The irreducibility of $X^3-X^2+1$ over $\mathbb{F}_9[X]$

Question

Let $K$ be the field with $9$ elements. I am asked to study the irreducibility of the polynomial $f:=X^3-X^2+1$ over $K[X]$.

I proceeded as follows. $\mathbb{Z}[i]/3\mathbb{Z}[i]$ is a field with $9$ elements, so $K\cong \mathbb{Z}[i]/3\mathbb{Z}[i]$. If $\phi$ is an isomorphism from $K$ to $\mathbb{Z}[i]/3\mathbb{Z}[i]$, then we may construct a ring homomorphism $\overline{\phi}: K[X] \to (\mathbb{Z}[i]/3\mathbb{Z}[i])[X]$ such that $\overline{\phi}(X)=X$ and $\overline{\phi}(a)=\phi(a)$ for any $a \in K$ via the universal property of the polynomial ring . It is pretty straightforward to show that $\overline{\phi}$ is an isomorphism.

Since $\overline{\phi}(f)=X^3-X^2+\hat{1}$ and $X^3-X^2+\hat{1}$ is irreducible over $(\mathbb{Z}[i]/3\mathbb{Z}[i])[X]$ (it has no roots in $\mathbb{Z}[i]/3\mathbb{Z}[i]$), it follows that $f$ is irreducible over $K[X]$: if it were reducible, we would have $f=gh$ for some nonconstant polynomials $g, h \in K[X]$. This would imply that $\overline{\phi}(f)=\overline{\phi}(g)\overline{\phi}(h)$, so since our $\overline{\phi}$ obviously preserves degree, we would get that $X^3-X^2+\hat{1}$ is reducible over $(\mathbb{Z}[i]/3\mathbb{Z}[i])[X]$, which would be a contradiction.

I would like to know if my solution is correct or if I am missing something.

Answer 1

Suppose it is reducibile over $\mathbb{F}_9[X]$. Then exists such $a\in \mathbb{F}_9$ that: $a^3=a^2-1;\;(*)$. Clearly $a^9=a$ and $a\ne 0$. So we have \begin{align} (a^2-1)^3 =a&\implies a^6-3a^4+3a^2-a=1\\ &\implies a^4-3a^3+a^2+2a=0\\ &\implies 2a^2=a+1\\ &\implies a^2 = a+2\\ &\implies a=-3\\ (*)&\implies 0=-1\\ \end{align} A contradiction.

Answer 2

Your solution is ok. If a polynomial factors over some field, its homomorphic image surely factors over the isomorphic image of the field.

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The taxing part in your solution was to check that the polynomial has no zeros in that field of nine elements. I mean, it was taxing, if you plugged in all the nine elements. A trick to get around that would be to observe that an element $a$ of $K$ is a root of either a linear polynomial or a quadratic polynomial $g(x)$ with coefficients of $f(x)$ in $\Bbb{Z}/3\Bbb{Z}$. All depending on whether $a$ is in the prime field or not (ask, if you don't know how to prove this).

Consequently $g(x)$ must be a factor of $f(x)$. Hence $f(x)=g(x)h(x)$, but this time with both $g(x)$ and $h(x)$ with coefficients in the prime field. One of the factors, $g$ or $h$, is linear. Hence has a zero in the prime field. Hence so does $f$. A contradiction.