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 $a_{1},\ldots,a_{n}\in k$ are different elements and $F(X)=(X-a_{1})\cdots(X-a_{n})$. Let $W_{i}(X)\in k[X]$ be given by $W_{i}(X)=(X-a_{1})\cdots(X-a_{i-1})\cdot(X-a_{i+i})\cdots(X-a_{n})$. Then there exist $\lambda_{1},\ldots,\lambda_{n}\in k$ such that $F(X)$ divides polynomial

 $U(X)=\bigg{(}\sum_{i=1}^{n}\lambda_{i}\cdot W_{i}(X)\bigg{)}-1.$

Proof. Note, that $W_{i}(a_{i})\neq 0$ for any $i$. Thus we may define $\lambda_{i}=\big{(}W_{i}(a_{i})\big{)}^{-1}$. 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. $\square$

Corollary. Under the same assumptions as in lemma, we have that ideal $\bigg{(}\overline{W_{1}(X)},\ldots,\overline{W_{n}(X)}\bigg{)}$ 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

 $\bigg{(}\sum_{i=1}^{n}\lambda_{i}\cdot W_{i}(X)\bigg{)}-1=F(X)\cdot V(X).$

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. $\square$

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.

Title explicit generators of a quotient polynomial ring associated to a given polynomial ExplicitGeneratorsOfAQuotientPolynomialRingAssociatedToAGivenPolynomial 2013-03-22 19:10:01 2013-03-22 19:10:01 joking (16130) joking (16130) 6 joking (16130) Derivation msc 11C08 msc 12E05 msc 13P05