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Saturday, May 11, 2013

Quotients of the Gaussian Integers (8.3.6-7)

Dummit and Foote Abstract Algebra, section 8.3, exercises 6-7:

MathJax TeX Test Page 6. (a) Prove Z[i]/(1+i) is a field of order 2.
(b) Let qZ be a prime with q3mod4. Prove Z[i]/(q) is a field of order q2.
(c) Let pZ be a prime with p1mod4 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 πZ[i] be irreducible.
(a) For n0 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 α0 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)(1i)+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+biZ[i]. Let a=a+vaq and b=b+vbq with 0a,b<q. We have a+bi=(a+vaq)+(b+vbq)i=a+bi+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 rZ[i]. We therefore have (ac)+(bd)i=qr, i.e. q divides (ac) and (bd). 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)(abi)¯0. Letting v1 denote the inverse of ¯(a+bi)(abi) modulo q, we have ¯(a+bi)(v1(abi))=¯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 rZ[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+πrπr=xyπn+1r=xπnyπnxπn+(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)πxx+(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 1i 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+biZ[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 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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