(b) Let $q∈\mathbb{Z}$ be a prime with $q ≡ 3 \mod{4}$. Prove $\mathbb{Z}[i]/(q)$ is a field of order $q^2$.
(c) Let $p∈\mathbb{Z}$ be a prime with $p ≡ 1 \mod{4}$ and let $p=\pi \overline{\pi}$ be its factorization in $\mathbb{Z}[i]$. Prove $\mathbb{Z}[i]/(p)≅\mathbb{Z}[i]/(\pi) \times \mathbb{Z}[i]/(\overline{\pi})$ as rings, $\mathbb{Z}[i]/(p)$ is of order $p^2$ and therefore $\mathbb{Z}[i]/(\pi)$ and $\mathbb{Z}[i]/(\overline{\pi})$ are fields of order $p$.
7. Let $\pi∈\mathbb{Z}[i]$ be irreducible.
(a) For $n \geq 0$ an integer, show that $(\pi^{n+1}) \subseteq (\pi^n)$ and that multiplication by $\pi^n$ induces an isomorphism $\mathbb{Z}[i]/(\pi)≅(\pi^n)/(\pi^{n+1})$ as additive abelian groups.
(b) Prove $|\mathbb{Z}[i]/(\pi^n)|=|\mathbb{Z}[i]/(\pi)|^n$.
(c) Prove for $\alpha≠0$ that $|\mathbb{Z}[i]/(\alpha)|=N(\alpha)$.
Proof: (6) (a) By the division algorithm, every element in $\mathbb{Z}[i]/(1+i)$ can be represented by an element of norm $< 2$. We have $-i=-(1+i)+1$,$i=(1+i)-1$, and $-1=-(1+i)(1-i)+1$, so that the order is $≤2$. Since $1+i$ is not a unit, we have $(1+i) \subset \mathbb{Z}[i]$, so that the order must be 2. Since $\mathbb{Z}[i]/(1+i)$ is now a finite integral domain by the primality of $1+i$, we have it is a field.
(b) Observe arbitrary $a+bi∈\mathbb{Z}[i]$. Let $a=a'+v_aq$ and $b=b'+v_bq$ with $0≤ a',b' < q$. We have $a+bi=(a'+v_aq)+(b'+v_bq)i=a'+b'i+q(v_a+v_bi)$, so that in this sense $a+bi$ can be reduced modulo $q$. This allows a maximum order of $q^2$. Now, assume reduced $\overline{a+bi} = \overline{c+di}$ in $\mathbb{Z}[i]/(q)$, so that $a+bi=c+di+qr$ for some $r∈\mathbb{Z}[i]$. We therefore have $(a-c)+(b-d)i=qr$, i.e. $q$ divides $(a-c)$ and $(b-d)$. Since $a,b,c,d < q$, we have $a=c$ and $b=d$, i.e. $a+bi=c+di$. Therefore there are precisely $q^2$ elements. This order counting holds when $q$ is an arbitrary integer.
Once again let $\overline{a+bi}$ be a typical (nonzero) element. Since $q$ is irreducible thus prime thus $\mathbb{Z}[i]/(q)$ is an integral domain, we have $\overline{(a+bi)(a-bi)}≠\overline{0}$. Letting $v^{-1}$ denote the inverse of $\overline{(a+bi)(a-bi)}$ modulo $q$, we have $\overline{(a+bi)(v^{-1}(a-bi))}=\overline{1}$, so that $\mathbb{Z}[i]/(q)$ is a field.
Alternatively, by a previous exercise $\mathbb{Z}[i]/I$ is finite for any nonzero ideal $I$, so these finite integral domains are clearly fields.
(c) Since $(\pi)$ is prime and thus maximal by a property of PIDs, and since $\pi$ is not associate to $\overline{\pi}$ i.e. $\pi r ≠ \overline{\pi}$ for any $r∈\mathbb{Z}[i]$, we have $(\pi,\overline{\pi})=\mathbb{Z}[i]$ so that the Chinese Remainder Theorem holds by $(\pi)(\overline{\pi})=(\pi\overline{\pi})=(p)$. Parallel to the proof above, $\mathbb{Z}[i]/(p)$ is of order $p^2$ and since $(\pi),(\overline{\pi})≠\mathbb{Z}[i]$ we have they are both finite integral domains (fields) of order $p$.
(7) (a) The containment should be evident, so we must demonstrate the group isomorphism by $φ(x+(\pi))=x\pi^n+(\pi^{n+1})$. Well-definedness:$$x+(\pi)=y+(\pi)⇒x=y+\pi r⇒\pi r = x-y ⇒ \pi^{n+1} r = x\pi^n-y\pi^n⇒$$$$x\pi^n+(\pi^{n+1})=y\pi^n+(\pi^{n+1})⇒φ(x+(\pi))=φ(y+(\pi))$$Homomorphism:$$φ(x+(\pi)+y+(\pi))=φ((x+y)+(\pi))=$$$$x\pi^n+y\pi^n+(\pi^{n+1})=φ(x+(\pi))+φ(y+(\pi))$$ Surjectivity: $$x∈(\pi^n)/(\pi^{n+1})⇒x=\pi^n r + (\pi^{n+1})⇒φ(r+(\pi))=x$$ Injectivity: $$φ(x+(\pi))=(\pi^{n+1})⇒x\pi^n∈(\pi^{n+1})⇒\pi \mid x⇒x+(\pi)=0+(\pi)$$ (b) Assume $\pi$ is not associate to $1+i$. Let $N(\pi)=p$. We claim $\mathbb{Z}[i]/(p^n)≅\mathbb{Z}[i]/(\pi^n) \times \mathbb{Z}[i]/(\overline{\pi^n})$. The greatest common divisor of $\pi^n$ and $\overline{\pi^n}$ according to proposition 13 is $1$, and since $\mathbb{Z}[i]$ is a Euclidean Domain it can written as a linear combination of $\pi^n$ and $\overline{\pi^n}$, implying $(\pi^n,\overline{\pi^n})=\mathbb{Z}[i]$. Since $(\pi^n)(\overline{\pi^n})=(p^n)$, the Chinese Remainder Theorem takes care of the claim.
Now, we have $\mathbb{Z}/(p^n)$ is of order $p^{2n}$, therefore $|\mathbb{Z}[i]/(\pi^n)| \cdot |\mathbb{Z}[i]/(\overline{\pi^n})| = p^{2n}$ and in particular these former two are orders of $p$ power. We are now in a position to prove our claim by induction; assuming the claim doesn't hold for some particular $n$, we have either $|\mathbb{Z}[i]/(\pi^n)| < p^n$ or $|\mathbb{Z}[i]/(\overline{\pi^n})| < p^n$. Without loss of generality we can assume the former, so that $|\mathbb{Z}[i]/(\pi^n)|=|\mathbb{Z}[i]/(\pi^m)|$ for some $m < n$ by induction. Write $\mathbb{Z}[i]/(\pi^m)≅(\mathbb{Z}[i]/(\pi^n))/((\pi^m)/(\pi^n))$ so that necessarily $(\pi^m) \subseteq (\pi^n)$, a contradiction by $\pi^m \not ∈ (\pi^n)$.
So assume $\pi=1+i$, which by the associativity of $1+i$ to $1-i$ implies this is the last case we have to consider. We see that this is the case when $n=2m$ is even, as $\mathbb{Z}[i]/(\pi^{2m})=\mathbb{Z}/((\pi^2)^m)=\mathbb{Z}/(2^m)$, which has $2^{2m}=2^n=|\mathbb{Z}[i]/(\pi)|^n$ elements as claimed.
So assume $n=2m+1$. We first show that $|\mathbb{Z}[i]/(\pi^{2m+1})|$ is of $2$ power; by way of contradiction, assume there is some prime $p$ dividing the order of this quotient. Since the quotient may be considered a group under addition, where exponentiation of addition is multiplication, by Cauchy's Theorem we have some $\overline{a+bi}∈\mathbb{Z}[i]/(\pi^{2m+1})$ such that $\overline{p}$ is the smallest positive integer for which $\overline{p(a+bi)}=\overline{0}$. In other words, $\pi^{2m+1}$ doesn't divide $a+bi$ so that there are less than $2m+1$ factors of $\pi$ in the decomposition of $a+bi$, yet since $\pi \not \mid p$ for any $p≠2$ we must also have $\pi^{2m+1} \not \mid p(a+bi)$, a contradiction by $\overline{p(a+bi)}=\overline{0}$.
We have $(\pi^{2m+2}) \subset (\pi^{2m+1}) \subset (\pi^{2m})$, so that $$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m})/(\pi^{2m+2}))| <$$$$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m+1})/(\pi^{2m+2}))| <$$$$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m+2})/(\pi^{2m+2}))|$$implying $$|\mathbb{Z}[i]/(\pi^{2m})| = 2^{2m} < |\mathbb{Z}[i]/(\pi^{2m+1})| < |\mathbb{Z}[i]/(\pi^{2m+2})| = 2^{2m+2}$$Since $|\mathbb{Z}[i]/(\pi^{2m+1})|$ is of $2$ power, we must have $|\mathbb{Z}[i]/(\pi^{2m+1})|=2^{2m+1}=|\mathbb{Z}[i]/(\pi)|^{2m+1}$.
(c) Let $\alpha = \pi_1^{a_1}...\pi_n^{a_n}$ be its decomposition, so that $|\mathbb{Z}[i]/(\alpha)|=|\mathbb{Z}[i]/((\pi_1^{a_1})...(\pi_n^{a_n}))|$, the latter of which we claim is equivalent to $|\mathbb{Z}[i]/(\pi_1^{a_1})| ... |\mathbb{Z}[i]/(\pi_n^{a_n})|$ by the Chinese Remainder Theorem; taking any $(\pi_x^{a_x})$ and $(\pi_y^{a_y})$ with $x≠y$ we can see $(\pi_x^{a_x},\pi_y^{a_y})=\mathbb{Z}[i]$ since the greatest common divisor of the pair is $1$. The original claim now follows by$$|\mathbb{Z}[i]/(\alpha)|=|\mathbb{Z}[i]/(\pi_1^{a_1})...(\pi_n^{a_n})|=|\mathbb{Z}[i]/(\pi_1^{a_1})| ... |\mathbb{Z}[i]/(\pi_n^{a_n})| =$$$$|\mathbb{Z}[i]/(\pi_1)|^{a_1} ... |\mathbb{Z}[i]/(\pi_n)|^{a_n}=N(\pi_1)^{a_1} ... N(\pi_n)^{a_n} = N(\alpha)~\square$$
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