explicit generators of a quotient polynomial ring associated to a given polynomial

Let $k$ be a field and consider ring of polynomials $k[X]$. If $F(X)\in k[X]$ and $W(X)\in k[X]$, then we will write $\overline{W(X)}$ to denote element in $k[X]/(F(X))$ represented by $W(X)$.

Lemma. Assume, that $a_1,\mathrm{\dots },a_n\in k$ are different elements and $F(X)=(X-a_1)\mathrm{\cdots }(X-a_n)$. Let $W_i(X)\in k[X]$ be given by $W_i(X)=(X-a_1)\mathrm{\cdots }(X-a_{i-1})\cdot (X-a_{i+i})\mathrm{\cdots }(X-a_n)$. Then there exist ${\lambda }_1,\mathrm{\dots },{\lambda }_n\in k$ such that $F(X)$ divides polynomial

$$U(X)=\left(\sum _{i=1}^n{\lambda }_i\cdot W_i(X)\right)-1.$$

Proof. Note, that $W_i(a_i)\ne 0$ for any $i$. Thus we may define ${\lambda }_i={\left(W_i(a_i)\right)}^{-1}$. Then for any $i$ we have ${\lambda }_i\cdot W_i(a_i)=1$, therefore

$$V(X)=\sum _{i=1}^n{\lambda }_i\cdot W_i(X)$$

is such that $V(a_i)=1$ for any $i$. In particular $U(a_i)=V(a_i)-1=0$ and thus $(X-a_i)$ divides $U(X)$ for any $i$. This completes the proof. $\mathrm{\square }$

Corollary. Under the same assumptions as in lemma, we have that ideal $(\overline{W_1(X)},\mathrm{\dots },\overline{W_n(X)})$ in $k[X]/(F(X))$ is equal to $k[X]/(F(X))$.

Proof. Indeed, all we need to show is that we can generate $\overline{1}$. Lemma implies, that there is $V(X)\in k[X]$ such that

$$\left(\sum _{i=1}^n{\lambda }_i\cdot W_i(X)\right)-1=F(X)\cdot V(X).$$

Now, after aplying quotient homomorphism to both sides we have

$$\sum _{i=1}^n{\lambda }_i\cdot \overline{W_i(X)}=\overline{1}.$$

This completes the proof. $\mathrm{\square }$

Remark. This gives us an explicit formula for generators of $k[X]/(F(X))$. In particular the dimension over $k$ of this ring is at most $\mathrm{deg}F(X)$. It can be shown that actualy it is equal, even if $F(X)$ is arbitrary.