Splitting of prime ideals in quadratic extensions of ℚ, part 1

Our discussion of algebraic number theory returns by popular demand. Way back last April we presented some generalities on factorization of prime ideals in extension fields. (For explanation of what that means, including other necessary concepts, you’ll have to review earlier installments of this series, which can be found here.)

If this is all Greek to you, I apologize, but that’s unavoidable at this stage of a rather technical subject. You may want to go back to the earliest parts of the series to see how the subject got its start and why it may be interesting.

The discussion in the previous installment probably seems rather dry and abstract, but when we look at simple examples, such as quadratic extensions, why it’s interesting becomes clearer.

Because of how the ramification indexes and inertial degrees are related, for any prime ideal (p) of ℤ there are only three different possibilities for how the ideal factors in the ring of integers of a quadratic extension:

  1. (p)=P1⋅P2, so (p) splits completely. (e=f=1, g=2)
  2. (p)=P is a prime ideal in ℚ(√d), so p is inert. (e=g=1, f=2)
  3. (p)=P2 where P is prime, and p is ramified. (f=g=1, e=2)

It turns out that there are simple criteria for each of these cases. But figuring out what the criteria are is tricky.

Recall that in ℚ(√3) we found (13)=(4+√3)⋅(4-√3) and (-11)=(8+5√3)⋅(8-5√3), so both (11) and (13) split completely. Clearly, (3)=(√3)2, so (3) is an example of a prime ideal if ℚ that is ramified. How about an example of a prime ideal that is inert in the extension? This is a little harder for a couple of reasons. (p) will be inert just in case it neither splits nor is ramified, but we don’t yet have simple criteria to rule out those cases.

So let’s back up a little and look at the details. We found examples where (p) splits in the integers of ℚ(√3) by solving the equation ±p=a2-3b2, because that gave elements a±b√3 whose norm was ±p. Being able to find such elements guaranteed that the prime split. But ℚ(√d) with d=3 is a special case, since here d≡3 (mod 4). In that case, and also if d≡2 (mod 4), the integers of the extension have the form a+b√d with a,b∈ℚ.

If d≡1 (mod 4), integers can also have the form (a+b√d)/2, with a,b∈ℚ, and we might have a factorization like (p) = ((a+b√d)/2)⋅((a-b√d)/2), so we would have also to consider solvability of ±4p=a2-3b2. If we were to look at solvability of the approriate equation, according as to whether or not d≡1 (mod 4), the solvability would be a sufficient condition for (p) to split (or ramify if a=0). Notice that this sufficient condition for (p) to split holds regardless of whether or not ℚ(√d) is a PID.

Now we need to find a convenient necessary condition for (p) to split. Unfortunately, solvability of one simple equation is not a necessary condition in general. It would be, as we’ll see in a minute, if the ring of integers of ℚ(√d) happens to be a PID, as is true when d=3. However, in quadratic extensions where the ring of integers isn’t a PID, being unable to solve the applicable equation doesn’t guarantee (p) cannot split, because there might be non-principal ideals that are factors of (p).

So let’s ignore that problem for a moment and just focus on the case where the ring of integers of ℚ(√d) is a PID. Can we then find a necessary condition on p for (p) to split or ramify, i. e. for (p) to not be a prime ideal of the integers of ℚ(√d)? That is, what must be true about p if (p) splits or ramifies?

If (p) splits or ramifies, then (p)=P1⋅P2 for nontrivial ideals Pi. (The ideals are the same or distinct according as (p) ramifies or splits.) Assuming ℚ(√d) is a PID, then P1 is generated by a+b√d where both a,b∈ℚ, if d≡2 or 3 (mod 4), or else by (a+b√d)/2 with a,b∈ℚ, if d≡1 (mod 4). Likewise, the conjugate ideal P2 is generated by a-b√d or (a-b√d)/2. Since p∈P1⋅P2, by definition of a product of ideals, p is of the form p = ε(a+b√d)(a-b√d) = ε(a2-db2) or p = ε(a+b√d)(a-b√d)/4 = ε(a2-db2)/4 for some integer ε of ℚ(√d).

Recall that the norm of an element of a Galois extension field is the product of all conjugates of the element. So for an element that is also in the base field, the norm (with respect to a quadratic extension, which is always Galois) is the square of the element. Taking norms of both sides of the possible equations, then either p2 = N(ε)(a2-db2)2 or 16p2 = N(ε)(a2-db2)2. For simplicity, consider just the first case. Then N(ε) is a positive integer that has to be 1, p, or p2. If N(ε)≠1 then N(a±b√d) = a2-db2 must be ±1, so a±b√d must be a unit, and both Pi must be non-proper ideals (i. e. equal to the whole ring). Hence N(ε)=1. This will be true also in the other case (when d≡1 (mod 4)), so ±p=a2-db2 or ±4p=a2-db2. Consequently, solvability of the appropriate equation (depending on d mod 4), is a necessary condition for (p) to split or ramify.

So we have a necessary and sufficient condition for (p) to split or ramify in ℚ(√d), in terms of solvability of Diophantine equations, provided Oℚ(√d) is a PID. Since the only other possibility is for (p) to be inert, we also have a necessary and sufficient condition for that.

However, still assuming that Oℚ(√d) is a PID, we can find a further necessary condition for (p) to split or ramify. Take those equations we just found and reduce them modulo p. Then both equations become a2≡db2 (mod p). Since p is prime, ℤ/pℤ is a field. Assume first that b≢0 (mod p). Then b has an inverse in the finite field. So we have d≡(a/b)2 (mod p). In other words, d is a square mod p. This is the additional necessary condition we were looking for on p in order for (p) to split or ramify. Since it’s a necessary condition, if d is not a square mod p, then (p) must not split or ramify, and thus p is inert. And so, for d to be a non-square mod p is a sufficient condition for p to be inert.

(What if b≡0 (mod p)? Then b=b1p. So ±p = a2 – (b1p)2d or else ±4p = a2 – (b1p)2d. Either way, p∣a, hence p2 divides the right side of either equation, and hence the left side also. But that’s not possible unless p=2 – which is always a special case.)

To summarize, then, let p≠2 be prime and d square-free and not 0 or 1. Then the solvability of ±tp=a2-db2 (where t is 4 or 1 according as d≡1 (mod 4) or not), is sufficient for (p) to split or ramify. And if the integers of ℚ(√d) are a PID, then solvability of the appropriate equation provides a necessary and sufficient condition for (p) to split or ramify. Further, in that case, d being a square mod p is necessary for (p) to split or ramify.

Remarkably, d being a square mod p is a necessary and sufficient condition for (p) to split or ramify, even if the integers of ℚ(√d) aren’t a PID, but that’s harder to prove. Since solvability of Diophantine equations is generally not obvious by inspection, it’s very convenient to have a necessary and sufficient conditions for (p) to split or ramify simply in terms of the properties of d mod p.

In the next installment, which hopefully will not be as long in coming as this one, we’ll show a much cleaner way to state necessary and sufficient conditions for (p) to split, ramify, or remain inert, in the case of any quadratic extension of ℚ, whether or not the ring of integers is a PID. This will be done in terms of what has long been called a “reciprocity law”.

However, that will be only the beginning. It turns out that there are far more general kinds of reciprocity laws for many other types of field extensions. That’s what “class field theory” is all about, and why the whole subject is so appealing, once you get the basic ideas.


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