Polynomial ring quotient：How to deal with polynomial quotient rings

Often you can use the universal properties (of quotients and polynomial algebras) in order to simplify the rings. For example,

$\mathbb{F}_2[x]/(x4+1) = \mathbb{F}_2[x]/((x+1)4) \cong \mathbb{F}_2[y]/(y4)$

and this cannot be simplified anymore. You just "cut" the polynomials in $y$ above $y4$. Of course $y5$ is still there, but it equals zero. And $1+y$ is a unit, since $1+y+y2+y3$ is inverse to it (more generally, unit + nilpotent = unit, this is the geometric series).

Often the Chinese Remainder Theorem helps to simplify the rings. It shows that if $f \in K[x]$ decomposes as $f_1{k_1} \cdot \dotsc \cdot f_n{k_n}$ with irreducible $f_i$, then $K[x]/(f) \cong \prod_{i=1}{n} K[x]/(f_i{k_i})$.

For example, $\mathbb{R}[x]/(x2-1) = \mathbb{R}[x]/((x+1)(x-1)) \cong \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x-1) \cong \mathbb{R} \times \mathbb{R}$. Another interesting example is $\mathbb{Q}[x]/(xn-1) \cong \prod_{d|n} \mathbb{Q}(\zeta_d)$.

Since th...