(b) Let q∈Z be a prime with q≡3mod4. Prove Z[i]/(q) is a field of order q2.
(c) Let p∈Z be a prime with p≡1mod4 and let p=π¯π be its factorization in Z[i]. Prove Z[i]/(p)≅Z[i]/(π)×Z[i]/(¯π) as rings, Z[i]/(p) is of order p2 and therefore Z[i]/(π) and Z[i]/(¯π) are fields of order p.
7. Let π∈Z[i] be irreducible.
(a) For n≥0 an integer, show that (πn+1)⊆(πn) and that multiplication by πn induces an isomorphism Z[i]/(π)≅(πn)/(πn+1) as additive abelian groups.
(b) Prove |Z[i]/(πn)|=|Z[i]/(π)|n.
(c) Prove for α≠0 that |Z[i]/(α)|=N(α).
Proof: (6) (a) By the division algorithm, every element in 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)⊂Z[i], so that the order must be 2. Since 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∈Z[i]. Let a=a′+vaq and b=b′+vbq with 0≤a′,b′<q. We have a+bi=(a′+vaq)+(b′+vbq)i=a′+b′i+q(va+vbi), so that in this sense a+bi can be reduced modulo q. This allows a maximum order of q2. Now, assume reduced ¯a+bi=¯c+di in Z[i]/(q), so that a+bi=c+di+qr for some r∈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 q2 elements. This order counting holds when q is an arbitrary integer.
Once again let ¯a+bi be a typical (nonzero) element. Since q is irreducible thus prime thus Z[i]/(q) is an integral domain, we have ¯(a+bi)(a−bi)≠¯0. Letting v−1 denote the inverse of ¯(a+bi)(a−bi) modulo q, we have ¯(a+bi)(v−1(a−bi))=¯1, so that Z[i]/(q) is a field.
Alternatively, by a previous exercise Z[i]/I is finite for any nonzero ideal I, so these finite integral domains are clearly fields.
(c) Since (π) is prime and thus maximal by a property of PIDs, and since π is not associate to ¯π i.e. πr≠¯π for any r∈Z[i], we have (π,¯π)=Z[i] so that the Chinese Remainder Theorem holds by (π)(¯π)=(π¯π)=(p). Parallel to the proof above, Z[i]/(p) is of order p2 and since (π),(¯π)≠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+(π))=xπn+(πn+1). Well-definedness:x+(π)=y+(π)⇒x=y+πr⇒πr=x−y⇒πn+1r=xπn−yπn⇒xπn+(πn+1)=yπn+(πn+1)⇒φ(x+(π))=φ(y+(π))Homomorphism:φ(x+(π)+y+(π))=φ((x+y)+(π))=xπn+yπn+(πn+1)=φ(x+(π))+φ(y+(π)) Surjectivity: x∈(πn)/(πn+1)⇒x=πnr+(πn+1)⇒φ(r+(π))=x Injectivity: φ(x+(π))=(πn+1)⇒xπn∈(πn+1)⇒π∣x⇒x+(π)=0+(π) (b) Assume π is not associate to 1+i. Let N(π)=p. We claim Z[i]/(pn)≅Z[i]/(πn)×Z[i]/(¯πn). The greatest common divisor of πn and ¯πn according to proposition 13 is 1, and since Z[i] is a Euclidean Domain it can written as a linear combination of πn and ¯πn, implying (πn,¯πn)=Z[i]. Since (πn)(¯πn)=(pn), the Chinese Remainder Theorem takes care of the claim.
Now, we have Z/(pn) is of order p2n, therefore |Z[i]/(πn)|⋅|Z[i]/(¯πn)|=p2n 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 |Z[i]/(πn)|<pn or |Z[i]/(¯πn)|<pn. Without loss of generality we can assume the former, so that |Z[i]/(πn)|=|Z[i]/(πm)| for some m<n by induction. Write Z[i]/(πm)≅(Z[i]/(πn))/((πm)/(πn)) so that necessarily (πm)⊆(πn), a contradiction by πm∉(πn).
So assume π=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 Z[i]/(π2m)=Z/((π2)m)=Z/(2m), which has 22m=2n=|Z[i]/(π)|n elements as claimed.
So assume n=2m+1. We first show that |Z[i]/(π2m+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 ¯a+bi∈Z[i]/(π2m+1) such that ¯p is the smallest positive integer for which ¯p(a+bi)=¯0. In other words, π2m+1 doesn't divide a+bi so that there are less than 2m+1 factors of π in the decomposition of a+bi, yet since π∤p for any p≠2 we must also have π2m+1∤p(a+bi), a contradiction by ¯p(a+bi)=¯0.
We have (π2m+2)⊂(π2m+1)⊂(π2m), so that |(Z[i]/(π2m+2))/((π2m)/(π2m+2))|<|(Z[i]/(π2m+2))/((π2m+1)/(π2m+2))|<|(Z[i]/(π2m+2))/((π2m+2)/(π2m+2))|implying |Z[i]/(π2m)|=22m<|Z[i]/(π2m+1)|<|Z[i]/(π2m+2)|=22m+2Since |Z[i]/(π2m+1)| is of 2 power, we must have |Z[i]/(π2m+1)|=22m+1=|Z[i]/(π)|2m+1.
(c) Let α=πa11...πann be its decomposition, so that |Z[i]/(α)|=|Z[i]/((πa11)...(πann))|, the latter of which we claim is equivalent to |Z[i]/(πa11)|...|Z[i]/(πann)| by the Chinese Remainder Theorem; taking any (πaxx) and (πayy) with x≠y we can see (πaxx,πayy)=Z[i] since the greatest common divisor of the pair is 1. The original claim now follows by|Z[i]/(α)|=|Z[i]/(πa11)...(πann)|=|Z[i]/(πa11)|...|Z[i]/(πann)|=|Z[i]/(π1)|a1...|Z[i]/(πn)|an=N(π1)a1...N(πn)an=N(α) ◻
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