set theoretic equivalence in quotient ring

Question

If I am given a ring $R$ and a 2-sided ideal $K\subseteq R$, I know that I have a well-defined quotient ring $R/K$. My question is the following: We know that if we have $a,b\in R$, then $(a+K)+(b+K)=(a+b)+K$. This is exactly a set-theoretic equivalence as well. But if we take $(a+K)\cdot (b+K)=a\cdot b+K$, this is not necessarily a set theoretic equivalence (do not know how to show that $(a+K)\cdot (b+K)\supseteq a\cdot b+K$ for general case). Therefore I claim that if we want it to be real set equivalence, we need to say that $[(a+K)\cdot (b+K)]+K=a\cdot b+K$. Is my observation correct?

Answer 1

Maybe you have in mind that we can define $AB = \{ab \mid a \in A, b \in B\}$ for $A,B \subseteq R$. Let denote this product by $A \ast B$, just to distinguish it from the product of cosets defined by $(a+K)\cdot (b+K)=ab+K$

Then, if I understand well, you want to compare $(a+K)\cdot (b+K)$ and $(a+K)\ast (b+K)$.

Since $K$ is a $2$-sided ideal of $R$: $$(a+K)\ast (b+K) = \{(a+k)(b+k') = ab+kb+ak'+kk' \mid k,k' \in K\} \subseteq \{ab+\tilde k \mid \tilde k \in K\} = ab+K$$

However, the reverse inclusion may fail in general. Take $R= \Bbb Z, K=3\Bbb Z,a=0=b$. Then $(a+K)\ast (b+K)=9\Bbb Z$, whereas $(a+K)\cdot (b+K) = 0 + 3\Bbb Z$.