Problem 26 Proof Monday Recitation

Adam Konkol

26. a. Proposition: Equality modulo an ideal $I$ of a ring $R$ is an equivalence relation.

Proof: It is sufficent to show that equality in $R/I$ is reflexive, symmetric, and transitive. For any element $a$, $a-a = 0 \in I$, and always $0 \in I$ because $I$ must be a subring and $0$ is in every subring. So $a \sim a$. Suppose $a \sim b$. Then $a-b \in I$, so $0 - (a-b) = -a +b = b-a \in I$, so $b \sim a$; equality is symmetric. We show that if $a \sim b$ and $b \sim c$, then $a \sim c$. This means $a-b \in I$ and $b-c \in I$, so $(a-b)+(b-c) = a-c \in I$ by the closure of ideals. So $a \sim c$. $\blacksquare$

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

Proof: Let $t,t' \in I$; we show that $(r+t)(s+t') \sim rs + I$.

\begin{align} (r+t)(s+t') &= rs + ts + rt' + tt' &= rs + (ts + rt' + tt') \end{align}

For $(r+t)(s+t') \sim rs + I$, $(ts + rt' + tt') \in I$. Since $t,t' \in I$, $ts, rt', tt' \in I$ by definition of an ideal; any element in $I$ can be multiplied by elements from $R$ (i.e. $r,s$) to give an element in $I$. Also, since $I$ is a group, it is closed under addition of elements in $I$, so $(ts + rt' + tt') \in I$. $\blacksquare$

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