Splitting field of $x^3 +x +1$ over $\mathbb F_{11}$

Question

This is a HW problem for an algebra course.

Determine the splitting field of $f(x)=x^3+x+1$ over $\mathbb F_{11}$.

I tried to use the answers from this question and this question to help me, but want to know if I'm doing it correctly.

Attempt: We have that $f(x)$ factors as $(x^2+2x+5)(x+9)$. Since $x+9$ is already a linear factor, we need only concern ourselves with $x^2+2x+5$. We know that $x^2+2x+5$ is irreducible over $\mathbb F_{11}$, so its splitting field will be $\mathbb F_{11^2}$, and hence the splitting field of $f$ is $\mathbb F_{11^2}$.

Is this correct?

Answer 1

It is not hard to show that$\newcommand{\f}{\mathbb{F}_{11}}$ the factor $g:=x^2+2x+5$ is irreducible over $\f$. Once you have that, let $\alpha$ be the root of $g$ (in a fixed algebraic closure of $\f$). Then the splitting field for $f$ over $\f$ is $\f(\alpha)$, which we know has degree $2$ over $\f$. This shows that $|\f(\alpha)|=11^2$. Since finite fields are unique, we know that $\f(\alpha)=\mathbb{F}_{11^2}$ (in the algebraic closure).

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I noticed that OP seems to be unsure of the reason why $g$ being quadratic implies the splitting field has degree $2$ over $\f$. If we have an algebraic closure of a field $F$, then the splitting field for a polynomial $f$ is simply $F(\alpha_1,\dots,\alpha_n)$, where $\alpha_1,\dots,\alpha_n$ are the $n$ possibly repeated roots of $f$ in the algebraic closure.

Now, in your case, the roots of $f$ are $-9$, $\alpha$, $-2-\alpha$, so the splitting field is $\f(-9, \alpha, -2-\alpha)$, but then $\f(-9, \alpha, -2-\alpha)=\f(\alpha)$, which we know is quadratic over $\f$ because of the following theorem:

Let $K/F$ be an extension. Let $\alpha\in K$ be a root of an irreducible polynomial $f$ of degree $n$. Then $F(\alpha)\cong F[x]/(f)$ and $[F(\alpha):F]=n$.