Past exam of the options II course of the University of Oxford / 2005 / B

Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.