The first part of the problem (regarding the degree and the uniqueness of the coset representative) is a consequence of the fact that F[x] is a Euclidean domain with degree as the norm.

Let \(\displaystyle{R}={F}\frac{{{x}}}{{p}}{\left({x}\right)}\), the quotient ring consider any coset \(\displaystyle{\left({p}{\left({x}\right)}\right)}+{q}{)}{x}{)}\in{F}{\left[{x}\right]}\)

By Euclidean algorithm, \(\displaystyle\exists{a}{\left({x}\right)},{b}{\left({x}\right)}\in{F}{\left[{x}\right]}\), with q(x)=a(x)p(x)+b(x) with eirher b(x)=0 or deg(b(x)) Thus, every coset (p(x))+q(x) is equal to some coset (p(x))+b(x), with deg(b(x)) We have already proved that every coset can be represented by a polynomial b(x) of degree less than the degree of p)(x). Here is the proof of the uniqueness of b(x) (for each coset). The main point is that for two choices of b(x) and c(x), both of degree < deg(p(x)), the difference is divisible by p(x). Now , a polynomial of lower degree is divisible by p(x) if and only if that polynomial is identically 0.

Claim: b(x) is unique.

Proof: (p(x))+b(x)=(p(x))+c(x)

\(\displaystyle\Leftrightarrow\) b(x)-c(x) is diviseble by p(x)

\(\displaystyle\Leftrightarrow\) b(x)-c(x)=0 (as deg(b(x)-c(x)) \(\displaystyle\Leftrightarrow\) b(x)=c(x)

Coming to the last part, now let F be a finite field.

Let \(\displaystyle{R}={F}\frac{{{x}}}{{p}}{\left({x}\right)}\), deg p(x)=d

Claim: R has \(\displaystyle{q}^{{d}}\) elements.

Proof: From the previous discussion,number of elements in R = number of distinct cosets in \(\displaystyle{F}\frac{{{x}}}{{{p}{\left({x}\right)}}}\) = number of polynomials of degree \(\displaystyle{<}{d}\in{F}{\left[{x}\right]}={q}^{{d}}\)</span>

(a polynomial of degree\(\displaystyle{\sum_{{0}}^{{{d}-{1}}}}{a}_{{i}}{x}^{{i}}{a}_{{i}}\in{F}\), each \(\displaystyle{a}_{{i}}\) can take q values from the field F)

Let \(\displaystyle{R}={F}\frac{{{x}}}{{p}}{\left({x}\right)}\), the quotient ring consider any coset \(\displaystyle{\left({p}{\left({x}\right)}\right)}+{q}{)}{x}{)}\in{F}{\left[{x}\right]}\)

By Euclidean algorithm, \(\displaystyle\exists{a}{\left({x}\right)},{b}{\left({x}\right)}\in{F}{\left[{x}\right]}\), with q(x)=a(x)p(x)+b(x) with eirher b(x)=0 or deg(b(x)) Thus, every coset (p(x))+q(x) is equal to some coset (p(x))+b(x), with deg(b(x)) We have already proved that every coset can be represented by a polynomial b(x) of degree less than the degree of p)(x). Here is the proof of the uniqueness of b(x) (for each coset). The main point is that for two choices of b(x) and c(x), both of degree < deg(p(x)), the difference is divisible by p(x). Now , a polynomial of lower degree is divisible by p(x) if and only if that polynomial is identically 0.

Claim: b(x) is unique.

Proof: (p(x))+b(x)=(p(x))+c(x)

\(\displaystyle\Leftrightarrow\) b(x)-c(x) is diviseble by p(x)

\(\displaystyle\Leftrightarrow\) b(x)-c(x)=0 (as deg(b(x)-c(x)) \(\displaystyle\Leftrightarrow\) b(x)=c(x)

Coming to the last part, now let F be a finite field.

Let \(\displaystyle{R}={F}\frac{{{x}}}{{p}}{\left({x}\right)}\), deg p(x)=d

Claim: R has \(\displaystyle{q}^{{d}}\) elements.

Proof: From the previous discussion,number of elements in R = number of distinct cosets in \(\displaystyle{F}\frac{{{x}}}{{{p}{\left({x}\right)}}}\) = number of polynomials of degree \(\displaystyle{<}{d}\in{F}{\left[{x}\right]}={q}^{{d}}\)</span>

(a polynomial of degree\(\displaystyle{\sum_{{0}}^{{{d}-{1}}}}{a}_{{i}}{x}^{{i}}{a}_{{i}}\in{F}\), each \(\displaystyle{a}_{{i}}\) can take q values from the field F)