Let $K$ be an associative commutative ring with identity. Let $R$ be an ideal of the ring $K$. Consider the factor ring.
Let $[\cdot]_1$, $[\cdot]_2$ and $[\cdot]_3$ - some residue classes that $[\cdot]_1 \cdot [\cdot]_2 = [\cdot]_3$.
Is it truth that $\forall c \in [\cdot]_3$ $\exists a \in [\cdot]_1, b \in [\cdot]_2$ that $ab = c$? If it is, how to prove it?
I tried to take advantage that there is a bijection between factors and further consider expressions $ab_1 = d_1$ and $ab_2 = d_1$ for $b_1 \ne b_2$. But next it does not work because the ring may contains zero divisors. Also I tried to express $c$ explicitly using that if $a_1b_1 = c_1$, then $c_1 - c \in R$. But it did not work too.
Thanks for the help!
No, this is not necessarily true. For instance, suppose $K=\mathbb{Z}$, $R=2\mathbb{Z}$, and $[\cdot]_1=[\cdot]_2=[\cdot]_3=2\mathbb{Z}$. Then $[\cdot]_1\cdot[\cdot]_2=[\cdot]_3$, but if $a\in[\cdot]_1$ and $b\in[\cdot]_2$, then $ab$ must be divisible by $4$, not just by $2$. In particular, $2\in[\cdot]_3$ but cannot be written as such a product $ab$.