Math 254A, UC Berkeley, Fall 2001 (Kedlaya) Lecture 23: Quadratic Fields and Quadratic Reciprocity October 26, 2001

Reference. Neukirch, Section 1.8.

Jargon watch. Recall that a rational integer D ≠ 1 is called a fundamental discriminant if one of the following holds:

(a) D is odd, squarefree, and congruent to 1 mod 4;

(b) D is divisible by 4, and D/4 is odd, squarefree, and congruent to 3 mod 4.

(c) D is divisible by 8, and D/8 is odd and squarefree.

The point: D is a fundamental discriminant if and only if the discriminant of the quadratic field ℚ(\sqrt{D}) is equal to D.

Outline of lecture.

1. Let D be a fundamental discriminant. We may apply the criterion from last time to test whether primes split in ℚ(\sqrt{D}), since the ring of integers is ℚ[α] for α = \sqrt{D/4}, if D ≡ 0 (mod 4), or α = (1 + \sqrt{D})/2, if if D ≡ 1 (mod 4).

2. We compute the splitting behavior of a prime p as follows:

(a) If p divides D, then p is ramified. That is, there is one prime 𝔭 above (p) with e(𝔭/(p)) = 2 and f(𝔭/(p)) = 1.

(b) If p is odd and does not divide D, then (p) is unramified, and splits into two primes if and only if the polynomial x^{2} − D factors modulo p.

(c) If p = 2 and D is odd, then (p) is unramified, and splits into two primes if and only if the polynomial x^{2} − x + (D − 1)/4 factors modulo 2, that is, if and only if D ≡ 1 (mod 8).

3. For quadratic fields, it turns out there is a beautiful condition that reduces the question of whether D has a square root modulo p to a question about p modulo the prime factors of D: the law of quadratic reciprocity. To state it, define the Legendre symbol (\frac{a}{p}), for integers a, p with p > 0 an odd prime, as

(\frac{a}{p}) = \begin{cases}0 & a ≡ 0 (mod p) \\ 1 & a ≢ 0 (mod p)anda ≡ x^{2} (mod p)for some x ∈ ℤ \\ −1 & a ≢ 0 (mod p)anda ≢ x^{2} (mod p)for all x ∈ ℤ. \end{cases}

In other words, (\frac{a}{p}) ≡ a^{(p−1)/2} (mod p) (because 𝔽_{p}^{∗} is cyclic).

4. State Gauss’s law of quadratic reciprocity: for p, q odd primes,

(\frac{p}{q}) (\frac{q}{p}) = (−1)^{(p−1)(q−1)/4},  (\frac{−1}{p}) = (−1)^{(p−1)/2},  (\frac{2}{p}) = (−1)^{(p^{2}−1)/8}.

Of these, the second is elementary, and the third follows from the splitting result above.

5. Sketch the proof from Neukirch Section 1.8, using the Gauss sum

τ = ∑_{a∈𝔽_{p}^{∗}} (\frac{a}{p}) ζ_{p}^{a},

where ζ_{p} is a primitive p-th root of unity. Namely, calculate explicitly that τ^{2} = (\frac{−1}{p}) p. We will revisit this proof when we study cyclotomic fields, several lectures from now, and give it a more conceptual interpretation.

6. Advertise class field theory, which can be viewed (among other things) as a generalization of quadratic reciprocity to a splitting criterion for any Galois extension of any number field with abelian Galois group. (I will be covering class field theory next semester.)

Addendum to previous lecture. A more conceptual way to see the relationship, in case 𝔬_{L} = 𝔬_{K}[α], between the factorization of 𝔭𝔬_{L} and the factorization of P (x) modulo 𝔭 (for 𝔭 a prime of 𝔬_{K} and P (x) the minimal polynomial of α) is as follows. Put 𝔭 = ∏_{i} 𝔟_{i}^{e_{i}} and suppose P (x) ≡ ∏_{j} P_{j}(x)^{c_{j}} (mod 𝔭), where each P_{j} is irreducible modulo 𝔭. Then on one hand, by the Chinese remainder theorem, 𝔬_{L}/(𝔭𝔬_{L}) ≅ ∏_{i} 𝔬_{L}/𝔟_{i}^{e_{i}}. On the other hand,

𝔬_{L}/(𝔭𝔬_{L}) ≅ 𝔬_{K}[x]/[(P (x)) + 𝔭] ≅ (𝔬_{K}/𝔭)[x]/(P (x)) ≅ ∏_{j}(𝔬_{K}/𝔭)[x]/(P_{j}(x)^{c_{j}}).

We have now written the ring 𝔬_{L}/(𝔭𝔬_{L}) as a finite product of local rings in two ways. It’s a standard fact from algebra that such a decomposition is unique if it exists, so we can pair off the β_{i} with the P_{j}.

In case you haven’t seen that standard fact before, here’s how it’s proved: call an element e of a ring an idempotent if e^{2} = e. In a local ring, this implies e(e − 1) = 0, and e and e − 1 cannot both be in the maximal ideal. Thus either e is a unit, so e − 1 = 0, or e − 1 is a unit, so e = 0. That is, the only idempotents in a local ring are 0 and 1.

Suppose the ring R is isomorphic both to ∏_{i} S_{i} and ∏_{j} T_{j}, where the S_{i} and T_{j} are local. Let e_{i} be the element of R which has 1 in S_{i} and 0 in the other S’s, and let f_{j} be the element of R which has 1 in T_{j} and 0 in the other T ’s. Now e_{i}x ∈ S_{i} for all x ∈ R (that is, e_{i}x has 0 in the S’s other than S_{i}), and likewise f_{j}x ∈ T_{j} for all x ∈ R. Also, e_{i}f_{j} is an idempotent for all i, j; since it’s in S_{i}, it’s either equal to e_{i} or 0. But the sum of the f_{j} is 1, so ∑_{j} e_{i}f_{j} = e_{i} · 1 = e_{i}, so the e_{i}f_{j} cannot all be 0. So there is some j such that e_{i}f_{j} = e_{i}; by a similar argument, e_{i}f_{j} = f_{j}. Thus the rings S_{i} and T_{j} are isomorphic (they’re both the image of multiplication by e_{i} = f_{j}), so we can factor them off and keep going.