\documentclass{article}
\usepackage{fullpage,amsmath,amssymb}

\DeclareMathOperator{\im}{im}
\renewcommand{\phi}{\varphi}
\newcommand{\ZZ}{\mathbb Z}
\newcommand{\QQ}{\mathbb Q}

\begin{document}

\noindent
Math 541 \\
Problem Set 9
\begin{enumerate}


\item[7.4.4.] This is immediate from Proposition 7.4.12, since $R \cong R/(0)$.


\item[7.4.13.]  
\begin{enumerate}
\item First observe that a subring of an integral domain is either an integral domain or 0, since any zero divisors in the subring would be zero divisors in the big ring.

Let $\psi: S \to S/P$ be the natural map and consider the composition $\psi \circ \phi: R \to S/P$.  We have
\[ \ker (\psi \circ \phi) = (\psi \circ \phi)^{-1}(0) = \phi^{-1}(\psi^{-1}(0)) = \psi^{-1}(\ker \phi) = \psi^{-1}(P), \]
so by the first isomorphism theorem, $\im(\phi\circ\psi) = R/\phi^{-1}(P)$.  Now $\im(\phi\circ\psi)$ is a subring of the integral domain $S/P$, hence is either 0 or is itself an integral domain, so $\phi^{-1}(P)$ is either $R$ or a prime ideal by Proposition 7.4.13.

For second statement, observe that if $R$ is a subring of $S$ and $\phi: R \to S$ is the inclusion, then $\phi^{-1}(P) = R \cap P$.

\item Again let $\psi: S \to S/M$ be the natural map and consider the composition $\psi \circ \phi: R \to S/M$.  Since both $\phi$ and $\psi$ are surjective, so too is $\psi \circ \phi$.  By the first isomorphism theorem, $R/\ker(\psi\circ\phi) \cong \im(\psi\circ\phi) = S/M$.  Since $S/M$ is a field, $\ker(\psi\circ\phi) = \phi^{-1}(M)$ is maximal by Proposition 7.4.12.

For a counterexample when $\phi$ is not surjective, let $\phi: \ZZ \to \QQ$ be the inclusion.  Since $\QQ$ is a field, $(0)$ is maximal, but $\phi^{-1}((0)) = (0)$ is not maximal in $\ZZ$; for example, it is contained in the ideal $(4)$.
\end{enumerate}


\item[7.4.27.] Since $a$ is nilpotent, there is an $n \ge 1$ such that $a^n = 0$.  Observe that
\[ (1-ab)(1 + ab + a^2b^2 + \dotsb + a^{n-1}b^{n-1}) = 1 - a^n b^n = 1, \]
so $1-ab$ is a unit.


\item[7.5.3.] Let $\phi: \ZZ \to F$ be the ring homomorphism determined by $\phi(1) = 1$.  Then $\im \phi$ is a subring of $F$ and is not 0 since $1 \ne 0$ in $F$, so $\im \phi$ is an integral domain.  Thus $\ker \phi$ is a prime ideal of $\ZZ$, so either $\ker \phi = p\ZZ$ for some prime number $p$ or $\ker \phi = (0)$.  In the former case, $\im \phi \cong \ZZ/p\ZZ$, which is a field, so we are done.  In the latter case, $\im \phi \cong \ZZ$, so take
\[ F_0 = \{ m/n \in F: m, n \in \im \phi \text{ and } n \ne 0 \} \]
which is isomorphic to $\QQ$.

We got $F_0$ by letting 1 generate a field, but any subfield of $F$ contains 1, hence contains $F_0$.

\vspace{1in}
(7.5.1. on reverse.)



\newpage
\item[7.5.1.] In the proof of Theorem 7.5.15, they observe that $r/d = re/rd$ for all $e \in D$.  We will use this fact freely.

Multiplication is well-defined: if $a/b = a'/b'$ and $c/d = c'/d'$ then $ab' = a'b$ and $cd' = c'd$, so $acb'd' = a'c'bd$, so $ac/bd = a'c'/b'd'$.

The additive identity is $0/d$: $a/b + 0/d = (ad+0b)/bd = ad/bd = a/b$.

The additive inverse of $a/d$ is $-a/d$: $a/d + -a/d = (ad-ad)/d^2 = 0/d^2$.

Addition is associative: $(a/b + c/d) + e/f = (ad + bc)/bd + e/f = (adf + bcf + bde)/bdf$, whereas $a/b + (c/d + e/f) = a/b + (cf + de)/df = (adf + bcf + bde)/df$, which is the same.

Addition is commutative: $a/b + c/d = (ad + bc)/bd$, whereas $c/d + a/b = (cb + da)/db$, which is the same since $R$ is commutative.

Multiplication is associative: $(a/b \cdot c/d) \cdot e/f = (ac/bd) \cdot e/f = (ac)e/(bf)d$, whereas $a/b \cdot (c/d \cdot e/f) = a/b \dot (ce/df) = a(ce)/b(df)$, which is the same since multiplication in $R$ is associative.

Multiplication is distributive: $(a/b + c/d) \cdot e/f = [(ad + bc)/bd] \cdot e/f = [(ad + bc)e]/bdf = (ade+bce)/bdf$ since multiplication in $R$ is distributive, whereas $a/b \cdot e/f + c/d \cdot e/f = ae/bf + ce/df = (aedf + cebf)/bdf^2 = (aed + ceb)f/bdf^2$, which is the same.  Distributivity on the other side is similar.

Multiplication is commutative: $a/b \cdot c/d = ac/bd$, whereas $c/d \cdot a/b = ca/db$, which is the same since multiplication in $R$ is commutative.

The multiplicative identity is $d/d$: $a/b \cdot d/d = ad/bd$, which equals $a/b$ since $adb = bda$.


\end{enumerate}
\end{document}