Skip to main content ☰ Contents You! < Prev ^ Up Next > \(\def\R{\mathbb{R}}
\def\C{\mathbb{C}}
\def\Q{\mathbb{Q}}
\def\N{\mathbb{N}}
\def\Z{\mathbb{Z}}
\def\mfa{\mathfrak{a}}
\def\mfb{\mathfrak{b}}
\def\mfc{\mathfrak{c}}
\def\mfp{\mathfrak{p}}
\def\mfq{\mathfrak{q}}
\def\mfm{\mathfrak{m}}
\def\ev{\mathrm{ev}}
\def\Im{\mathrm{Im}}
\def\End{\mathrm{End}}
\def\Aut{\mathrm{Aut}}
\def\Int{\mathrm{Int}}
\def\Hom{\mathrm{Hom}}
\def\Iso{\mathrm{Iso}}
\def\GL{\mathrm{GL}}
\def\ker{\mathrm{ker }}
\def\Span{\mathrm{Span}}
\def\tr{\mathrm{tr}}
\def\Rings{\mathrm{Rings}}
\def\Alg{{F\text{-Alg}}}
\def\Ann{{\rm Ann}}
\newcommand{\unit}{1\!\!1}
\usepackage{tikz}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 3.2 Isomorphism theorems
Theorem 3.2.1 . (Fundamental theorem of homomorphisms of rings).
Let \(\varphi\colon R\to S\) be a ring homomorphism, and let \(\pi\colon R\to R/\ker(\varphi)\) be the canonical ring homomorphism (see Lemma 3.1.2 ). There is a unique ring monomorphism from \(\overline{\varphi}\colon R/\ker(\varphi)\to S\) such that \(\overline{\varphi}\circ\pi=\varphi\text{.}\) Moreover, there is a ring isomorphism
\begin{equation*}
R/\ker(\varphi)\simeq\varphi(R)\text{.}
\end{equation*}
Theorem 3.2.2 . (Correspondence between ideals of a ring and its quotient ring).
Let \(I\) be an ideal of a ring \(R\text{.}\) There is one-one map between ideals of \(R\) containing \(I\) and ideals of \(R/I\text{.}\)
\begin{equation*}
\{J\supseteq I:J\text{ is an ideal of }R\}\longleftrightarrow\{\overline{J}:\overline{J}\text{ an ideal of }R/I\}
\end{equation*}
Theorem 3.2.3 . (First Isomorphism Theorem).
Let \(\varphi\colon R\to S\) be an epimorphism of rings.
If \(I\) is an ideal in \(R\) then \(\varphi(I)\) is an ideal in \(S\text{.}\)
[First Isomorphism Theorem]. If
\(I\) is an ideal in
\(R\) containing
\(\ker(\varphi)\) then
\begin{equation*}
\overline{\varphi}\colon R/I\to S/\varphi(I)\quad\text{given by}\quad r+I\mapsto\varphi(r)+\varphi(I)
\end{equation*}
is an isomorphism of rings.
Theorem 3.2.4 . (Second Isomorphism Theorem).
Let \(R\) be a ring, \(S\) be a subring of \(R\text{,}\) and let \(I\) be an ideal in \(R\text{.}\) The set
\begin{equation*}
S+I=\{s+x:s\in S\text{ and }x\in I\}
\end{equation*}
is a subring of \(R\text{,}\) and the mapping
\begin{equation*}
\varphi\colon (S+I)/I\to S/(S\cap I)\quad\text{defined by}\quad s+I\mapsto s+(S\cap I),\;\text{where }s\in S;
\end{equation*}
is a ring isomorphism.
Theorem 3.2.5 . (Third isomorphism theorem).
Let \(R\) be a ring and let \(I\) and \(J\) be two-sided ideals of \(R\) with \(I\subseteq J\text{.}\) Then \(J/I\) is an ideal of \(R/I\) and
\begin{equation*}
(R/I)\big/(J/I)\simeq R/J\text{.}
\end{equation*}
