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\noindent{\Large{\textsc{Modern Algebra II\hfill Assignment 8.2  \\ Fall 2019 \hfill Due November 6}}}

\large

\vspace{1cm}

\begin{exercise}
Let $R$ be a commutative ring and $S \subset R^{\times}$ a multiplicative set.  Show that the identity map $I: R \to R$ lifts to an isomorphism $\widehat{I} : S^{-1} R \xrightarrow{\sim} R$.  That is, localizing with elements that are already units in $R$ has no effect on $R$.  [{\em Suggestion:}  Show that $\widehat{I}^{-1} = \vp_S$.]
\end{exercise}

\begin{exercise}
Let $R$ be a commutative ring, $A \subset R$ an ideal, and $S \subset R$ a multiplicative set.  Let $\pi : R \to R/A$ be the canonical surjection and $S/A = \pi(S)  = \{s+A \, | \, s \in S \}$.  Recall that since $\pi$ is a homomorphism, $S/A$ is a multiplicative subset of $R/A$.
\begin{itemize}

\item[\bf a.]  Show that $\psi = \vp_{S/A} \circ \pi : R \to (S/A)^{-1}(R/A)$ maps $S$ into the unit group of $(S/A)^{-1}(R/A)$.

\item[\bf b.]  Let $\widehat{\psi} : S^{-1} R \to (S/A)^{-1}(R/A)$ denote the induced map.  Prove that $\widehat{\psi}$ is surjective and $\ker \widehat{\psi} = S^{-1}A$.  Conclude that $(S^{-1}R)/(S^{-1} A) \cong (S/A)^{-1} (R/A)$.

\end{itemize}

\noindent This shows that the operations ``localize at $S$'' and ``divide by $A$'' commute.

\end{exercise}

\begin{exercise}
Let $R$ be a commutative ring and $S$ a multiplicative set.  Let $P \subset R$ be a prime ideal such that $P \cap S = \varnothing$.

\begin{itemize}

\item[\bf a.]  Prove that $P = \vp_S^{-1}(S^{-1} P)$.

\item[\bf b.]  Prove that $S^{-1}P$ is a prime ideal in $S^{-1}R$.  [{\em Remark:}  Don't forget to prove it's proper!]

\end{itemize}

\end{exercise}


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