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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 3 Solutions}
\begin{document}

\maketitle

\begin{enumerate}
    \item \begin{enumerate}
    \item $\Z(x)$, the field of rational functions in $x$.
    \item $\Q[\sqrt{2}]=\Q(\sqrt{2})$ is a field.
    \item $\C(x,y)=\C(x)(y)$.
    \end{enumerate}

    \item Let $F$ denote the field of fractions of $R=k[[x]]$ and for any $f\in k((x))$ let  $v(f)$ be the unique integer
    such that $x^{v(f)} f=c_0+c_1 x+\cdots$.  Define
    $$\phi:k((x))\longrightarrow F$$ by
    $$\phi(f)=\begin{cases}\frac{f}{1} &\text{if}\ f\in k[[x]]\\
    x^{v(f)} f/x^{v(f)} &\text{otherwise}\end{cases}.$$
    It is clear that $\phi$ is an injective homomorphism.  To show surjectivity, let $f/g \in F$.  Then
    $x^{v(g)} g$ is a unit in $k[[x]]$ (from lecture) so there exists $h\in k[[x]]$ such that $gh=1$.
    Thus,
    $$\frac{f}{g}=x^{v(g)}f h.$$  Clearly $fh\in k[[x]]$ and since $v(g)\in\Z,$ $f/g$ is in the image
    of $\phi$.

    \item For any maximal ideal $\M$, it is clear that $1\not\in \M$ and hence $a=a/1 \in A_{\M}$ for any $a\in A$ and any $\M$.
    It follows that
    $$A\subset \bigcap_{\substack{\M\subseteq A\\\M\ \text{maximal}}} A_{\M}.$$
    For the reverse inclusion, let $x\in A_{\M}$ for all maximal $\M\subset A$.  Then we can write $x=a_{\M}/b_{\M}$
    for each $\M$ with $a_{\M}\in A,b_{\M}\in A\setminus \M.$  Observe that the ideal $B$ generated by the $b_{\M}$ is all of $A$,
    for if not, then $B\subset \M_0$ for some maximal ideal $\M_0$, but $b_{\M_0}\in B$ and $b_{\M_0}\not\in \M_0\supset B,$
    a contradiction.  Hence we can write
    $$1=\sum_{\M\in T} c_{\M}b_{\M}$$ for some finite set $T$ where $c_{\M}\in A$.  It follows that
    $$x=\sum_{\M\in T} x c_{\M}b_{\M}=\sum_{\M\in T} c_{\M}a_{\M}\in A,$$
    and we are done.

    \item
    \begin{enumerate}

    \item We use the result from class that a ring $R$ is local with unique maximal ideal $\M$ if any $f\in R\setminus \M$ is a unit.
    So let $\M\subset \CC_0$ be the ideal of germs of functions vanishing at $0$ (this is obviously an ideal).  If $[(U,f)]\in \CC_0\setminus \M,$
    then $f(0)\neq 0$ so that by continuity, there exists a neighborhood $V\subset U$ of $0$ with $f$ nonvanishing on $V$.  Then
    $[(V,1/f)]\in \CC_0\setminus\M$ and
    $$[V,1/f][U,f]=[V,1]=\id\in \CC_0,$$ so $f$ is invertable, as required.

    \item   Now let $\CC(\R)$ denote the ring of all continuous functions on $\R$ and let $\M\subset \CC(\R)$ be the maximal ideal
    of all functions vanishing at $0$.  Define
    $$\varphi:\CC(R)_{\M}\longrightarrow \CC_0$$ by
    $$\varphi([f/g])=[U_g,f/g],$$ where $[f/g]$ denotes the equivalence class of $f/g$ and $U_g$ is a neighborhood of $0$ on which $g$ does not
    vanish (such a neighborhood always exists since $g(0)\neq 0$ and $g$ is continuous).  We claim that $\varphi$ is an
    isomorphism.
    There are several things to check:
    \begin{enumerate}
    \item \textbf{The map $\varphi$ is well defined:}  That is, we must show that it does not depend on the choice of representative $f/g$.  So suppose
    that $f/g,h/k$ are any two representatives of the same equivalence class in $\CC(\R)$.  Then there exists a continuous function $s\not\in\M$
    such that $s(fk-gh)\equiv 0$ on $\R$.  Since $s\not\in \M,$ there exists a neighborhood $U$ of $0$ such that $s$ does not vanish on $U$.
    It follows that $fk-gh$ must vanish on $U$ and hence that $f/g=k/h$ on $U\cap U_g\cap U_h$ so that $[(U_g,f/g)]=[(U_h,k/h)]$, so $\varphi$
    is well defined.

    \item \textbf{Injectivity:}  Suppose that $\varphi([f/g])=\varphi([k/h])$.  Then there exists some neighborhood $U$
    of $0$ such that $f/g-k/h\equiv 0$ on $U$.  Now there exists some $\epsilon>0$ with $(-\epsilon,\epsilon)\subset U$.  Let
    $$\chi_{\epsilon}(t)=\begin{cases}0 &\text{if}\ |t|>\epsilon\\ 1 &\text{if}\ |t|<\epsilon/2\\ 2\left(1-\frac{|t|}{\epsilon}\right)
     &\text{otherwise}\end{cases}.$$
     Then clearly $\chi_{\epsilon}\in\CC(\R)\setminus\M$ and
     $$\chi_{\epsilon}(fh-gk)\equiv 0$$ on all of $\R$, in other words, $[f/g]=[k/h]$ in $\CC(\R)_{\M}$, as required.

     \item \textbf{Surjectivity:} Let $[(U,f)]\in \CC_0$.  Then for some $\epsilon>0$ we have $(-\epsilon,\epsilon)\subsetneq U$.  Now define
     $$\tilde{f}(t)=\begin{cases}f(t) &\text{if}\ t\in (-\epsilon,\epsilon)\\ f(\epsilon) &\text{if}\ t\geq \epsilon\\
     f(-\epsilon) &\text{if}\ t\leq -\epsilon\end{cases}.$$
     Then it is not hard to see that $\tilde{f}\in\CC(\R)\subset \CC(\R)_{\M}$ and that $\varphi([\tilde{f}])=[(U,f)].$

     \item \textbf{The map $\varphi$ is a homomorphism:}  This follows directly from the definition of addition and multiplication
     in $\CC_0$.
    \end{enumerate}
    \end{enumerate}
\end{enumerate}

\end{document}