In a worked example in my textbook, we are describing the cosets in $R/A$, where $R=\mathbb{Z}[i]$, the Gaussian integers, and $A = (2+i)R$, the ideal of all multiples of $2 + i$.
It starts by stating that a typical coset $x$ in $R/A$ will have the form $x=(m + ni) + A$, with integers $m,n$. This makes sense to me, but I do not understand the following part:
"Since $2 + i \in A$, we have $i + A = -2 + A$"
Can someone walk me through how we arrive at $i+A = -2+A$?
Well, as you have said, an element in $R/A$ looks like $(m+ni) + A$. So $2+i \in A$ means $2+i +A = A$ (this is an equality of elements in $R/A$). So $(2+i)+A - (2+A) = A - (2+A)$, giving $i+A = -2+A$. (If you are unsure how to get the last equality, think about what addition (and hence subtraction) means in $R/A$.)