\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
deepdream of
          a sidewalk
Show Answer
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\groupAdd}[1] { +_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \require{mathtools} \)
Math and science::Algebra::Aluffi

Rings. Third isomorphism theorem, an example.

Let \( a, b \in R \) be elements in a commutative ring \( R \). Use \( ( \bar{b} ) \) to denote the equivalence class of \( b \) in \( R/(a) \). Then firstly we have:

[\[ (\bar{b}) = \frac{?}{(a)} \]]
<p>and with that result it can be seen that:</p>
[ \[ \frac{R/(a)}{(\bar{b})} \cong \quad ? \]]

Recap of ring generators:

Ring generators

Let \( a \in R \) be an element of a ring. The subset \( Ra \) is a left-ideal and the subset \( aR \) is a right-ideal. If \( R \) is a commutative ring, the ideals coincide, and the syntax \( (a) \) is used to denote the ideal.