Reducible polynomial over $\mathbb{F}_9$, but irreducible over $\mathbb{F}_{27}$

Question

Show that $P=X^4+X+2$ is reducible over $\mathbb{F}_9$, but irreducible over $\mathbb{F}_{27}$.

I would appreciate any hints. (I know $P$ is irreducible over $\mathbb{F}_3$ by brute force, and that it factors completely in $\mathbb{F}_{81}$ by noting $\overline{X}, \overline{X^3}, \overline{X^9}, \overline{X^{27}}$ are distinct roots in the isomorphic field $\mathbb{F}_3[X]/(P)$ (is there a nicer way of showing this, other than looking for roots in this quotient?))

Thanks in advance.

Answer 1

The excelent hint of @Jyrki Lahtonen: takes care of the $\mathbb{F}_9$ case.

HINT: ( for $\mathbb{F}_{27}$)

The following result is easy to prove: Let $P$ an irreducible polynomial over $K$ and $K \subset L$ a Galois extension. Then the group $Gal(L/F)$ acts transitively on the irreducible $P_1$, $\ldots$ , $P_m$ factors of $P$ over $L$. In particular, $P$ decomposes over $L$ in $m$ irreducible polynomials of equal degree and $m \mid [L:K]$.

As a consequence: if $P(X)$ of degree $n$ is irreducible over $K$ and $L/K$ is Galois of degree relatively prime to $n$ then $P(X)$ remains irreducible over $L$.

${\bf Added:}$ In fact, the same result takes care of the $\mathbb{F}_9$ case. We know that $P$ decomposes (completely) over $\mathbb{F}_{81}$ into $4$ factors (of degree $1$). If $P$ were irreducible over $\mathbb{F}_9$ the Galois group $Gal({F}_{81}/ \mathbb{F}_9)$ ( of order $2$) could not act transitively on the factors. Now the group $Gal(\mathbb{F}_{9}/ \mathbb{F}_3)$ acts transitively on the factors of $P$ over $ \mathbb{F}_9$, and there are more than one, so there are two, of equal degree.

Answer 2

The facts that $\mathbb F_{q'}$ is a field extension of $ \mathbb F_{q}$ if and only if $q'$ is a power of $q$, that there is a 'unique' extension with given cardinality or degree, and therefore that all extensions are normal (irreducible poly with a root in the extension factors completely) make this problem, as asked, pretty straight-forward almost without calculation, though I get less than you did with calculation.

By calculation (!), $P$ has no roots over ${\mathbb F}_3$. Hence it cannot have an irreducible cubic factor, which means that the polynomial cannot factor over $\mathbb F_{27}$, as otherwise it would have to do completely.

Hence $P$ must factor completely over $\mathbb F_{81}$ because either a root of $P$ generates the 'unique' extension of degree 4 over $\mathbb F_3$, or $P$ factored over $\mathbb F_3$ as quadratic polynomials, and so factored completely in $\mathbb F_9$ (which you ruled you by calculation - but...)

Either way $P$ has to factor in $\mathbb F_9$, because we know that it factors completely in $\mathbb F_{81}$, which is the 'unique' extension of degree 2 over $\mathbb F_9$.

Answer 3

It has no root in base field. Therefore roots of $f$ cannot be in $\mathbb F_{27}$. Since it is a polynomial over $\mathbb F_3$, all roots are in $\mathbb F_{9}$ or $\mathbb F_{81}$. Therefore it is reducible over $\mathbb F_9$.