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 $$a|e' \implies (g_i\geq e_i \forall i) \and u|w \ $$, and $$b|e' \implies (g_i\geq f_i\forall i)\and v|w \ $$. Together, these two things imply $$(g_i\geq \text{max}(e_i,f_i)\forall i)\and (u,v|w) \ $$. 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 $$\phi(a+bi)\phi(c+di)= \begin{cases} 0, & 2|(a+b) \ \or \ 2|(c+d) \\ 1, & 2\not | (a+b) \and 2\not|(c+d) \end{cases} \ $$

=$$\phi(a+bi)\phi(c+di)= \begin{cases} 0, & 2|(a+b)(c+d) \\ 1, & 2\not | (a+b)(c+d) \end{cases} \ $$

=$$\phi(a+bi)\phi(c+di)= \begin{cases} 0, & 2|(ac+ad+bc+bd) \\ 1, & 2\not |(ac+ad+bc+bd) \end{cases} \ $$

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 $$\phi(a+bi)+\phi(c+di)=\begin{cases} 0, & 2|(a+b) \\ 1, & 2\not | (a+b) \end{cases} + \begin{cases} 0, & 2|(c+d) \\ 1, & 2\not | (c+d) \end{cases} \ $$

$$=\begin{cases} 0, & (2|(a+b) \and 2|(c+d))\or(2\not|(a+b) \and 2\not|(c+d)) \\ 1, & (2|(a+b) \and 2\not|(c+d))\or(2\not|(a+b) \and 2|(c+d)) \end{cases} \ $$

$$=\begin{cases} 0, & 2|(a+b+c+d) \\ 1, & 2\not|(a+b+c+d) \end{cases} \ $$

$$=\phi((a+c)+(b+d)i) \ $$.

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 $$\mathbb{Z}[i]/(1+i) = \left\{{0,1}\right\} \ $$.