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.