User:J D Bowen/Math850

8.3.2) Given $a,b\in R \ $ and units $u,v \ $, by Proposition 13, page 287, we are given a greatest common divisor of $a=u\Pi p_i^{e_i} \ $ and $b=v\Pi p_i^{f_i} \ $, namely $d=\Pi p_i^{\text{min}(e_i,f_i)} \ $. Define $e=\frac{ab}{d}=\frac{uv\Pi p_i^{e_i+f_i}}{\Pi p_i^{\text{min}(e_i,f_i)}} =uv\Pi p_i^{\text{max}(e_i,f_i)} \ $. Define $S=\left\{{m\in \mathbb{N}:1\leq m \leq n, e_m<f_m}\right\} \ $, and $T=\left\{{m\in \mathbb{N}:1\leq m \leq n, e_m\geq f_m}\right\} \ $. Clearly $S\cup T = \left\{{1,\dots,n}\right\} \ $. Note that $\frac{e}{a}=\frac{uv\Pi p_i^{\text{max}(e_i,f_i)}}{u\Pi p_i^{e_i}} = v \left({ \prod_S 1}\right)\left({\prod_T p_i^{e_i-f_i} }\right) \ $. Since for all $i\in T, \ e_i-f_i\geq 0 \ $, this is a well-defined ring element and so $a|e \ $. Similarly, observe that $\frac{e}{b}=\frac{uv\Pi p_i^{\text{max}(e_i,f_i)}}{v\Pi p_i^{f_i}} = u \left({ \prod_T 1}\right)\left({\prod_S p_i^{f_i-e_i} }\right) \ $. Since for all $i\in S, \ f_i-e_i>0 \ $, this is a well defined ring element and so $b|e \ $. Hence, $e \ $ is a common multiple of $a, b \ $. Now we must only show it is the least such multiple, that is, $a,b|e' \implies e|e' \ $. So suppose $a,b | e' \ $, where $e' = w \Pi p_i^{g_i}, \ w \ $ a unit. Then $a|e' \implies (g_i\geq e_i \forall i) \ \& \ u|w \ $, and $b|e' \implies (g_i\geq f_i\forall i) \ \& \ v|w \ $. Together, these two things imply $(g_i\geq \text{max}(e_i,f_i)\forall i) \ \& \ (u,v|w) \ $. This then implies $e|e' \ $.

8.3.6a) We aim to show $\mathbb{Z}[i]/(1+i) = \left\{{0,1}\right\} \ $. It behooves us to determine precisely what $J=(1+i) \ $ is, so notice that if we let $z=a+bi, \ a,b\in\mathbb{Z} \ $, then $z\in J \iff \exists s \in \mathbb{Z}[i] : s(1+i)=a+bi \ $ Solving this equation for $s \ $ gives us $s=\frac{a+bi}{1+i}=\frac{(a+bi)(1-i)}{2}=\frac{a+bi-ai+b}{2}=\frac{(a+b)+(b-a)i}{2} \ $. Observe that $a+b=2k\iff a=2k-b \iff b-a=b-2k+b=2(b-k) \ $, so $2|(a+b)\iff 2|(b-a) \ $. Hence, $2|(a+b)\iff a+bi\in J \ $. Define the map $\phi:\mathbb{Z}[i]\to\left\{{0,1}\right\} \ $ as $\phi(a+bi)=\begin{cases} 0, & 2|(a+b) \\ 1, & 2\not | (a+b) \end{cases} \ $. Observe that $\phi(a+bi)\phi(c+di)= \begin{cases} 0, & 2|(a+b) \ \text{or} \ 2|(c+d) \\ 1, & 2\not | (a+b) \ \& \ 2\not|(c+d) \end{cases} \ $ =$\phi(a+bi)\phi(c+di)= \begin{cases} 0, & 2|(a+b)(c+d) \\ 1, & 2\not | (a+b)(c+d) \end{cases} \ $ =$\phi(a+bi)\phi(c+di)= \begin{cases} 0, & 2|(ac+ad+bc+bd) \\ 1, & 2\not |(ac+ad+bc+bd) \end{cases} \ $ As we showed above, $2|(x+y)\iff 2|(y-x) \ $, so this is $=\begin{cases} 0, & 2|(ac-bd+ad+bc) \\ 1, & 2\not | (ac-bd+ad+bc) \end{cases} \ $ $=\phi(ac-bd+(ad+bc)i)=\phi((a+bi)(c+di)) \ $. Further observe that $\phi(a+bi)+\phi(c+di)=\begin{cases} 0, & 2|(a+b) \\ 1, & 2\not | (a+b) \end{cases} + \begin{cases} 0, & 2|(c+d) \\ 1, & 2\not | (c+d) \end{cases} \ $ $=\begin{cases} 0, & (2|(a+b) \ \& \ 2|(c+d))\lor(2\not|(a+b) \ \& \ 2\not|(c+d)) \\ 1, & (2|(a+b) \ \& \ 2\not|(c+d)) \ \text{or} \ (2\not|(a+b) \ \& \ 2|(c+d)) \end{cases} \ $ $=\begin{cases} 0, & 2|(a+b+c+d) \\ 1, & 2\not|(a+b+c+d) \end{cases} \ $ $=\phi((a+c)+(b+d)i) \ $. Therefore, $\phi \ $ is a homomorphism $\mathbb{Z}[i]\to\left\{{0,1}\right\} \ $. Since $\phi \ $ is a homomorphism with $\text{ker}(\phi)=(1+i) \ $, the first isomorphism theorem tells us $\mathbb{Z}[i]/(1+i) = \left\{{0,1}\right\} \ $.