User:J D Bowen/Math735 HW7
8.2.3)
Let $P \ $ be a principal ideal domain, and $I \ $ a prime ideal in $P \ $. Then by Proposition 7, pg 280, $I \ $ is also a maximal ideal, or $I=(0) \ $.
If it is a maximal ideal, then, by Proposition 12, pg 254, this means $P/I \ $ is a field. Since all fields are principal ideal domains, $P/I \ $ is a principal ideal domain.
If $I=(0) \ $, then $P/I=P \ $ which is a principal ideal domain by assumption.
8.2.4)
Let $ R \ $ be an Integral Domain. Prove that if the following two conditions hold, then $ R \ $ is a Principal Ideal Domain: (1) Any two nonzero a, b $ \in \ $ R have a gcd which can be written as $ ra+sb \ $ for some r,s $ \in \ $ R, and (2) If $ a_1, a_2, a_3, \dots \ $ are nonzero elements of $ R \ $ s.t. $ a_i + 1 \ $ divides $ a_i \ $ $ \forall \ $ i , then $ \exists \ $ a N > 0 s.t. $ a_n \ $ is a unit times $ a_N \ $ $ \forall \ $ n $ \ge \ $ N.
Let $a,b\in R \ $ be nonzero and $d=gcd(a,b) \ $. Then since $d \ $ is a common divisor, $(d)\supset (a), \ (d)\supset (b) \ $, and thus $(d)\supset (a,b) \ $ is an ideal contained in $(a,b) \ $.
By condition (1), $d=ra+sb \ $, which implies conversely that $(d) \subset (a,b) \ $. Note that if $ I \ $ is an ideal of $ R \ $ and $ R \ $ is generated by two elements, then $ I \ $ is generated by at most one element.
Then by using induction, we deduce that if $ I \ $ is generated by $ n \ $ elements $ a_1, a_2, a_3, \dots, a_n \ $, then $ I \ $ is actually a principal ideal.
Now suppose that n=3, where $ I \ $ is generated by $(a, b) \ $ and $c \ $. Then $ I \ $ is the smallest ideal contained in $(a,b),(c) \ $, so it's the smallest ideal containing $d, c \ $, since $d=gcd(a,d) \ $.
Therefore $ R \ $ is a principal ideal domain once we know that every ideal $ I \ $ $ \in \ $ R is generated by a finite number of elements. The finite generation follows from condition (2). By arguing by contradiction, assume $ I \ $ is an ideal $ \subset \ $ $ R \ $ that cannot be generated by a finite set of its elements. Take a nonzero a $ \in \ $ $ I \ $, then $ (a_1) \subset (a_1, a_2) \subset I \ $ with strict inclusions. It follows that we get $ (a_1) \subset (a_1, a_2) \subset (a_1, a_2, a_3) \subset \dots \subset I \ $. Now, we know each of the ideals $ (a_1, a_2, \dots , a_n) \ $ is principal. Let $ (a_1, a_2, \dots , a_n)= (r_n) \ $. Then $ r_2 | r_1 \ $ and $ r_3 | r_2 \ $ and so on. The quotients $ r_n | r_n+1 \ $ are non-units because of the strict inclusions. This is in contradiction to condition (2), which says that $ \exists \ $ N s.t. $ r_N | r_n \ $ is a unit for n $ \ge \ $ N.
8.2.8) Start with observing that $R \ $ is a principal ideal domain implies that $R \ $ is an integral domain, and by homework 7.5.2, we have $D^{-1}R \ $ is an integral domain. Therefore, all we must show is that $D^{-1}R \ $ is principally generated.
Let $\phi:R\to D^{-1}R \ $ be defined $a\mapsto \tfrac{ad}{d} \ $. Observe that
$\phi(a+b)=\tfrac{(a+b)d}{d} = \tfrac{ad+bd}{d} = \tfrac{ad}{d}+\tfrac{bd}{d}=\phi(a)+\phi(b) \ $,
and
$\phi(ab)=\tfrac{abd}{d} = \left({ \tfrac{ad}{d} }\right) \left({ \tfrac{bd}{d} }\right) = \phi(a)\phi(b) \ $,
since $dd \ $ is just another $d \ $. So $\phi \ $ is a homomorphism.
Let $J \ $ be an ideal in $D^{-1}R \ $. Then since $\phi \ $ is a ring homomorphism, $\phi^{-1}(J) \ $ is an ideal in $R \ $. Since $R \ $ is a principal ideal domain, $\exists x : (x)=\phi^{-1}(J) \ $.
We aim to show $(\phi(x))=J \ $. Hence $J \ $ is principally generated, and since $J \ $ was an arbitrary ideal, $D^{-1}R \ $ is a principal ideal domain.
Let $y\in J \ $. Then $\phi^{-1}(y)\in\phi^{-1}(J)=(x) \ $. Since $(x) \ $ is an ideal, then we have $\phi^{-1}(y)x\in(x) \ $. Phiing both sides, we have $\phi(\phi^{-1}(y)x)=y\phi(x)\in J \ \forall y\in J \ $...
8.3.1)
a)
Let $\phi:\mathbb{Q}^\times \to \mathbb{Q}^\times \ $ be defined as follows. Suppose $a/b \in \mathbb{Q} \ $ is in lowest terms. Then we have $a=2^{a_1}3^{a_2}5^{a_3}\dots, \ b=2^{b_1}3^{b_2}5^{b_3}\dots, \ a_i b_i = 0 \forall i \ $. Define
$\phi\left({\frac{a}{b}}\right) = \frac{2^{a_2}3^{a_1}\dots}{2^{b_2}3^{b_1}\dots} \ $.
Observe that this is a group homomorphism of $\mathbb{Q}^\times \ $:
$\phi\left({\frac{a}{b}\cdot\frac{c}{d}}\right) = \phi\left({ \frac{2^{a_1+c_1}3^{a_2+c_2}\dots}{2^{b_1+d_1}3^{b_2+d_2}\dots} }\right) \ $
This is not in lowest terms, but we can still phi it because there IS some fraction in lowest terms equivalent to it: call that fraction $e/f \ $ and set $g=ac/e=bd/f \ $. Then
$\phi(e/f)=\phi\left({\frac{e}{f}\frac{g}{g}}\right)=\phi\left({ \frac{ac}{bd}}\right) \ $,
so it is acceptable to perform the "swapping" operation on a fraction not in lowest terms; we will receive the same result. Therefore,
$\phi\left({\frac{a}{b}\cdot\frac{c}{d}}\right)=\frac{2^{a_2+c_2}3^{a_1+c_1}\dots}{2^{b_2+d_2}3^{b_1+d_1}\dots} = \frac{2^{a_2}3^{a_1}\dots}{2^{b_2}3^{b_3}\dots} \cdot \frac{2^{c_2}3^{c_1}\dots}{2^{d_2}3^{d_1}\dots} = \phi\left({\frac{a}{b}}\right) \phi\left({\frac{c}{d}}\right) \ $.
Since $\phi \ $ is bijective (note $\phi=\phi^{-1} \ $), it is an automorphism.
b) Suppose $a/b \in \mathbb{Q} \ $ is in lowest terms. Then we have $a=p_1^{a_1}p_2^{a_2}p_3^{a_3}\dots, \ b=p_1^{b_1}p_2^{b_2}p_3^{b_3}\dots, \ a_i b_i = 0 \forall i \ $, where $p_k \ $ is an ennumeration of the primes. Define
$\phi_{jk}\left({\frac{a}{b}}\right) = \frac{\dots p_j^{a_k}\dots p_k^{a_j}\dots}{\dots p_j^{b_k}\dots p_k^{b_j}\dots} \ $.
Observe that this is a group homomorphism of $\mathbb{Q}^\times \ $:
$\phi_{jk}\left({\frac{a}{b}\cdot\frac{c}{d}}\right) = \phi_{jk}\left({ \frac{\dots p_j^{a_j+c_j}\dots p_k^{a_k+c_k}\dots}{\dots p_j^{b_j+d_j}\dots p_k^{b_k+d_k}\dots} }\right) = \frac{\dots p_j^{a_k+c_k}\dots p_k^{a_j+c_j}\dots}{\dots p_j^{b_k+d_k}\dots p_k^{b_j+d_j}\dots} = \phi_{jk}\left({\frac{a}{b}}\right) \phi_{jk}\left({\frac{c}{d}}\right) \ $.
c) Observe that
$\underbrace{\phi_{jk}\left({\frac{p_j}{p_k}}\right) +\dots+\phi_{jk}\left({\frac{p_j}{p_k}}\right)}_{p_j \ \text{times}}=p_j\frac{p_k}{p_j}=p_k \ $,
but
$\phi_{jk}\left({\underbrace{\frac{p_j}{p_k}+\dots+\frac{p_j}{p_k}}_{p_j \ \text{times}}}\right)=\phi_{jk}\left({\frac{p_j^2}{p_k}}\right)= \frac{p_k^2}{p_j} \ $,
and so $\phi_{jk} \ $ is not a ring isomorphism.
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) \land u|w \ $, and $b|e' \implies (g_i\geq f_i\forall i)\land v|w \ $. Together, these two things imply $(g_i\geq \text{max}(e_i,f_i)\forall i)\land (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) \ \lor \ 2|(c+d)
\\ 1, & 2\not | (a+b) \land 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) \land 2|(c+d))\lor(2\not|(a+b) \land 2\not|(c+d))
\\ 1, & (2|(a+b) \land 2\not|(c+d))\lor(2\not|(a+b) \land 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\} \ $.