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\begin{center} {\Large\bf Algebra , Assignment 10} \\ Solutions
\end{center}
\ben
\item Let $R$ be a ring and $M\subset R$ an ideal. Assume $M\neq R$
(but do {\em not} assume $M$ is maximal). Let $a\in R$.
\ben
\item {\em Assume} there exists an ideal $N$ with $M\subset N \subset R$ and
$N\neq M,R$ and $a\in N$ Prove that $\ol{a}$ has no multiplicative inverse in $R/M$
\So For any
$r\in R$, $\ol{a}\cdot\ol{r} = \ol{ar}$. Setting $n=ar$, as $N$ is an ideal
$n\in N$. We claim $\ol{n}\neq \ol{1}$. For if $\ol{n}=\ol{1}$ then
$n-1=m\in M\subset N$ so $1=n-(n-1)\in N$ and $N=R$.
\item {\em Assume} there does {\em not} exist an ideal $N$ with $M\subset N \subset R$ and
$N\neq M,R$, and $a\in N$. Prove that $\ol{a}$ has an multiplicative inverse in $R/M$.
\So Set \[ M^+=\{m+ar: m\in M, r\in R\} \]
Then $a\in M^+$ so $M\neq M^+$ and $M\subset M^+$ and therefore (as there is no
intermediate size ideal) $M^+=R$ so $1\in M^+$ and $1=m+ar$ for some $r\in R$
and so $\ol{1}=\ol{a}\ol{r}$.
\een
\item Let $R$ be a ring and let $a,b\in R$. Set
\[ M = \{ar+bs: r,s\in R\} \]
Prove that $M$ is an ideal.
\So
\ben \item Identity: $0=a(0)+b(0)\in M$
\item Closure under Addition. Let $\ah,\beta\in M$ so that
$\ah=ar_1+bs_1$, $\beta=ar_2+bs_2$. Then $\ah+\beta=a(r_1+r_2)+b(s_1+s_2)\in M$.
\item Closure under Additive Inverse. Let $\ah\in M$ so that $\ah=ar+bs$.
Then $-\ah = a(-r)+b(-s)\in M$.
\item Closure under Multiplication by Anything. Let $\ah\in M$ so that
$\ah=ar+bs$. For any $t\in R$, $t\ah = a(rt)+b(st) \in M$.
\een
\item Let $Z[i]=\{a+bi: a,b\in Z\}$ where $i=\sqrt{-1}$, the usual
{\em Gaussian Integers}.
\ben
\item For $\ah\in Z[i]$ define (this is called a {\em norm})
$N(\ah)=|\ah|^2$, where $|\cdot|$ is the
usual complex number absolute value, that is
$|c+di|=\sqrt{c^2+ d^2}$.
Give a formula for $N(\ah)$ for $\ah\in R$.
Show $N(\ah\beta)= N(\ah)N(\beta)$.
\So
$N(a+bi)=a^2+b^2$. We need show
\[ (a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2 \]
This follows from elementary (high-school) algebra. Or we may think
of this as the writing of complex numbers $\ah=a+bi=Re^{i\theta}$, so
that $|\ah|=R$. When we multiply complex numbers the $R$ values multiply
so that $|\ah\beta|=|\ah|\cdot|\beta|$ and, squaring,
$N(\ah\beta)=N(\ah)N(\beta)$.
\item Precisely which elements of $Z[i]$ have multiplicative inverses?
(Use the norm to show that you have everything.)
\So $+1,-1,i,-i$. To be a unit you must have norm one (as
norms can't go down under multiplication by nonzero elements) and
so $a^2+b^2=1$ which has these four solutions.
\item Define $\phi:Z[i]\ra Z[i]$ by $\phi(a+bi)=a-bi$. (This is
generally known as {\em complex conjugation}.) Show that $\phi$ is
a homomorphism by showing $\phi(\ah+\beta)=\phi(\ah)+\phi(\beta)$
and $\phi(\ah\beta)=\phi(\ah)\phi(\beta)$. Show that $\phi$
has kernel $\{0\}$.
\So Straightforward calculation.
\item A number $\ah\in Z[i]$ is called {\em composite} if we can
write $\ah=\beta\gamma$ where neither $\beta$ not $\gamma$ have
multiplicative inverses. (That last condition is to avoid
``trivial" factorizations like $23 = i(-23i)$.) If it is nonzero,
not a unit, and not composite it is called {\em prime}.
Show that
$2$ is composite. Show that $41$ is composite.
Show $7+2i$ is prime. (Idea: Use the norm)
\So $2=(1+i)(1-i)$, $41=(5+4i)(5-4i)$, but $7+2i$ has norm $53$.
If $7+2i=\ah\beta$ then $53=N(\ah)N(\beta)$ and the norm is a
positive integer and $53$ is a prime in the integers so one of
$N(\ah),N(\beta)=1$, say $\ah$, and thus $\ah$ is a unit.
\een
\item Set
\[ R = \{ a + b\sqrt{-5}: a,b\in Z \} \]
\ben
\item
Give a formula for $N(\ah)$ (as defined above) for $\ah=a+b\sqrt{-5}\in R$.
\So \[ N(a+b\sqrt{-5})=
(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2 \]
\item Precisely which elements of $R$ have multiplicative inverses?
(Use the norm to argue that you have all of them.)
\So Only $+1,-1$. As norm goes up under multiplication by
a nonzero element the norm must be one and these are the only two
solutions to $a^2+5b^2=1$.
\item Set \[ I = \{2\ah+(1+\sqrt{-5})\beta: \ah,\beta\in R\} \]
Plot those $(a,b)$ with $-4\leq a,b\leq +4$ so that
$a+b\sqrt{-5}\in I$.
\So It will be a checkerboard pattern with every other
element. That is, every $(a,b)$ with $a+b$ even.
\item Show that I is an ideal.
\So Special case of general result in second problem above.
\item Show that $1\not\in I$.
\So For any $\ah=a+b\sqrt{-5}$ and $\beta=c+d\sqrt{-5}$ we
have $2\ah+(1+\sqrt{-5})\beta = x+y\sqrt{-5}$ where
$x=2a+c-5d$ and $y=2b+c+d$. But then $x+y=2a+2b+2c-4d$ is even.
That is, every element $x+y\sqrt{-5}\in I$ must have $x+y$ even.
So $1=1+0\sqrt{-5}\not\in I$.
\item Show that $I$ is {\em not} a principal ideal. (Idea: Assume
$I=(\kappa)$ and use the properties of $N(\cdot)$ above.)
\So If $I=(\kappa)$ then $\kappa|2$ so $N(\kappa)|N(2)=4$ and
$\kappa|(1+\sqrt{-5})$ so $N(\kappa)|N(1+\sqrt{-5})=6$. So
$N(\kappa)|2$. But {\em no} $\kappa\in R$ has norm two so
$N(\kappa)=1$ so $\kappa=\pm 1$ but we know $1,-1\not\in I$.
\item Find representatives of $R/I$. What well known field is it
isomorphic to?
\So $R/I = \{ \ol{0},\ol{1}\}$ and is isomorphic to $Z_2$.
\een
\een
\end{document}