## Tag Archives: quadratic integer ring

### Exhibit an algebraic integer in a quadratic integer ring having a given norm and trace

Find an algebraic integer $\alpha \in \mathbb{Q}(\sqrt{D})$ having norm 31 and trace 17.

Let $\alpha = a+b\sqrt{D}$ and $\overline{\alpha} = a-b\sqrt{D}$. Now $N(\alpha) = \alpha\overline{\alpha} = a^2 - Db^2$ and $\mathsf{tr}(\alpha) = \alpha + \overline{\alpha} = 2a$. We wish to find $a$, $b$, and $D$ such that $a+b\sqrt{D}$ is an algebraic integer, $a^2-Db^2 = 31$, and $2a = 17$. If such an integer exists, then $a = \frac{17}{2}$ is a half-integer- in particular, we must have $D \equiv 1$ mod 4. Substituting, and letting $b = \frac{b_0}{2}$, we have $Db_0^2 = 165 = 3 \cdot 5 \cdot 11$. Thus $D = 165$ and $b_0 = \pm 1$, so that $b = \pm \frac{1}{2}$. Indeed, we can verify that $N(\frac{17}{2} + \frac{1}{2}\sqrt{165}) = 31$ and $\mathsf{tr}(\frac{17}{2} + \frac{1}{2}\sqrt{165}) = 17$.

### Factor the principal ideals in an algebraic integer ring which are generated by ramified rational primes

Let $K = \mathbb{Q}(\sqrt{-5})$ and let $\mathcal{O}$ be the ring of integers in $K$. Factor $(p)$ in $\mathcal{O}$, where $p$ is a ramified rational prime.

By Theorem 9.6, if $p$ does not divide the discriminant $d = -20$ of $K$ (using Theorem 6.11), then $p$ is not ramified. Now $20 = 2^2 \cdot 5$, so that if $p$ is ramified in $K$, it is either 2 or 5.

We claim that $(2) = (2,1+\sqrt{-5})^2$. Indeed, the $(\supseteq)$ direction is clear, and we have $2 = -(1+\sqrt{-5})^2 + 2 \cdot (1+\sqrt{-2}) - 2 \cdot 2$. We claim also that $P = (2,1+\sqrt{-5})$ is maximal. To this end, let $a+b \sqrt{-5} \in \mathcal{O}$, and say $a - b \equiv c$ mod 2 where $c \in \{0,1\}$. Evidently, $a+b\sqrt{-5} \equiv c$ mod $P$; if $c \equiv 0$ mod 2 then $a+b\sqrt{-5} \equiv 0$ mod $P$, and if $c \equiv 1$ mod 2 then $a+b\sqrt{-5} \equiv 1$ mod $P$. Now suppose $1 \in P$; then $1 = 2(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5})$. Comparing coefficients mod 2, we have $0 \equiv h+k \equiv 1$ mod 2, a contradiction. So $\mathcal{O}/(2,1+\sqrt{-5}) = \{ \overline{0}, \overline{1}\}$, and thus $\mathcal{O}/(2,1+\sqrt{-5}) \cong \mathbb{Z}/(2)$ is a field. Hence $P$ is maximal, and $(2) = (2,1+\sqrt{-5})^2$ is the prime factorization of $(2)$.

Certianly $(5) = (\sqrt{-5})^2$. We claim that $Q = (\sqrt{-5})$ is prime. To see this, note that $5 \in (\sqrt{-5})$. If $a+b\sqrt{-5} \in \mathcal{O}$, then $a+b\sqrt{-5} \equiv a \equiv a_0$ mod $Q$, where $a_0 \in \{0,1,2,3,4\}$. Suppose $t \in Q \cap \mathbb{Z}$. Then $t = (a+b\sqrt{-5})\sqrt{-5}$ for some $a,b \in \mathbb{Z}$. Comparing coefficients, we have $t \equiv 0$ mod 5. In particular, $r \not\equiv s$ mod $Q$, where $r,s \in \{0,1,2,3,4\}$ are distinct. Thus $\mathcal{O}/(\sqrt{-5}) \cong \mathbb{Z}/(5)$ is a field, so that $(\sqrt{-5})$ is maximal. Thus $(5) = (\sqrt{-5})^2$ is the prime factorization of $(5)$.

### Factor (10) in QQ(sqrt(-6))

Find two distinct factorizations of 10 in $\mathbb{Q}(\sqrt{-6})$. Factor 10 as a product of four ideals in $\mathbb{Q}(\sqrt{-6})$.

Evidently $10 = 2 \cdot 5 = (2 + \sqrt{-6})(2 - \sqrt{-6})$; moreover, these factorizations are distinct since, as the only units in $\mathbb{Q}(\sqrt{-6})$ are $\pm 1$, 2 is not associate to either of $2 \pm \sqrt{-6}$.

We claim that $(10) = (2,2+\sqrt{-6})(2,2-\sqrt{-6})(5,2+\sqrt{-6})(5,2-2\sqrt{-6})$. To see the $(\supseteq)$ direction, note that this ideal product is generated by all $2^4$ possible selections of one generator from each factor. Name the generators $\alpha_1,\alpha_2,\beta_1,\beta_2$ such that $\alpha_1\alpha_2 = \beta_1\beta_2 = 10$. Now any selection which includes $\alpha_1$ (without loss of generality) and does not include $\alpha_2$ must include both $\beta_1$ and $\beta_2$, and so this ideal product is in $(10)$.

Note also that $5 \cdot 5 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) - 6 \cdot 2 \cdot 2 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) = 10$.

### Show that two ideals are distinct

Let $\mathcal{O}$ be the ring of integers in $\mathbb{Q}(\sqrt{-5})$. Show that the ideals $(3,1+2\sqrt{-5})$ and $(7,1-2\sqrt{-5})$ are distinct.

Suppose to the contrary that these ideals are equal. Then in particular, we have $3 = 7(a+b\sqrt{-5}) + (1-2\sqrt{-5})(c+d\sqrt{-5})$. Comparing coefficients yields the two integer equations $7a+c+10d = 0$ and $7b+d-2c = 0$. Mod 7, we have $c+3d\equiv 3$ and $d+5c \equiv 0$. Solving this system by substitution yields $0 \equiv 3$ mod 7; hence the system has no solution.

So these ideals are distinct.

### Compute the discriminant of a basis for a given ideal in a quadratic integer ring

Let $\mathcal{O}$ be the ring of integers in $\mathbb{Q}(\sqrt{D})$. Find a basis for the ideal $(4)$ over $\mathbb{Z}$. Then compute the discriminant of this basis.

Suppose first that $D \not\equiv 1$ mod 4.

Note that $(4) = \{4a + 4b\sqrt{D} \ |\ a,b \in \mathbb{Z} \}$. We claim that $\{4,4\sqrt{D}\}$ is a basis for $(4)$. Certainly $(4,4\sqrt{D})_\mathbb{Z} = (4)$. Now if $4a+4b\sqrt{D} = 0$ and $b \neq 0$, then $\sqrt{D} = -a/b$ is rational, a contradiction. So $4a = 0$, and $a = b = 0$. So $\{4,4\sqrt{D}\}$ is a basis for $(4)$ over $\mathbb{Z}$. Evidently, the discriminant of this basis is $\Delta[4,4\sqrt{D}] = 1024D$. Recall from Theorem 6.11 in TAN that the discriminant of $\mathbb{Q}(\sqrt{D})$ is $\Delta_\mathcal{O} = 4D$. Thus $\Delta[4,4\sqrt{D}] = 16^2 \Delta_\mathcal{O}$.

Now suppose $D \equiv 1$ mod 4.

Now $(4) = \{ 4a + 2b(1+\sqrt{D}) \ |\ a,b \in \mathbb{Z}\}$. We claim that $\{4,2+2\sqrt{D}\}$ is a basis. Certainly $(4,2+2\sqrt{D})_\mathbb{Z} = (4)$. If $4a + 2b + 2b\sqrt{D} = 0$ and $b \neq 0$, then $-(2a+b)/b = \sqrt{D}$, a contradiction. So $b = 0$, and thus $a = b = 0$. Hence $\{4,2+2\sqrt{D}\}$ is a basis for $(4)$. Evidently the discriminant of this basis is $\Delta[4,2+2\sqrt{D}] = 256D$. Again recall from Theorem 6.11 in TAN that $\Delta_\mathcal{O} = D$; hence $\Delta[4,2+2\sqrt{D}] = 16^2 \Delta_\mathcal{O}$.

### Find a basis for a given ideal in a quadratic integer ring

Let $I = (3+i,7+i)$ be considered as an ideal in $\mathbb{Z}[i]$. Find a generator and a basis for $I$. Draw a diagram to visualize the elements of $I$ on the complex plane.

Note that $7+i = (1+i)(2-i)$ and $3+i = (1+i)(4-3i)$, so that $I \subseteq (1+i)$. Conversely, since $-2i(3+i) + i(7+i) = 1+i$, we have $(1+i) \subseteq I$. Hence $I = (1+i)$.

We claim that $B = \{1+i, -1+i\}$ is a basis for $I$ over $\mathbb{Z}$. To see this, note that $-1+i = i(1+i)$, so that $(1+i,-1+i)_\mathbb{Z} \subseteq (1+i)$; certainly $(1+i) \subseteq (1+i, -1+i)_\mathbb{Z}$, so these sets are equal. Now suppose $a(1+i) + b(-1+i) = 0$. Comparing coefficients, we have $a-b = 0$ and $a+b = 0$, so that $a = b = 0$. Thus $B$ is free as a generating set for $I$ over $\mathbb{Z}$.

We can visualize this ideal in the complex plane as in the following diagram.

The ideal (1+i) in the Gaussian integers

### Show that a given generating set is a basis for an ideal in a quadratic integer ring

Show that $B = \{5+i, 2+3i\}$ is a basis for the ideal $(2+3i)$ in $\mathbb{Z}[i]$.

First we wish to show that $(5+i,2+3i)_\mathbb{Z} = (2+3i)$; the $(\supseteq)$ direction is clear. Now suppose $\zeta = a(5+i) + b(2+3i) = (5a+2b) + (a+3b)i$; note that $\zeta = (2+3i)(a+b-ai) \in (2+3i)$, as desired.

Now suppose we have integers $a,b \in \mathbb{Z}$ such that $(5+1)a + (2+3i)b = 0$. Comparing coefficients, we have $5a+2b = 0$ and $a+3b = 0$. Evidently this system has only the solution $a = b = 0$, so that $B$ is $\mathbb{Z}$-linearly independent. Hence $B$ is a basis for $(2+3i)$.

### Find a generator and a basis for a given ideal in a quadratic integer ring

Show that $A = \{4a+2\sqrt{2}b \ |\ a,b \in \mathbb{Z}\}$ is an ideal in the ring of algebraic integers in $\mathbb{Q}(\sqrt{2})$. Find a generator and a basis for this ideal.

First, since $2 \not\equiv 1$ mod 4, the elements of $A$ are certainly algebraic integers. (Integers in $\mathbb{Q}(\sqrt{2})$ have the form $m+n\sqrt{2}$ with $m,n \in \mathbb{Z}$.) We claim that $A = (2\sqrt{2})$. To see this, note that if $4a+2\sqrt{2}b \in A$, then $4a+2\sqrt{2}b = 2\sqrt{2}(b+a\sqrt{2}) \in (2\sqrt{2})$. Conversely, if $2\sqrt{2}(a+b\sqrt{2}) \in (2\sqrt{2})$, then $2\sqrt{2}(a+b\sqrt{2}) = 4b+2\sqrt{2}a \in A$. So $A = (2\sqrt{2})$, and more generally, $A$ is an ideal in the ring of integers in $\mathbb{Q}(\sqrt{2})$.

Now we claim that $B = \{4, 2\sqrt{2}\}$ is a basis of $A$. It is certainly a generating set over $\mathbb{Z}$. Suppose now that $4a + 2\sqrt{2}b = 0$. If $b \neq 0$, then $\sqrt{2} = -4a/2b$, so that $\sqrt{2}$ is rational, a contradiction. So $b = 0$, and we have $4a = 0$, so $a = 0$. Thus $B$ is $\mathbb{Z}$-linearly independent, and hence a basis for $A$.

### Factor some given algebraic integers in a quadratic field

Factor $5 + \sqrt{3}$ in $\mathbb{Q}(\sqrt{3})$ and $7 + \sqrt{-5}$ in $\mathbb{Q}(\sqrt{-5})$.

Note that $N(5+\sqrt{3}) = 5^2 - 3 = 22 = 2 \cdot 11$. If $\alpha = \beta\gamma$, then $N(\beta)N(\gamma) = 2 \cdot 11$, so (without loss of generality) $N(\beta) = 2$. Suppose $\beta = h+k\sqrt{3}$; then $h^2-3k^2 = 2$. Mod 3, we have $h^2 \equiv 2$. This is a contradiction since 2 is not a square mod 3. That is, no algebraic integer in $\mathbb{Q}(\sqrt{3})$ has norm 2. (Similarly, no integer has norm 11.) Thus $5+\sqrt{3}$ is irreducible as an integer in $\mathbb{Q}(\sqrt{3})$.

Evidently, $(1+\sqrt{-5})(2-\sqrt{-5}) = 7+\sqrt{-5}$. Now $N(1+\sqrt{-5}) = 6$ and $N(2-\sqrt{-5}) = 9$. We claim that no integer in $\mathbb{Q}(\sqrt{-5})$ has norm 3; to that end, suppose $h^2+5k^2 = 3$. We must have $k = 0$, so that $h^2 = 3$, a contradiction. In particular, both $1+\sqrt{-5}$ and $2-\sqrt{-5}$ are irreducible in $\mathbb{Q}(\sqrt{-5})$.

### The polynomial x² – √2 is irreducible over ZZ[√2]

Prove that $p(x) = x^2 - \sqrt{2}$ is irreducible over $\mathbb{Z}[\sqrt{2}]$.

In this previous exercise, we showed that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain, and thus a unique factorization domain.

If $p(x)$ is reducible, then it has a root; say $\alpha = a+b\sqrt{2}$. Evidently, then, we have $(a^2 + 2b^2) + 2ab \sqrt{2} = \sqrt{2}$. Comparing coefficients, $2ab = 1$ for some integers $a$ and $b$, a contradiction. Thus $p(x)$ is irreducible over $\mathbb{Z}[\sqrt{2}]$.