Lesson 07 · Gaussian integers and characters · 50–65 minutes
The Quartic Character
When 4 divides p − 1 there are exactly two characters of order 4, a conjugate pair, and nothing in Lesson 06 prefers one to the other. For the primes of P* one can be singled out: exactly one of them sends 2 to i. That choice, χ4(2) = i, is the normalization used for the rest of the course. It splits the units into four cosets of the fourth powers, H, 2H, 4H and 8H, which Part III adds up.
Where this lesson is headed
χ4(2) = i, Cj = {n : χ4(n) = ij} = 2jH, n(p−1)/4 ≡ (2(p−1)/4)j for n ∈ Cj
In words: for p in P*, the normalization χ4(2) = i picks one quartic character, independent of the primitive root. Its level sets are the four cosets 2jH of the fourth powers, each of size (p − 1)/4, and a single power, n(p−1)/4, tells you which coset n is in. Negation swaps Cj with Cj+2.
p: a prime in P* (p ≡ 5 mod 8, p > 5)
H = {x4 mod p}: the fourth powers, a subgroup of size (p − 1)/4
χ4: the quartic character with χ4(2) = i
χ̄4 = χ43: its conjugate, with χ̄4(2) = −i
Cj: the units with χ4(n) = ij, j = 0, 1, 2, 3
u = 2(p−1)/4 mod p
01. Choosing one of two characters
Start with p = 13 and the primitive root g = 2. The row iind n is one quartic character; its conjugate is the other. Which one sends 2 to i? Now change g to 7: the row iind n changes, but the pinned χ4 does not. The four colored groups are its level sets.
Guided experiments
Do these in order. "Set this up" chooses p and g. For values that are powers of i, answer with the exponent k in ik; for cosets, with j.
Why a normalization is needed
The two quartic characters are complex conjugates: one is obtained from the other by swapping i and −i. Any formula built from χ has a twin built from χ̄, and they differ exactly by that swap. Without a rule to choose, a quantity like "β = k + it" would only be defined up to the sign of t.
The rule χ4(2) = i works because, for p ∈ P*, 2 is not even a square. So χ(2) is ±i rather than ±1, and the two characters disagree there. Lesson 09 returns to this: trace and norm cannot see the choice, only the pin can.
02. The pin and the cosets
Each step has the general statement on the left and your p and g from Stage 01 on the right.
1
The two quartic characters are conjugates
For 4 | p − 1, the characters of order 4 are χj with j = (p − 1)/4 and 3(p − 1)/4 (Lesson 06, φ(4) = 2). The first is ψ(n) = iind n; the second is ψ3 = ψ̄, since i3 = −i = ī. Conjugation swaps i and −i and fixes ±1, and ψ·ψ̄ = χ0.
2
The fourth powers are the kernel
By Lesson 02 with ℓ = 4, x ↦ x4 is 4-to-one onto H = {n : 4 | ind n}, of size (p − 1)/4. And ψ(n) = 1 exactly when 4 | ind n, so
H = {n : χ(n) = 1}
for either quartic character χ.
3
The pin χ4(2) = i
For p ∈ P*, 2 is not a square, so ind 2 is odd and ψ(2) = iind 2 is i or −i. Then ψ̄(2) is the other. So exactly one quartic character has χ(2) = i; call it χ4. Its definition mentions only 2, not g, so it does not depend on the choice of primitive root: a different g may swap which of ψ, ψ̄ is χ4, but χ4 itself is the same function.
4
The cosets are H, 2H, 4H, 8H
The level sets Cj = {n : χ4(n) = ij} are cosets of H: n and m lie in the same one exactly when χ4(n/m) = 1, i.e. n/m ∈ H. Since χ4(2j) = ij,
Cj = 2jH, |Cj| = (p − 1)/4
and multiplying adds coset labels: Ci·Cj = Ci+j mod 4.
5
A quartic Euler criterion
Let u = 2(p−1)/4. Then u2 = 2(p−1)/2 ≡ (2/p) = −1 (Euler's criterion), so 1, u, −1, −u are the four fourth roots of 1. For n ∈ Cj, n = 2jh with h ∈ H, and h(p−1)/4 ≡ 1 (as h = x4 and xp−1 ≡ 1). So
n(p−1)/4 ≡ uj (mod p)
One modular power places n in its coset, with no primitive root. This is how the rest of the course computes χ4.
6
Negation swaps opposite cosets
For p ≡ 5 (mod 8), χ4 is odd: χ4(−1) = −1 = i2, so −1 ∈ C2. Therefore
n ∈ Cj ⇔ p − n ∈ Cj+2
The pairing n ↔ p − n matches C0 with C2 and C1 with C3. Lesson 08 uses exactly this to show S0 + S2 = S1 + S3.
7
Why exactly P*
p ≡ 3 (mod 4): no quartic character at all (4 ∤ p − 1). p ≡ 1 (mod 8): 2 is a square, so χ(2) = ±1 for both quartic characters, and they agree at 2: the pin cannot choose. Also χ(−1) = +1, so step 6 fails. p = 5: H = {1} and the cosets are single points, too small for the sums of Part III. p ∈ P*: every step above works.
03. The pinned quartic character
Assemble the argument, then test it on every prime of P* below 1000 and every primitive root.
Theorem. Let p ∈ P*. There is exactly one quartic character χ4 with χ4(2) = i. It does not depend on a choice of primitive root, it is odd, and its level sets are Cj = 2jH (j = 0, 1, 2, 3), each of size (p − 1)/4, with −Cj = Cj+2. For every unit n, χ4(n) = ij where n(p−1)/4 ≡ (2(p−1)/4)j (mod p).
Complete proof
Existence and uniqueness. The quartic characters are ψ = iind and ψ̄. For p ≡ 5 (mod 8), (2/p) = −1, so ψ(2)2 = −1 and ψ(2) = ±i; ψ̄(2) is the opposite sign. Exactly one of ψ, ψ̄ takes i at 2. Since the condition χ(2) = i refers to no primitive root, the character it selects is independent of g.
Cosets. The kernel of χ4 is H = {n : 4 | ind n} = {fourth powers}, of size (p − 1)/4. Its level sets are cosets of H, and since χ4(2j) = ij, Cj = 2jH. As χ4(−1) = (−1)(p−1)/4 = −1, −Cj = Cj+2.
Quartic Euler criterion. For h = x4 ∈ H, h(p−1)/4 = xp−1 ≡ 1, so for n = 2jh, n(p−1)/4 ≡ uj with u = 2(p−1)/4. Since u2 ≡ −1, the four values u0, …, u3 are distinct, so j is determined. ∎
Computer check
For each of the 42 primes of P* below 1000 and every primitive root g: builds ψ = iind, picks the one of ψ, ψ̄ with value i at 2, and checks it is the same function for every g; checks the cosets 2jH have equal size, −Cj = Cj+2, and the quartic Euler criterion for every unit.
Not run yet.
Exit problem
Use p = 29.