 Solution

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.