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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?

R. H. Vellstra
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    How did you find the factorisation into irreducible factors? Did you check that $x^2+2x+5$ is irreducible? You just write "we know that $x^2+2x+5$ is irreducible". You could also give an argument, what the splitting field $\Bbb F_{11}(a)$ for a root $a$ of it is isomorphic to $\Bbb F_{11^2}$. – Dietrich Burde Apr 16 '23 at 18:25
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    @DietrichBurde I agree, that is a gap. I just checked by hand the elements of $\mathbb F_{11}$ to see that $x^2+2x+5$ has no zeroes there. – R. H. Vellstra Apr 16 '23 at 18:52
  • I think that you have the correct solution. The question "Is this correct" depends on the grading, i.e., how much detail is required from you. – Dietrich Burde Apr 16 '23 at 18:55
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    You can use that $x^2+2x+5=(x+1)^2+4,$ so it is irreducible because $-1$ is not a square modulo $11.$ – Thomas Andrews Apr 16 '23 at 19:36
  • In @ThomasAndrews's comment, I think it's clearer to say that $-4$ is not a square modulo $11$. While they're equivalent, there's another $-1$ in the expression that might be misleading. – Greg Martin Apr 16 '23 at 19:45
  • I guess this should'v been in the original question, but does it follow that this extension is of degree $2$ over $\mathbb F_{11}$, because $x^2+2x+5$ is irreducible of degree $2$? – R. H. Vellstra Apr 16 '23 at 19:56

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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).

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 $[F(\alpha):F]=n$.

Paprika7191
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