> [!NOTE] Theorem (Quotient Ring Addition is Well-Defined) > Let $(R,+,\times)$ be a [[Rings|ring]]. Let $I$ be an [[Ideal of Ring|ideal]] of $R$. Let $(R/I,+,\times)$ be the [[Quotient Ring|quotient ring]] of $R$ by $I.$ Then $+$ is [[Well-defined Function with Respect to Equivalence Relation|well-defined]] on $R/I,$ that is: if $a,a',b,b'\in R$ such that $a+I=a'+I$ and $b+I=b'+I$ then $(a+b)+I=(a'+b')+I$and$(ab)+I = (a'b')+I$ **Proof**: By [[Necessary Condition for Equality of Cosets]], $a-a' \in I$ and $b-b' \in I$. Then $(a+b)-(a'+b')=(a-a')+(b-b')\in I$since $I$ is closed under $+.$ Thus by the same lemma, $(a+b)+I=(a'+b')+I.$ Now for multiplication, we have: $ab-a'b'=ab-ab'+ab'-a'b'=a(b-b')+(a-a')b' $Since $I$ is an ideal and $(a-a'),(b-b')\in I$ we have $a(b-b'),-b'(a-a')\in I$ and so $ab-a'b'\in I.$ Thus by Equality of Cosets, $(ab)+I=(a'b')+I.$ # Applications Note [[Quotient Ring is Ring]].