# Irreducible of $\mathbb Z/p\mathbb Z$.

What are the irreducible number of $\mathbb Z/p\mathbb Z$ ? It looks strange since in a field it looks complicate to talk about irreducible since all element are invertible. So if my question has no sense, how can I prove using Eisenstein that $X^5+2X+2$ is irreducible over $\mathbb Z/3\mathbb Z$ ?

• You mean the irreducible polynomials in $\mathbb Z/3\mathbb Z[X]$? Or are you actually asking about irreducibles of $\mathbb Z/3\mathbb Z$? Commented Dec 19, 2015 at 13:08
• In your specific case, a polynomial of fifth degree over $\mathbb Z/p\mathbb Z$ is irreducible if it has no roots and divided $x^{p^5}-x$. That's because $5$ is prime - for polynomials of non-prime degree, you need to use the more complicated cyclotomic polynomials. Commented Dec 19, 2015 at 13:12
• @ThomasAndrews. Evidently the OP has been told to do this exercise using Eisenstein's criterion. But that criterion involves irreducible elements of some UFD which has the base field as its quotient field. Obviously, $Z/3Z$ has no subrings, so the criterion is meaningless in this case. Commented Dec 19, 2015 at 13:32
• No evidence OP was asked to use Eisenstein; only evidence is that this is the approach OP knows. It's possible that OP was asked to us Eisenstein, but that is not in the text of the question. @David Commented Dec 19, 2015 at 13:36
• @MSE What led you to believe this should be done with Eisenstein's criterion? Commented Dec 19, 2015 at 14:09

Let $F_9$ be the field $9$ elements. If the polynomial is reducible, it will have a factor of degree $\leq 2$, hence a root in $F_9$. Therefore it is enough to show that it has no roots in $F_9$. Assume $x$ is such a root. Clearly, $x \ne 0,1$.

The group $F_9^{*}$ has eight elements. So by Lagrange's theorem, $x^8 = 1$, hence $x^4 = \pm 1$. Therefore, we have $0 = x^5 + 2x + 2 = \pm x + 2x + 2.$ But this implies $x = 1$, a contradiction.

• This is very nice to the point of elegance, and allows me to discard my similar but much longer answer. Commented Dec 20, 2015 at 2:23
• Nice................+1 @David Commented Dec 22, 2015 at 7:26

First of all, $f(X)=X^5+2X+2$ does not have roots in $\mathbb Z/ 3\mathbb Z$ ($f(0)=2$, $f(1)=2$, $f(-1)=2$). Suppose by contradiction that the polynomial can be factored. If the polynomial is reducible, then its factors have degree higher than one; the only possibility is that

$$X^5+2X+2=(X^2+aX+b)(X^3+cX^2+dX+e) = \\ X^5+(a+c)X^4+(d+ac+b)X^3+(e+ad+bc)X^2+(ae+bd)X+eb$$

for some $a,\ldots,e\in \mathbb{Z}/3\mathbb Z$. So

$$\begin{cases} a+c=0 \\ ac+b+d=0 \\ ad+bc+e=0 \\ ae+bd=2 \\ eb=2 \end{cases}$$

From the last equation, either $e=1,b=2$ or $e=2,b=1$. Suppose the latter:

$$\begin{cases} c=-a \\ -a^2+d+1 =0 \\ ad-a+2=0 \\ 2a+d=2\end{cases} \implies \begin{cases}c=-a \\a(a+2)=0 \\-2a^2+a+2=0 \\d=2-2a \end{cases}$$

but the third equation has no solutions in $\mathbb{Z}/3\mathbb{Z}$. The other case is similar.

Trying to explain why Eisenstein's criterion cannot be used.

In a field, such as $\Bbb{Z}/p\Bbb{Z}$, all the non-zero elements are units. Therefore a field has no irreducible elements in the sense needed for Eisenstein's criterion.

Eisenstein's criterion can be used to conclude that a polynomial in $K[x]$, $K$ a field, is irreducible only when $K$ is the field of fractions of a suitable integral domain $R$. By suitable I mean an integral domain, where we can prove Gauss's lemma, e.g. a GCD-domain (see the latter Wikipedia article for a description).

The simplest example is the familiar case, when $K=\Bbb{Q}$ and $R=\Bbb{Z}$. Other examples where Eisenstein's criterion can be used to good effect are:

• $K$ is the field of $p$-adic numbers $\Bbb{Q}_p$ and $R$ is the ring of $p$-adic integers (denoted $\Bbb{Z}_p$, but should not be confused with the residue class ring of integers $\Bbb{Z}/p\Bbb{Z}$ which is often denoted the same way). Here Eisenstein's criterion is superceded by the use of Newton's polygon, which is more general.
• $R$ is the polynomial ring $L[t]$ over some field $L$, and $K=L(t)$ is the corresponding field of fractions, i.e. the field of rational functions over $L$. For example the linear polynomial $t$ is clearly irreducible in $R=\Bbb{Q}[t]$. Therefore over the field $K=\Bbb{Q}(t)$ Eisenstein allows us to conclude that polynomials of the form $x^n-t$ are irreducible in $K[x]$. The irreducible polynomial $t$ takes the role of the prime $p$.

I don't think Eisenstein's the way to go (I don't even know if Eisenstein works when the coefficients are in a field, like $\Bbb Z/3\Bbb Z$). I don't know whether there is a clever way, but you can brute force it:

First, show that there are no linear factors by checking that there are no zeroes. That's easy. Next, assume there is a factorization into a second-degree and third-degree polynomial, like this: $$X^5 + 2X + 2 = (X^3 + aX^2 + bX + c)(X^2 + dX + e)\\ = X^5 + (a + d)X^4 + (b + ad + e)X^3 + (ae + db + c)X^2 + (be + cd)X + ce$$ and show that this cannot be true by some contradiction (i.e. try to solve it, and find for instance that $a = 1$ and $a = 0$ must both be true, or something like that).