Sachchidanand Prasad

25-01-2023

Problem: Determine if each of the following is irreducible.

  1. $x^{2} +x+1\in \mathbb{Z} _2[x]$.
  2. $x^{3} +x+1\in \mathbb{Z} _3[x]$.
  3. $x^{4} +10x^{2} + 5\in \mathbb{Z}[x]$.
  4. $x^5 + 2x + 4 \in \mathbb{Q} [x]$.
  5. $x^{5} + 2x + 1 \in \mathbb{Z}[x]$.

Solution:

Some important results involving the irreducibility checks are as follows:

  • Let $\mathbb{F} $ be a field and let $f(x)\in \mathbb{F} [x]$ be a polynomial with $\deg f(x)$ is either $2$ or $3$, then $f(x)$ is irreducible over $\mathbb{F} $ if and only if it has a zero in $\mathbb{F} $.
  • Let $f(x)\in \mathbb{Z} [x]$. If $f(x)$ is reducible over $\mathbb{Q} $, then it is reducible over $\mathbb{Z} $.
  • [Mod $p$ Irreducibility Test] Let $p$ be a prime and suppose that $f(x)\in \mathbb{Z} [x]$ with $\deg f(x) \geq 1$. Let $\bar{f} (x)$ denotes the polynomial in $\mathbb{Z} _p[x]$ obtained from $f(x)$ by reducing all the coefficients of $f(x)$ modulo $p$. If $f(x)$ is irreducible over $\mathbb{Z} _p$ and $\deg \bar{f}(x) = \deg f(x)$, then $f(x)$ is irreducible over $\mathbb{Q} $.
  • [Eisenstein's Criterion] Let \[ f(x) = a_nx^n + a_{n-1}x^{n-1} +\cdots+ a_0\in \mathbb{Z}[x]. \] If there is a prime $p$ such that $p$ does not divide $a_n$, $p$ divides $a_{n-1},\cdots,a_0$ and $p^2$ does not divide $a_0$, then $f(x)$ is irreducible over $\mathbb{Q} $.
  • [Irreducibility of $p^{\text{th}}$ cyclotomic polynomial] For any prime $p$, the $p$th cyclotomic polynomial \[ \Phi (x) = \frac{x^p - 1}{x-1}= x^{p-1} + x^{p-2} + \cdots + x + 1 \] is irreducible over $\mathbb{Q} $.

1. The polynomial is $f(x)=x^{2} +x+1\in\mathbb{Z}_2 [x]$, so it will be irreducible if and only if it does not have a zero. Observe that \[ f(0) = 1~\mathrm{mod}~2 \text{ and } f(1) = 1~\mathrm{mod}~2. \] Thus, the polynomial does not have a root in $\mathbb{Z} _2$ and hence, it is irreducible.

2. The polynomial is $f(x)=x^{3} +x+1\in\mathbb{Z}_3 [x]$, so it will be irreducible if and only if it does not have a zero. Note that \[ f(1) = 0~\mathrm{mod}~3. \] Thus, the polynomial is reducible over $\mathbb{Z} _3$.

3. The polynomial is $f(x)=x^{4} +10x^{2} +5\in\mathbb{Z}[x]$. We will use the Eisenstein's criterion for this. Take $p=5$. Note that \[ p \mid 5,~p \mid 10, ~p\nmid 1,\text{ and } p^2\nmid 5. \] So using the Eisenstein's criterion, the polynomial is irreducible over $\mathbb{Z} $.

4. The polynomial is $f(x)=x^5 + 2x + 4\in\mathbb{Q}[x]$. Note that the Eisenstein's criterion will not work here. So we will apply the mod $p$ test. If we take $p=2$, then this is not helpful, so we will try with $p=3$. \[ \bar{f}(x) = x^5 + 2x + 1. \] Substitution of $0, 1,$ and $2$ into $\bar{f}$ does not yield $0$, so there are no linear factors.

We will now check about the quadratic factors. Let $x^2 + ax +b \in \mathbb{Z} _3[x]$ is a factor. So we have nine choices for $a$ and $b$. Note that \begin{align*} & (a,b) = (0,0)\implies \bar{f}(x) = x^2, \\ & (a,b) = (0,1)\implies \bar{f}(x) = x^2 + 1, \\ & (a,b) = (0,2)\implies \bar{f}(x) = x^2 + 2, \\ & (a,b) = (1,0)\implies \bar{f}(x) = x^2 + x, \\ & (a,b) = (1,1)\implies \bar{f}(x) = x^2 + x + 1, \\ & (a,b) = (1,2)\implies \bar{f}(x) = x^2 + x + 2, \\ & (a,b) = (2,0)\implies \bar{f}(x) = x^2 + 2x, \\ & (a,b) = (2,1)\implies \bar{f}(x) = x^2 + 2x + 1, \\ & (a,b) = (2,2)\implies \bar{f}(x) = x^2 + 2x + 2. \end{align*} We can immediately rule out each of the nine that has a zero over $\mathbb{Z} _3$, since $\bar{f}$ does not have one.

This leaves only $x^2+1,~x^{2} +x+2$, and $x^{2} +2x+2$ to check. We can use long division to check these. We have \begin{align*} &x^5 + 2x + 1 = \left( x^3 + 2x \right)\left( x^{2} +1 \right) + 1\\ &x^5 + 2x + 1 = \left( x^3 + 2x^{2} +2x \right)\left( x^{2} +x + 2\right) + (x+1)\\ &x^5 + 2x + 1 = \left( x^3 + x^{2} +2x \right)\left( x^{2} +2x + 2 \right) + (x+1). \end{align*} Thus, the given polynomial is irreducible over $\mathbb{Z} _3$ and hence irreducible over $\mathbb{Z} $.

The polynomial is $f(x)=x^5 + 2x + 1\in\mathbb{Q}[x]$. Note that from previous problem we conclude that this is irreducible over $\mathbb{Z} _3$ and hence irreducible over $\mathbb{Q} $.