Find an irreducible polynomial of degree 5 over the field \(\mathbb Z/2\) of two elements and use it to construct a field of order 32 as a quotient of the polynomial ring \(\mathbb Z/2[x]\).
Using the polynomial found in part (a), find a \(5\times5\) matrix \(M\) over \(\mathbb Z/2\) of order 31, so that \(M^{31}=I\) but \(M\neq I\).
Solution
1. The polynomial\[ f(x)=x^5+x^2+1\in\mathbb F_2[x] \]is irreducible.
A reducible polynomial of degree \(5\) must have an irreducible factor of degree \(1\) or \(2\). We have \[ f(0)=1, \qquad f(1)=1+1+1=1, \] so \(f\) has no linear factor. The only monic irreducible quadratic over \(\mathbb F_2\) is \[ q(x)=x^2+x+1. \] Modulo \(q\), we have \(x^3=1\), hence \(x^5=x^2\), so \[ f(x)\equiv x^2+x^2+1=1\pmod q. \] Thus \(q\nmid f\). Therefore \(f\) is irreducible.
2. Consequently\[ F=\mathbb F_2[x]/(f(x)) \]is a field with \(2^5=32\) elements.
Since \(f\) is irreducible of degree \(5\), the quotient is a field and has \(2^5\) elements as a \(5\)-dimensional vector space over \(\mathbb F_2\).
3. Let \(\alpha\) denote the image of \(x\) in \(F\). Then \(\alpha\) has multiplicative order \(31\).
The multiplicative group \(F^\times\) has \[ |F^\times|=32-1=31 \] elements. Since \(31\) is prime, every nonidentity element of \(F^\times\) has order \(31\). The element \(\alpha\) is nonzero because \(f(0)\ne0\), and \(\alpha\ne1\) because \(f(1)\ne0\). Hence \(\alpha\) has order \(31\).
4. With respect to the basis\[ 1,\alpha,\alpha^2,\alpha^3,\alpha^4 \]of \(F\) over \(\mathbb F_2\), multiplication by \(\alpha\) is represented by\[ M= \begin{pmatrix} 0&0&0&0&1\\ 1&0&0&0&0\\ 0&1&0&0&1\\ 0&0&1&0&0\\ 0&0&0&1&0 \end{pmatrix}. \]
The first four columns record \[ \alpha\cdot1=\alpha, \quad \alpha\cdot\alpha=\alpha^2, \quad \alpha\cdot\alpha^2=\alpha^3, \quad \alpha\cdot\alpha^3=\alpha^4. \] Since \(f(\alpha)=0\), \[ \alpha^5+\alpha^2+1=0, \] so in characteristic \(2\), \[ \alpha^5=\alpha^2+1, \] which gives the last column.
5. The matrix \(M\) has order \(31\).
The linear transformation represented by \(M\) is multiplication by \(\alpha\). Therefore \(M^k\) represents multiplication by \(\alpha^k\). By 3, \[ \alpha^{31}=1 \quad\text{and}\quad \alpha\ne1, \] so \[ M^{31}=I \quad\text{and}\quad M\ne I. \] As \(31\) is prime, the order of \(M\) is exactly \(31\).