$f = X^{3} + 2$ irreducible in $\mathbb{F}_{49}[X]$.

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Question: Prove that $f = X^{3} + 2$ is irreducible in $\mathbb{F}_{49}[X]$. Is $f$ irreducible in $\mathbb{F}_{7^{n}}$ for all even $n$?

My attempt: Notice that $\deg(f) = 3$, therefore $f$ is irreducible in $\mathbb{F}_{49}[X]$ if and only if $f$ has no roots. We will first look at $f$ in $\mathbb{F}_{7}[X]$, since if $f$ would have roots in $\mathbb{F}_{7}$ then it would also have roots in $\mathbb{F}_{49}$ because it is a subfield. With checking we find that $f$ has no roots in $\mathbb{F}_{7}$. Now let $\alpha \in \overline{\mathbb{F}_{7}}$ be a root of $f$. Then, it follows that $\mathbb{F}_{7}(\alpha)$ is a finite field extension of $\mathbb{F}_{7}$ of degree 3. Then it follows that $\# \mathbb{F}_{7}(\alpha) = 7^{3}$ and by uniqueness of this field we have $\mathbb{F}_{7}(\alpha) \cong \mathbb{F}_{7^{3}}$. Now, we clearly see that a field containing a root of $f$ must contain $\mathbb{F}_{7^3}$, it follows directly that this root is not contained in $\mathbb{F}_{49}$, and therefore $f$ is irreducible in $\mathbb{F}_{49}[X]$.

The polynomial $f$ is not irreducible over $\mathbb{F}_{7^{n}}$ for all even $n$. Consider for example $n = 6$, then since $3$ is a divisor of $6$ we have that $\mathbb{F}_{7^{3}} \subset \mathbb{F}_{7^{6}}$ is a subfield, and therefore the root of $f$ is contained in $\mathbb{F}_{7^{6}}$.

My concern: My main concern with my proof is if my approach of assuming that $\mathbb{F}_{7}(\alpha) \cong \mathbb{F}_{7^{3}}$ is correct, and whether I can directly say that such a root of $f$ can never be in $\mathbb{F}_{49}$. If anybody can help me out or has suggestions I would appreciate that.

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It is correct, since the minimal polynomial $m_\alpha \in \mathbb{F}_7[X]$ of $\alpha$ divides $X^3+2$, because $\alpha^3+2 = 0$. It has to be $\deg m_\alpha = 3$, since otherwise $X^3 +2 = m_\alpha g$ for some $g \in \mathbb{F}_7[X]$ with $\deg g = 1$ or $\deg m_\alpha = 1$, a contradiction to the non-existence of roots of $X^3+2$ in $\mathbb{F}_7$. Therefore $\mathbb{F}_7(\alpha) \cong \mathbb{F}_{7^3}$.

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For the first part, suppose $u$ is a root of $f$ in $\mathbb{F}_{49}$.

Then $-2=u^3$ and so $(-2)^{16}=u^{48}=1$.

But $(-2)^{16} \equiv 2 \bmod 7$, contradiction.