them, let's call it $p(x)$. So $$F_{11}(\alpha,\beta)=F_{11}(\alpha+\beta)\simeq\frac{F_{11}[x]}{(p(x))}$$
The polynomial $x^2+5x+1$ has no roots in $F_{11}$: indeed $$ x^2+5x+1=x^2-6x+9-8=(x-3)^2-8 $$ and $8$ is not a square modulo $11$.
Similarly, $x^2-5x+2=x^2+6x+9-7=(x+3)^2-7$ and $7$ is not a square modulo $11$.
Adding a root of $x^2+5x+1$ is the same as adding a square root of $2$. The elements are of the form $a+b\sqrt{2}$, and we must look whether $7$ is a square here: try and solve $$ (a+b\sqrt{2})^2=7 $$ and you'll have your answer.