26. a. Statement: The equality modulo an ideal I of a ring R is an equivalence relation.

Proof: (Reflexivity)

An ideal by definition is nonempty and any number times 0 is 0 so $0 \in I$

(Symmetry)

0 is in the ideal and it is closed under subtraction.

(Transitivity)

The ideal is closed under addition.

b. Statement: Multiplication on the quotient ring $R/I$ is well defined: $(r+I)(s+I)= rs+I$.

Proof: Let $r, s \in I$, $r \sim_I r’ \rightarrow r-r’ \in I, s \sim_I s’ \rightarrow s-s’ \in I$, and $r+I=r’+I, s+I=s’+I$. Suppose $a=r-r’$ and $b=s-s’$. Then:

$(r + I)(s + I) = rs + I = (r' + a)(s' + b) + I = r's' + r'b + as' + ab + I$. It follows that, because I is an ideal $r'b + as' + ab \in I$ and because it is closed under addition $r'b + as' + ab + I = I$. Thus $r's' + r'b + as' + ab + I = r's' + I$.