The question is specifically asking: Consider $f(x) = [1]x^{3} – [1]x + [2] \in \mathbb{Z}_{5}[x]$. Prove that $f$ is irreducible.

**Attempt at a proof**

A previously completed exercise states that a $3^{rd}$ degree polynomial is irreducible in $\mathbb{F}[x]$ if it does not have any roots in $\mathbb{F}$. So we will show that $f$ has no roots in $\mathbb{Z}_{5}[x]$.

A corollary states that any polynomial $f \in \mathbb{F}[x]$ and $a \in \mathbb{F}$ can be written in the following form:

- $f = (x – a)q + f(a)$ where $q \in \mathbb{F}[x]$

Given this, we can rewrite $f = [1]x^{3} – [1]x + [2] = (x)([1]x^{2} – [1]) + [2] = (x – [0])([1]x^{2} – [1]) + [2]$ where $q = ([1]x^{2} – [1]) \in \mathbb{Z}_{5}[x]$. Then by that same corollary we can say $a$ is a root of $f$ iff $(x-a) \mid f$ but we see that $(x-[0]) \nmid f$ thus $f$ has no root in $\mathbb{Z}_{5}[x]$ and consequently is irreducible in $\mathbb{Z}_{5}[x]$. $\square$

I'm unconvinced that this is a sufficient proof for the question. Namely, I'm wondering about the fact that I've expressed $f$ as a product of polynomials $deg(k) = deg(x-[0]) = 1$ and $deg(q) = deg([1]x^{2} – [1]) = 2$ with a constant $deg(f([0])) = deg([2]) = 0$ so that the degrees work out $3 = deg(f) = deg((x-a)q + f(a)) = deg(x-a) + deg(q) + deg(f(a)) = 1 + 2 + 0$

However, an irreducible polynomial is one for which when expressed as a factor of $2$ polynomials ($f = kq$, where $k,q \in \mathbb{F}[x]$) either $k$ or $q$ has degree $0$, i.e., $f$ is comprised of a constant times a polynomial of the same degree as $f$. So it would seem that I've shown, using the corollary and facts given to me through the texts and exercises that $f$ is irreduicble, however I've expressed it as a product of $2$ polynomials with neither being a constant.

Is it the case that the definition of 'irreducibility' that I'm working with only applies when there isn't an extra constant summed on to the product? Or am I missing something else entirely?

## Best Answer

Another fun way to prove this: put $f=x^3-x+2 \in \mathbb{F}_5[x]$. From $$f=x^3+x^2-2x-x^2+x+2=x(x-1)(x+2)-(x-2)(x+1)$$ one sees that $f$ can be written as the sum of two polynomials with the property that each element of $\mathbb{F}_5=\{0,1,-2,2,-1\}$ is a zero of

oneof them butnotof the other. Hence $f$ has no zero in $\mathbb{F}_5$, and since its degree is $3$ it must be irreducible.In fact, it can be shown (a theorem of H.W. Lenstra (2010)) that

anyirreducible cubic in $\mathbb{F}_5[x]$ admits an irreducibility proof as above. See here.