The degree of a field extension $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower law states that for finite extensions $F\subseteq E\subseteq K$,If the left side is prime, one factor is one. Degree one forces equality of the corresponding fields, so there is no proper intermediate field.

Let $K$ have 32 elements. Its prime field is ${\mathbb{F}}_2$, and its dimension over that field is five, since a dimension-$d$ vector space over ${\mathbb{F}}_2$ has $2^d$ elements. Every element outside ${\mathbb{F}}_2$ has minimal polynomial of degree five: its generated intermediate field has degree greater than one and dividing five.

Every element of $K$ satisfies $x^{32}=x$, by the multiplicative group of a field of order 31 for nonzero elements. Hence $X^{32}-X$ splits completely over $K$ and has no repeated roots, since its derivative is minus one. Conversely any irreducible polynomial $q$ over ${\mathbb{F}}_2$ gives a quotient ring ${\mathbb{F}}_2[X]/(q)$ with 32 elements. Its defining element satisfies $X^{32}-X$, so $q$ divides that polynomial and therefore has all five distinct roots in $K$.

The 30 elements of $K\setminus {\mathbb{F}}_2$ are thus partitioned into disjoint five-element root sets of the irreducible polynomials. Their number isNo classification of the six individual polynomials is needed; the tower degree and separability already count them all.

For a finite group action, Burnside's lemma givesPair each vertex with its opposite face. This correspondence is equivariant under every tetrahedral symmetry. Each pair has six possible combined color states: three face colors and two vertex colors. Thus we are coloring four positions with six symbols; the two kinds of coloring must be counted together.

The group of rotations is the alternating group $A_4$. The identity has four cycles and fixes $6^4$ colorings. Its three half-turns have cycle type $(2,2)$ and each fix $6^2$. Its eight third-turns have cycle type $(3,1)$ and each also fix $6^2$. Consequently the rotation count isAllowing mirror equivalence gives the full tetrahedral symmetry group $S_4$. Now all permutations of the four paired positions are allowed, so a coloring is determined by a multiset of four symbols chosen from six. stars and bars givesFor an independent Burnside check, the five cycle types of $S_4$ give numerator $6^4+6\cdot 6^3+3\cdot 6^2+8\cdot 6^2+6\cdot 6=3024$, whose quotient by 24 is 126. This is paired face and vertex colorings of a tetrahedron.

The update defines a time-homogeneous Markov process: given its present value, the next value depends only on independent fresh input. Part (b)(i) identifies an invariant law: initializing with the rate-one exponential distribution preserves that distribution at every time. Part (b)(iii) shows convergence of the one-time laws to this invariant law from every deterministic initial value.

For a countable state space, the analogous countable-state Markov chain convergence theorem says that an irreducible Markov chain that is aperiodic and positive recurrent converges to its unique stationary distribution from every initial state. Here the state space is continuous, so that countable-state theorem does not directly supply the proof; the explicit coupling does. The two conclusions have the same equilibrium interpretation.