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?



1 Answer 1


$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 $q_1(x)=q_2(x)$

  • $\begingroup$ The proof of the lemma is clear to me. What I don't understand is how to use the lemma to get to equality of cosets g(x)+(f(x)) and r(x)+(f(x)) $\endgroup$ Aug 21, 2020 at 2:38
  • $\begingroup$ @Yevhen Melnyk: $g(x) - r(x) \in (f(x)) $ so this would imply equality of the cosets. $\endgroup$
    – Koro
    Aug 21, 2020 at 2:52
  • $\begingroup$ Can't believe it was so embarrassingly simple. Thank you $\endgroup$ Aug 21, 2020 at 2:59

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