# Why does uniqueness of quotient and remainder of g(x) by f(x) in a polynomial ring R[x] imply g(x)+(f(x))=r(x)+(f(x)) as cosets of (f(x)) in R[x]?

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

$$f(x)q_1(x)+r_1(x)=f(x)q_2(x)+r_2(x)\implies r_1(x)-r_2(x)=f(x)(q_2(x)-q_1(x))\implies$$ 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 $$q_1(x)=q_2(x)$$
• @Yevhen Melnyk: $g(x) - r(x) \in (f(x))$ so this would imply equality of the cosets.