# Show that $f \in \mathbb{Z}_{5}[x]$ is irreducible

abstract-algebrairreducible-polynomialspolynomialssolution-verification

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?

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 one of them but not of 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 any irreducible cubic in $$\mathbb{F}_5[x]$$ admits an irreducibility proof as above. See here.