Prove that \(K = (\mathbb{Z}/2\mathbb{Z})[y]/(y^4 + y^3 + y^2 + y + 1)\) is a field.
Prove that \(x^4 + x + 1\) and \(x^2 + x + 1\) factor completely over the field \(K\) above.
Prove that \(x^3 + x + 1\) is irreducible over the field \(K\) above.
Solution
All polynomial computations below are in characteristic two.
1. The polynomial \(h(y)=y^4+y^3+y^2+y+1\) is irreducibleover \(\mathbb F_2\), so \(K\) is a field with \(16\) elements.
Neither \(0\) nor \(1\) is a root of \(h\), so \(h\) has no linear factor. If it were reducible, its factorization into monic irreducibles would therefore consist of two quadratics. The only monic irreducible quadratic over \(\mathbb F_2\) is \(q(y)=y^2+y+1\): a root-free monic quadratic must have constant term one, and of \(y^2+1\) and \(y^2+y+1\) only the latter is nonzero at \(1\).
Modulo \(q\) one has \(y^2=y+1\), \(y^3=1\), and \(y^4=y\). Consequently \(h\equiv y+1\not\equiv0\pmod q\). Thus no irreducible quadratic divides \(h\), proving its irreducibility. The quotient by an irreducible polynomial over a field is a field [@DF04]. Polynomial division gives the basis \(1,\bar y,\bar y^2,\bar y^3\) of \(K\) over \(\mathbb F_2\), so \([K:\mathbb F_2]=4\) and \(|K|=2^4=16\).
2. The polynomial \(x^{16}-x\) factors as\[ x^{16}-x=\prod_{a\in K}(x-a) \]in \(K[x]\).
The multiplicative group \(K^\times\) has order \(15\), so Lagrange’s theorem gives \(a^{15}=1\) for each nonzero \(a\) [@DF04]. Hence every \(a\in K\), including zero, satisfies \(a^{16}=a\). The sixteen distinct linear factors on the right therefore divide the monic degree-sixteen polynomial on the left. Equality follows from their degrees and leading coefficients.
3. Both polynomials in part (b) divide \(x^{16}-x\), andtherefore split completely over \(K\).
In the quotient \(\mathbb F_2[x]/(x^4+x+1)\), write \(t\) for the class of \(x\). The identity \(t^4=t+1\) gives \[ t^{16}=(t+1)^4=t^4+1=t. \] Here \((u+v)^4=u^4+v^4\) follows by squaring twice in characteristic two. Thus \(x^{16}-x\) belongs to the ideal \((x^4+x+1)\), which is precisely the required divisibility.
In \(\mathbb F_2[x]/(x^2+x+1)\) one has \(t^2=t+1\) and \[ t^4=(t+1)^2=t^2+1=t, \qquad t^{16}=(t^4)^4=t^4=t. \] This proves the other divisibility. Both remain valid in \(K[x]\). By step 2, every irreducible factor of a divisor of \(x^{16}-x\) in \(K[x]\) is linear: unique factorization in the Euclidean domain \(K[x]\) applies [@DF04]. Thus each polynomial in part (b) factors completely over \(K\).
4. The polynomial \(g(x)=x^3+x+1\) is irreducible over \(K\).
The values \(g(0)=g(1)=1\) show that \(g\) has no root in \(\mathbb F_2\). A reducible cubic over a field has a linear factor, so \(g\) is irreducible over \(\mathbb F_2\). If \(b\in K\) were a root, its minimal polynomial over \(\mathbb F_2\) would be \(g\), and \([\mathbb F_2(b):\mathbb F_2]=3\). The tower law would then imply \[ 4=[K:\mathbb F_2] =[K:\mathbb F_2(b)]\,[\mathbb F_2(b):\mathbb F_2] =3[K:\mathbb F_2(b)], \] which is impossible for a positive integer degree. Hence \(g\) has no root in \(K\). Since its degree is three, it is irreducible over \(K\).