## 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.