I have been reading Algebra Chapter $0$ by Aluffi and I'm struggling to understand the following:
First, the author proves the lemma:
Let $f(x)$ be a monic polynomial, and assume
$$f(x)q_1(x)+r_1(x)=f(x)q_2(x)+r_2(x)$$
with both $r_1(x)$ and $r_2(x)$ polynomials of degree $< \deg f(x)$. Then $q_1(x) = q2(x)$ and $r_1(x) = r_2(x).$
Then it is claimed that this lemma can be summarised as follows:
Assume then that $R$ is a commutative ring. If $f(x)$ is monic then for every $g(x)\in R$ there exists a unique polynomial $r(x)$ of degree $<\deg f(x)$ and such that
$$g(x)+(f(x))=r(x)+(f(x))$$
as cosets of principal ideal $(f(x))$ in $R[x]$.
How can I see that the latter statement follows from the lemma?
Thanks
Best Answer
$f(x)q_1(x)+r_1(x)=f(x)q_2(x)+r_2(x)\implies r_1(x)-r_2(x)=f(x)(q_2(x)-q_1(x))\implies $ degree of $r_1(x)-r_2(x)$ is atleast equal to degree of $f(x)$ as degree of $r_1(x) $ is either less than that of $f(x) $ or $r_1(x)=0$ . Similarly for $r_2$.
So we must have $r_1(x)-r_2(x)=0$ whence by the equality above, it follows that $q_1(x)=q_2(x)$