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

Question

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 $\mathbb Z/3\mathbb Z$ ?

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

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 $p$.