# Let $f \in \mathbb F_3[X]$ be reducible, degree 4 or 5 and no roots, then a monic irreducible polynomial exists of degree 2 dividing $f$

Let $f \in \mathbb F_3[X]$ be reducible, degree 4 or 5 and no roots, then a monic irreducible polynomial exists of degree 2 dividing $f$.

I've proved that $X^2 + 1, X^2 + X + 2, X^2 + 2X + 2$ are the only monic irreducible polynomials of degree 2 in $\mathbb F_3[X]$.

In proving that one of these divide $f$, I know that $f = f_1 f_2$ for two polynomials of degree $\ge 1$. $\mathbb F_3[X]$ so every polynomial of degree $0$ is a unit. I also know that $\mathbb F_3[X]$ is a Euclidean domain, so there exists a unique factorization of $f = q_1 ... q_n$ of irreducible polynomials.

However how can I prove that either $f_1$ or $f_2$ is a monic irreducible polynomial of degree 2, or $q_j$ is such a polynomial in the unique factorization ?

Thanks.

• Hint: No do not need this to be over $\mathbb{F}_3$ for the statement to be true. It is just a matter of the possible ways of writing $4$ or $5$ as a sum of numbers which are strictly greater than $1$. – Tobias Kildetoft Nov 27 '13 at 13:49
• Why strictly greater than $1$ ? If $\deg k = 1$ then $k$ is irreducible and could be used in a factorization of $f$. – Shuzheng Nov 27 '13 at 13:52
• But then $f$ would have a root. – Tobias Kildetoft Nov 27 '13 at 13:52
• Ohh you are right, sorry. Could not imagine the answer would be so simple. Been sitting and trying to fit togheter theorems in order to solve the problem. – Shuzheng Nov 27 '13 at 13:56
• Do you have any idea how to conclude that $L = R/<X^5 - X + 1>$ consists of $243$ elements ? – Shuzheng Nov 27 '13 at 14:16