Problem 26 Proof Wed Recitation: Maria Acevedo

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

\begin{align} \forall x \sim _I x \end{align}
\begin{align} x-x \in I \end{align}
\begin{equation} x-x=0 \end{equation}

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

\begin{align} x\sim _Iy \rightarrow y \sim _Ix \end{align}
\begin{align} x-y \in I \rightarrow y-x \in I \end{align}
\begin{align} 0 \in I \end{align}
\begin{equation} 0 - (x-y) = y-x \end{equation}

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

\begin{align} x \sim _Iy, y \sim _Iz \rightarrow x \sim _Iz \end{align}
\begin{align} x-y \in I, y-z \in I \rightarrow x-z \in I \end{align}
\begin{equation} (x-y) + (y-z) = x-y+y-z = x-z \end{equation}

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 $r's' + r'b + as' + ab + I = r's' + I$.

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