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I want to calculate the degree of splitting field over $\mathbb{F}_{2^k}$ and $\mathbb{F}_{3^k}$ of $f(x)=x^4+3x^3+x+1$, when $k\ge 1$.

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    Have you checked for factors when $k=1$? For starters observe that $f(1)=6=0$ in all the fields of interest. – 2017-01-20

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I'll show how it works over $\mathbb{F}_{2^k}$.

  1. Observe that $x=1$ is a root of the given polynomial. Then we can write $$ x^4+3x^3+x+1=(x+1)(x^3+1). $$ Therefore, it is enough to answer the question for $x^3+1$.

  2. Observe that $x=1$ is a root of $x^3+1$. Then, we can write $$ x^3+1=(x+1)(x^2+x+1). $$ Therefore, it is enough to consider the question for $x^2+x+1$.

  3. Observe that $x^2+x+1$ is irreducible over $\mathbb{F}_2$ since it is of degree $2$ and neither $0$ nor $1$ is a root.

  4. Therefore, since $x^2+x+1$ is of degree $2$, in $\mathbb{F}_{2^k}$, it is either irreducible or splits completely. Since $x^2+x+1$ splits completely in $\mathbb{F}_{2^2}$, it splits in $\mathbb{F}_{2^k}$ iff $\mathbb{F}_{2^k}$ contains $\mathbb{F}_{2^2}$. In other words, exactly when $2\mid k$.

Therefore, if $k$ is even, $x^2+x+1$ splits in $\mathbb{F}_{2^k}$. If $k$ is odd, then $x^2+x+1$ splits in the degree $2$ extension $\mathbb{F}_{2^{2k}}$.

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    I apply the algorithm in the case of $\mathbb{F}_{3^k}$, so I can check my comprehension. $f(x)=(x-1)(x^3+x^2+x+2)$ over $\mathbb{F}_2[x]$, product or irreducible. So $f(x)$ splits in $\mathbb{F}_{3^3}$; if $\mathbb{F}_{3^3} \subset \mathbb{F}_{3^k}$, i.e. $3\mid k$, the degree is one over $\mathbb{F}_{3^k}$. If $3\nmid k$, the splitting field over $\mathbb{F}_{3^k}$ is $\mathbb{F}_{3^{3k}}$. Is this right? – 2017-01-20
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    @G.Cantisani That is the factorization in $\Bbb{F}_3[x]$ (probably a typo from you). And, yes, that is correct. Good job! – 2017-01-21