## Category Archives: TAN:PD

### Analyze a proof of a simplification of Kummer’s Theorem

Theorem 11.10 in TAN is the following simplification of Kummer’s Theorem: if $p$ is a regular odd prime, then $x^p + y^p = z^p$ has no solutions such that $p|xyz$. The proof itself, however, does not explicitly appeal to the regularity of $p$. Where is this required?

The regularity of $p$ is implicitly used when we appeal to Corollary 10.5 to show that $[x+\zeta y] = [\delta^p]$. Essentially, we show that $A^p$ is principal for some ideal $A$, but since $p$ does not divide the order of the class group, it must be that $A$ is itself principal.

### A bound on potential positive solutions of x^n + y^n = z^n

Suppose $(x,y,z)$ is a solution in $\mathbb{N}^+$ of the equation $x^n + y^n = z^n$, where $n > 2$. Show that $x,y > n$.

Note that $z > y$.

If $x^n + y^n = z^n$, then $x^n = z^n - y^n$ $= (z-y)(\sum_{k=0}^{n} {n \choose k} z^{n-k}y^k)$ $\geq \sum_{k=0}^{n} {n \choose k} z^{n-k}y^k$ $\geq {n \choose {n-1}} z y^{n-1}$ $= nzy^{n-1} > ny^{n-1}$. Thus $x^n > ny^{n-1}$. Likewise, $y^n > nx^{n-1}$.

Suppose, without loss of generality, that $x > y$. Then $y^n > nx^{n-1} > ny^{n-1}$, and so $y > n$. Thus $x > n$.

### If the roots of a monic irreducible polynomial over ZZ all have modulus at most 1, then the roots are roots of unity

Let $p(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial and suppose that for each root $\zeta$ of $p(x)$, we have $||\zeta || \leq 1$, where $|| \cdot ||$ denotes complex modulus. Show that the roots of $p(x)$ are roots of unity.

Note that $p(x)$ factors as $p(x) = \prod (x - \zeta_i)$, where $\zeta_i$ ranges over the roots of $p$. Now the constant coefficient of $p(x)$ is $\prod \zeta_i$, and thus $||\prod \zeta_i|| = \prod ||\zeta_i|| \leq 1$. On the other hand, the constant coefficient is an integer, and so is either 0 or 1. If the constant coefficient is 0, then $x$ divides $p(x)$. So the constant coefficient of $p$ is 1, and we have $||\zeta_i|| = 1$ for all $i$.

By Lemma 11.8, the $\zeta_i$ are all roots of unity.

### The conjugates of a root of unity are roots of unity

Let $\zeta$ be a root of unity. Show that the conjugates of $\zeta$ are also roots of unity.

If $\zeta$ is a root of 1, then $\zeta$ is a root of $p(x) = x^n - 1$ for some $n$. Thus the minimal polynomial of $\zeta$ over $\mathbb{Q}$ is also a root of $p(x)$, so the conjugates of $\zeta$ are roots of unity.

### Find all the Pythagorean triples with one parameter fixed

Find all of the positive Pythagorean triples $(x,y,z)$ where $y = 60$.

First we find all of the positive primitive solutions. If $(x,60,z)$ is a primitive Pythagorean triple, then by Theorem 11.1, we have integers $r$ and $s$ such that $x = r^2 - s^2$, $60 = 2rs$, and $z = r^2 + s^2$, where $0 > s > r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $rs = 30$; there are only four possibilities for $(r,s)$ with these constraints: $(r,s) \in \{(30,1),(15,2),(10,3),(6,5)\}$. These yield the triples $(899,60,901)$, $(221,60,229)$, $(91,60,109)$, and $(11,60,61)$.

If $(x,60,z)$ is not primitive, then $x$, $z$, and $60$ have a greatest common factor $d \neq 1$. There are 11 possibilities: $d \in \{2,3,4,5,6,10,12,15,20,30,60\}$. Then $(x/d,60/d,z/d)$ is primitive. First, we have a lemma.

Lemma: Consider the triple $T = (a,b,c)$ of positive integers. $T$ is not primitive Pythagorean if any of the following hold: (1) if $b$ is odd, (2) if $b = 2m$ where $m$ is odd, and (3) if $b = 2$. Proof: If $(a,b,c)$ is a primitive Pythagorean triple, then we have integers $r$ and $s$ such that $b = 2rs$, $0 < s < r$, $(r,s) = (1)$, and $r+s\equiv 1$ mod 2. If $b$ is odd, we contradict $b = 2rs$. If $b = 2m$ where $m$ is an odd, then $r$ and $s$ are both odd, but then $r+s \equiv 0$ mod 2. If $b = 2$, then $r = s = 1$, a contradiction. $\square$

By the lemma, no primitive Pythagorean triples $(x/d,60/d,z/d)$ exist for $d$ in $\{2,4,6,10,12,20,30,60\}$. We handle each remaining case in turn.

$(d = 3)$ Let $a = x/3$ and $c = z/3$, and suppose $(a,20,c)$ is a primitive Pythagorean triple. By Theorem 11.1, we have integers $r$ and $s$ such that $a = r^2 - s^2$, $20 = 2rs$, $c = r^2+s^2$, $0 < s < r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $rs = 10$; there are two possibilities for $(r,s)$ with these constraints: $(r,s) \in \{(10,1),(5,2)\}$. These yield the solutions $297^2 + 60^2 = 303^2$ and $63^2 + 60^2 = 87^2$.

$(d = 5)$ Let $a = x/5$ and $c = z/5$, and suppose $(a,12,c)$ is a primitive Pythagorean triple. By Theorem 11.1, we have integers $r$ and $s$ such that $a = r^2 - s^2$, $12 = 2rs$, $c = r^2+s^2$, $0 < s < r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $6 = rs$; the only such pair is $(r,s) = (3,2)$, which yields the solution $25^2 + 60^2 = 65^2$.

$(d = 15)$ Let $a = x/15$ and $c = z/15$, and suppose $(a,4,c)$ is a primitive Pythagorean triple. By Theorem 11.1, we have integers $r$ and $s$ such that $a = r^2 - s^2$, $4 = 2rs$, $c = r^2+s^2$, $0 < s < r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $rs = 2$, so that $(r,s) = (2,1)$. This yields the solution $45^2 + 60^2 = 75^2$.

### There are only finitely many Pythagorean triples (x,y,z) with any one parameter fixed

Suppose $(x,y,z)$ is a solution to $x^2 + y^2 = z^2$ in $\mathbb{N}^+$. Show that $x^2 \geq 2y+1$. Deduce that if we fix any of $x$, $y$, or $z$, then there are only finitely Pythagorean triples $(x,y,z)$.

Suppose $x^2 + y^2 = z^2$. Now $x^2 = z^2 - y^2$, and $z > y$. (We may assume that $xyz \neq 0$.) Now $x^2 = (z+y)(z-y) \geq z+y > 2y$. Since $x^2$ and $2y$ are integers, $x^2 \geq 2y+1$.

With $x$ fixed, there are only finitely many possible $y$, as they must satisfy $2y+1\leq x^2$. With $x$ and $y$ fixed, $z$ is determined, so there are only finitely many solutions with fixed $x$. Symmetrically, there are finitely many solutions with fixed $y$.

With $z$ fixed, we have $z > x,y$, so there are certainly only finitely many solutions.

### Exhibit an extended principal generator for an ideal in an algebraic integer ring

Let $A = (3,1+2\sqrt{-5})$ be an ideal in the ring of integers of $K = \mathbb{Q}(\sqrt{-5})$. Find an algebraic integer $\kappa$ such that $A = (\kappa) \cap \mathcal{O}_E$, where $E = K(\kappa)$.

We saw in the text that the class number of $K$ is 2. Note that $A^2 = (3)$, as indeed $3 = (1+2\sqrt{-5})^2 - 2 \cdot 3^2$. Let $\kappa = \sqrt{3}$. By our proof of Theorem 10.6, $A = \{ \tau \in \mathcal{O}_K \ |\ \tau/\kappa \in \mathcal{O}_K \}$.

### If A² is principal over an algebraic number field, then A² is principal over any extension

Let $K_1$ and $K_2$ be algebraic number fields with $K_1 \subseteq K_2$, and having integer rings $\mathcal{O}_1$ and $\mathcal{O}_2$, respectively. Suppose $(A)_{\mathcal{O}_1} \subseteq \mathcal{O}_1$ is an ideal such that $(A)^2$ is principal. Show that $(A)_{\mathcal{O}_2}^2$ is principal in $\mathcal{O}_2$.

We will let $G_1$ and $G_2$ denote the ideal class groups of $K_1$ and $K_2$, respectively. For brevity, if $S \subseteq \mathcal{O}_1$ is a subset, then we will denote by $(S)_1$ the ideal generated by $S$ in $\mathcal{O}_1$ and by $(S)_2$ the ideal generated by $S$ in $\mathcal{O}_2$.

We claim that if $(A)_1 \sim (B)_1$, then $(A)_2 \sim (B)_2$. To see this, suppose we have $\alpha$ and $\beta$ such that $(\alpha)_1(A)_1 = (\beta)_1(B)_1$. Then $(\alpha A)_1 = (\beta B)_1$. Now if $x = \sum r_i \alpha a_i \in (\alpha A)_2$, then each $\alpha a_i$ is in $(\beta B)_1$, and so has the form $\sum s_{i,j} \beta b_j$. So $x \in (\beta B)_2$. Conversely, $(\beta B)_2 \subseteq (\alpha A)_2$, and so $(A)_2 \sim (B)_2$.

The mapping $A \mapsto (A)_2$ then induces a well-defined mapping $\Psi : G_1 \rightarrow G_2$ given by $[A] \mapsto [(A)_2]$. Moreover, since $\Psi([A][B]) = \Psi([AB])$ $= [(AB)_2]$ $= [(A)_2][(B)_2]$ $= \Psi(A) \Psi(B)$, $\Psi$ is a group homomorphism.

In particular, if $A^2 \sim (1)$ in $\mathcal{O}_1$, then $\Psi(A) \sim (1)$ in $\mathcal{O}_2$, as desired.

### The exponent of the smallest power of an ideal which is principal divides the class number

Let $\mathbb{O}$ be the ring of integers in an algebraic number field $K$ of class number $k$. Let $A$ be an ideal. Show that if $k$ is minimal such that $A^k$ is principal, then $k|h$.

This $k$ is precisely the order of $[A]$ in the class group of $K$. The result then follows by Lagrange’s Theorem.

### The class number of QQ(sqrt(-6)) is 2

Compute the class number of $\mathbb{Q}(\sqrt{-6})$.

Recall that the ring of integers in this field is $\mathcal{O} = \mathbb{Z}[\sqrt{-6}]$, and that $\{1,\sqrt{-6}\}$ is an integral basis. We will now give an upper bound on the constant $C$ which is proven to exist in Theorem 10.2.

Let $A$ be an ideal in $\mathcal{O}$, and let $t = \lfloor \sqrt{N(A)} \rfloor$. That is, $t$ is the largest integer such that $t^2 \leq N(A)$. Now consider $\{b_0 + b_1\sqrt{-6} \ |\ 0 \leq b_0,b_1 \leq t\}$. Since $(t+1)^2 < N(A)$, there exist some $\beta_1,\beta_2$ in this set such that $\beta_2 - \beta_1 \in A$. Say $\alpha = \beta_2 - \beta_1 = a_0+a_1\sqrt{-6}$, where (clearly) $|a_0|,|a_1| \leq t$. By the triangle inequality, $|N(\alpha)| = |a_0^2 + 6a_1^2| \leq |a_1|^2 + 6|a_1|^2 \leq t^2 + 6t^2 = 7t^2$ $\leq 7 \cdot N(A)$. By our proof of Theorem 10.3, every class of ideals in $\mathcal{O}$ contains an ideal whose norm is at most 7. We will now construct all of the ideals having norm $k$ where $k \in \{1,2,3,4,5,6,7\}$.

$(k=1)$ There is only one ideal of norm 1, namely $A_1 = (1)$.

$(k=2)$ If $A$ is an ideal of norm 2, then $A$ divides $(2)$. Note that $(2,\sqrt{-6})^2 = (4,2\sqrt{-6},-6)$ $= (2)$. Now $4 = N((2)) = N((2,\sqrt{-6}))^2$, so that $N((2,\sqrt{-6})) = 2$. By Corollary 9.15, $(2,\sqrt{-6})$ is prime. Thus $A_2 = (2,\sqrt{-6})$ is the only ideal of norm 2.

$(k=3)$ If $A$ is an ideal of norm 3, then $A$ divides $(3)$. Note that $(3,\sqrt{-6})^2 = (9,3\sqrt{-6},-6) = (3)$. Now $9 = N((3)) = N((3,\sqrt{-6}))^2$, so that $N((3,\sqrt{-6})) = 3$. By Corollary 9.15, $(3,\sqrt{-6})$ is prime. Thus $A_3 = (3,\sqrt{-6})$ is the only ideal of norm 3.

$(k=4)$ Suppose $A$ is an ideal of norm 4. If $A$ is not prime, then it is a product of two prime ideals of norm 2; the only such ideal is $(2)$. If $A$ is prime, then by Theorem 9.19, $A$ divides $(2)$. But no prime ideals dividing $(2)$ have norm 4. So the only ideal of norm 4 is $A_4 = (2)$.

$(k=5)$ If $A$ is an ideal of norm 5, then $A$ divides $(5)$. Note that $(5) = (5,2+\sqrt{-6})(5,2-\sqrt{-6})$, since $5 = 5^2 - 2(2+\sqrt{-6})(2-\sqrt{-6})$. We claim that both of these factors are proper. Indeed, if $1 \in (5,2+\sqrt{-6})$, then $1 = 5(a_b\sqrt{-6}) + (2+\sqrt{-6})(h+k\sqrt{-6})$ for some $a,b,h,k \in \mathbb{Z}$. Comparing coefficients mod 5, we have $1 \equiv 0$, a contradiction. Thus $(5,2+\sqrt{-6})$ is proper. Similarly, $(5,2-\sqrt{-6})$ is proper. Now $25 = N((5)) = N((5,2+\sqrt{-6}))N((5,2-\sqrt{-5}))$, and neither factor is 1. So $N((5,2+\sqrt{-6})) = N((5,2-\sqrt{-6})) = 5$. Thus $A_5 = (5,2+\sqrt{-6})$ and $A_6 = (5,2-\sqrt{-6})$ are the only possible ideals of norm 5 in $\mathcal{O}$. Note that the discriminant of this field is not divisible by 5, so that these factors are distinct by Theorem 9.6. (Though this last step is not necessary.)

$(k=6)$ If $A$ is an ideal of norm 6, then $A$ divides $(6) = (2)(3)$. Thus the prime factors of $A$ come from the set $\{(2,\sqrt{-6}), (3,\sqrt{-6})\}$. The only such ideal is $A_7 = (\sqrt{-6})$.

$(k=7)$ If $A$ is an ideal of norm 7, then $A$ divides $(7)$. Note that $(7) = (1+\sqrt{-6})(1+\sqrt{-7})$. Moreover, by Corollary 9.11, $N((1+\sqrt{-6})) = N((1-\sqrt{-6})) = 7$, and by Corollary 9.15, these ideals are prime. So the only ideals of norm 7 are $A_8 = (1+\sqrt{-6})$ and $A_9 = (1-\sqrt{-6})$.

Certainly we have $A_1 \sim A_4 \sim A_7 \sim A_8 \sim A_9 \sim (1)$.

Note that $A_2A_3 = A_7 \sim (1)$ and that $A_2 = A_4 \sim (1)$. Thus $A_2 \sim A_3$.

We also have $A_5A_6 = (5)$ and $A_5^2 = (1-2\sqrt{-6})$, since $1-2\sqrt{-6} = 25 + 2(2+\sqrt{-6})^2 - 2 \cdot 5 \cdot (2+\sqrt{-6})$. So $A_5A_6 \sim A_5^2$, and thus $A_5 \sim A_6$.

We also that $A_5A_2 = (2+\sqrt{-6})$ since $5 \cdot 2 + 5 \cdot \sqrt{-6} - 2(2)(2+\sqrt{-6}) = 2+\sqrt{-6})$. So $A_5A_2 \sim A_5^2$, and thus $A_5 \sim A_2$. So $A_2 \sim A_3 \sim A_5 \sim A_6$.

Finally, we claim that $A_2$ is not principal. If $(2,\sqrt{-6}) = (\alpha)$, then $N(\alpha)$ divides both 4 and 6, and so is either 1 or 2. But no element of $\mathbb{Z}[\sqrt{-6}]$ has norm 2, and $A_2$ is proper; so $A_2$ is not principal, and $A_2 \not\sim (1)$.

Thus $\mathbb{Q}(\sqrt{-6})$ has two ideal classes.