Tag Archives: gaussian integers

Use the Dedekind-Hasse criterion to show that ZZ[i] is a unique factorization domain

Use the Dedekind-Hasse criterion to show that \mathbb{Z}[i] is a unique factorization domain.

Let \alpha,\beta \in \mathbb{Z}[i] be nonzero such that \beta does not divide \alpha and N(\alpha) > N(\beta). By the division algorithm in \mathbb{Z}[i], there exist \gamma,\delta such that \alpha = \gamma\beta + \delta and 0 < N(\delta) < N(\beta). So 0 < N(\alpha - \gamma\beta) < N(\delta). So \mathbb{Z}[i] satisfies the Dedekind-Hasse criterion, and thus is a unique factorization domain.

Note that this is a somewhat silly way to prove this result; the division algorithm makes \mathbb{Z}[i] a Euclidean domain and thus a unique factorization domain via some more general theorems.

Every ideal in ZZ[i] is principal

Prove that every ideal in \mathbb{Z}[i] is principal.

We showed here that \mathbb{Z}[i] is a Euclidean domain. Thus it is a principal ideal domain.

The ring of algebraic integers in QQ(i) is ZZ[i]

Prove that the ring of algebraic integers in \mathbb{Q}(i) is \mathbb{Z}[i].

Recall that an element in an algebraic extension of \mathbb{Q} is called an algebraic integer if its minimal polynomial over \mathbb{Q} has only integer coefficients.

First we will show that the minimal polynomial of every element of \mathbb{Z}[i] has only integer coefficients. To that end, let z = a+bi \in \mathbb{Z}[i]. If b = 0, then certainly z = a \in \mathbb{Z} is an algebraic integer, since its minimal polynomial is simply x - a. Suppose b \neq 0; in this case z \notin \mathbb{Q}, and so its minimal polynomial has degree at least 2. Note that z is a root of p(x) = x^2 - 2ax + (a^2 + b^2), and p(x) \in \mathbb{Z}[x]; so z has degree exactly 2, and in fact p(x) is its minimal polynomial. So z is an algebraic integer.

Conversely, suppose z = s+ti \in \mathbb{Q}(i) is an algebraic integer. If t = 0, then s \in \mathbb{Q} is an algebraic integer, and so s \in \mathbb{Z}. Certainly then z \in \mathbb{Z}[i]. Suppose now that t \neq 0. If s = 0, then z = ti is a root of x^2 + t^2, which is irreducible over \mathbb{Q} and hence the minimal polynomial of z. So t^2 \in \mathbb{Z}. If t is not an integer, then neither is t^2; hence t \in \mathbb{Z}, and so z \in \mathbb{Z}[i]. So we assume t \neq 0.

Now z \notin \mathbb{Q}, so the minimal polynomial of z over \mathbb{Q} has degree at least 2. Since z is (evidently) a root of p(x) = x^2 - 2sx + s^2+t^2, this must be the minimal polynomial of z. Since z is an algebraic integer, we have 2s, s^2+t^2 \in \mathbb{Z}.

Let us say s = a/b and t = c/d, where a,b,c,d \in \mathbb{Z} and \mathsf{gcd}(a,b) = \mathsf{c,d} = 1. Then (rewriting) we have \frac{2a}{b} \in \mathbb{Z}, so that b|2a. Let p be a prime factor of b; then p|2a. Since a and b are relatively prime, p|2, and in fact p = 2. So b = 2^k for some k \geq 0. However, if k \geq 2, then we have 2|a, a contradiction since a and b are relatively prime. So b = 1 or b = 2. Suppose b = 2.

We also have that s^2+t^2 \in \mathbb{Z}, so that b^2d^2|(a^2d^2 + c^2b^2). Say 4d^2k = a^2d^2 + 4c^2; rearranging, we have d^2(4k - a^2) = 4c^2. Let p be a prime factor of d. Again because c and d are relatively prime, we have p|4, so that p = 2. So d = 2^k for some k \geq 0. If k \geq 2, then we have 2|c$, a contradiction. So d = 1 or d = 2. Suppose d = 2.

Now (a^2+c^2)/4 \in \mathbb{Z}, so that a^2 + c^2 = 4k for some k \in \mathbb{Z}. In particular, a^2 + c^2 \equiv 0 mod 4. Note that the squares mod 4 are precisely 0 and 1, and the only solutions of this equivalence are (a,c) \in \{(0,0), (2,0), (0,2), (2,2)\}. In any case, a and c are either 0 (a contradiction) or divisible by 2 (a contradiction since they are relatively prime to b and d, respectively.) Now d = 1. Again, we have a^2 + 4c^2 = 4k for some integer k, so that a^2 \equiv 0 mod 4, a contradiction. So b = 1.

Thus z = s+ti \in \mathbb{Z}[i] as desired.

Hence, the ring of algebraic integers in \mathbb{Q}(i) is precisely \mathbb{Z}[i].

As a ring, the ZZ-tensor product of ZZ[i] and RR is isomorphic to CC

Let \mathbb{Z}[i] and \mathbb{R} be \mathbb{Z}-algebras via the inclusion map. Prove that as rings, \mathbb{Z}[i] \otimes_\mathbb{Z} \mathbb{R} and \mathbb{C} are isomorphic.

Define \varphi : \mathbb{Z}[i] \times \mathbb{R} \rightarrow \mathbb{C} by \varphi(z,x) = zx. Certainly this mapping is \mathbb{Z}-bilinear, and so induces a \mathbb{Z}-algebra homomorphism \Phi : \mathbb{Z}[i] \otimes_\mathbb{Z} \mathbb{R} \rightarrow \mathbb{C} such that \Phi(z \otimes x) = zx.

Note that every simple tensor (hence every element) of \mathbb{Z}[i] \otimes_\mathbb{Z} \mathbb{R} can be written in the form 1 \otimes x + i \otimes y, where x,y \in \mathbb{R}. Now \Phi(1 \otimes x + i \otimes y) = x+iy, so that \Phi is surjective. Suppose now that 0 = \Phi(1 \otimes x + i \otimes y) = x+iy; then x = y = 0, so that 1 \otimes x + i \otimes y = 0. Thus \mathsf{ker}\ \Phi = 0, and so \Phi is injective. Thus \Phi is a ring isomorphism.

QQ[i] is a field

Verify that \mathbb{Q}[i] is a subfield of \mathbb{C}.

It suffices to show that \mathbb{Q}[i] is closed under subtraction and division.

To that end, suppose \alpha = p_1+p_2i,\beta = q_1+q_2 i \in \mathbb{Q}[i].

\alpha - \beta = (p_1-q_1) + (p_2-q_2)i \in \mathbb{Q}[i].

If \beta \neq 0, then q_1^2+q_2^2 > 0.

\frac{\alpha}{\beta} = \frac{p_1q_1-p_2q_2}{q_1^2+q_2^2} + \frac{p_1q_2 + p_2q_1}{q_1^2+q_2^2}i \in \mathbb{Q}[i].

So \mathbb{Q}[i] is a subfield of \mathbb{C}.

Exhibit a residue system for a given quotient of the Gaussian integers

Exhibit a complete residue system for \mathbb{Z}[i]/(2+i). Plot the points \{a+bi \ |\ 0 \leq |a|,|b| \leq 4 \} in the complex plane, coloring each point according to which coset of (2+i) it is in.

Note that N(2+i) = 5. Now each residue class mod 2+i has a representative whose norm is strictly less than 5 by the division algorithm. It is easy to see then that the distinct residue classes have representatives in the set \{(0,0), (\pm 1,0), (\pm 2,0), (\pm 1, \pm 1), (0, \pm 1), (0,\pm 2) \}. We claim that all but 5 of these are redundant. Indeed:

  • (2) - (-i) = 2+i and (-i) - (-1+i) = (2+i)(-i).
  • (-2) - (i) = -(2+i) and (i) - (1-i) = (2+i)i
  • (-1) - (1+i) = -(2+i) and (-2i) - (-i) = (2+i)(-1)
  • (1) - (-1-i) = 2+i and (2i) - (1) = (2+i)i

Thus a complete residue system is contained in the set \{0, 1, -1, i, -i\}. We claim that this set is itself a residue system. To see this, we need to verify that no two are congruent mod 2+i. First, since 2+i is not a unit, none of 1, -1, i, and -i is congruent to 0 mod 2+i. Now (1) - (i) = 1-i has norm 2, and thus cannot be divisible by 2+i which has norm 5. Similarly, (1) - (-1) = 2 has norm 4, (1) - (-i) = 1+i has norm 2, (-1) - (i) = -1-i has norm 2, (-1) - (-i) = -1+i has norm 2, and (i) - (-i) = 2i has norm 4. Thus these representatives are distinct mod 2+i. Hence \{0,1,-1,i,-i\} is a complete residue system mod 2+i.

Note that \alpha \equiv \alpha + (2+i) \mod (2+i). Likewise, we may add any associate to \alpha and remain in the same congruence class. If we color the elements congruent to 0 green, to 1 purple, to i red, to -1 blue, and to -i orange, then a small region around the origin in the complex plane appears as follows.

Solve a linear congruence over the Gaussian integers

Solve the following congruence over \mathbb{Z}[i]: (3-2i)x \equiv 1 \mod (1-2i).

As a consequence of Fermat’s Little Theorem for \mathbb{Z}[i], we have x \equiv (3-2i)^{N(1-2i)-2} \mod (1-2i). Now N(1-2i) = 5, and (3-2i)^3 = -9-46i \equiv i \mod (1-2i). Indeed, (3-2i)i - 1 = (1-2i)(-1+i).

The complete set of solutions in \mathbb{Z}[i] is \{ (1-2i)\delta + i \ |\ \delta \in \mathbb{Z}[i] \}.

Show that one Gaussian integer divides another

Prove that -1+2i divides 7^16-1 in \mathbb{Z}[i].

First, we claim that 5|(7^16-1). To see this, note that 7^{16}-1 \equiv 2^{16}-1 \equiv (2^4)^4-1 \mod 5. By Fermat’s Little Theorem, 2^4 \equiv 1 \mod 5. Thus 7^{16}-1 \equiv 0 \mod 5 as desired. Note also that -1+2i divides 5 in \mathbb{Z}[i]; in particular, (-1+2i)(-1-2i) = 5. Since divisibility is transitive, -1+2i divides 7^{16}-1.

Gaussian integers having relatively prime norms are relatively prime

Let \alpha,\beta \in \mathbb{Z}[i]. Prove that if N(\alpha) and N(\beta) are relatively prime integers, then \alpha and \beta are relatively prime in \mathbb{Z}[i].

Suppose \gamma is a common factor of \alpha and \beta. Say \gamma\xi = \alpha and \gamma\eta = \beta. In particular, note that N(\gamma)|N(\alpha) and N(\gamma)|N(\beta). Since N(\alpha) and N(\beta) are relatively prime, N(\gamma) = 1. Thus \gamma is a unit, and so \alpha and \beta are relatively prime in \mathbb{Z}[i].

Gaussian integers which are relatively prime need not have relatively prime norms

Exhibit two Gaussian integers \alpha and \beta which are relatively prime in \mathbb{Z}[i] but such that N(\alpha) and N(\beta) are not relatively prime integers.

Note that 1+2i and 1-2i have norm 5. Using this previous exercise, 1+2i and 1-2i do not divide each other, and thus are relatively prime. In particular, 1+2i = (-1+i)(1-2i) + (-i). However, these elements both have norm 5, and so their norms are not relatively prime.