# Intuition in studying splitting and ramification of prime ideals

I am trying to learn Algebraic Number Theory alone and I'm having serious trouble understanding the ramification and splitting of primes ideals in Galois extensions of a number field $L/K$.

Some of my friends tell me lots of facts about these and define several structures and invariants, but I feel like the logical sequence of facts and some motivation for definitions are missing.

Some questions I have (among a lot of others):

1. For example, I would really like to know more about the Decomposition Group and the Inertia Group of an ideal $\mathfrak{q}$ over $\mathfrak{p}$. I get the definitions but it is not really clear if these groups are simply subgroups of the Galois group that have arithmetic information encoded in them or if they really help computationally to find some invariants (e,f,g) which would be hard to find without them.

2. Again about the Decomposition and Inertia Groups; It is clear that we are interested in the Decomposition group because we want to find the ideals that get fixed (not pointwise, I know) by the automorphisms of the Galois group. However, It is not clear why would we look at the automorphisms which induce the identity in the quotient ($B/\mathfrak{q})/(A/\mathfrak{p}$) (the elements of the Inertia Group). Is there an intuition for it?

3. What does ramification intuitively mean? I've asked this before but people always try to explain with an example from algebraic geometry (which I don't know nothing about) and I end up without getting it.

If someone can answer any of these questions and/or provide me with a reference that goes for intuitive views rather than encyclopedic description, I would appreciate it very much. Thank you.


For (1), the definition goes like this: Suppose you have a galois extension of number fields $L/K$, then you have an extension of rings of integers $\oO_L/\oO_K$. The galois group $\Gal(L/K)$ acts as automorphisms on $L$ that fix $K$. Hence, it acts on $\oO_L$ and fixes $\oO_K$. Thus, for a prime $\mf{p}$ in $\oO_K$, and a prime $\mf{q}$ in $\oO_L$ containing $\mf{p}$, any $\sigma\in\Gal(L/K)$ must fix $\mf{p}$, and hence it must send $\mf{q}$ to another prime $\mf{q}'$ in $\oO_L$ which also contains $\mf{p}$ - ie, lies above $\mf{p}$. Thus, you may consider the subgroup of $\Gal(L/K)$ which fixes $\mf{q}$ - this is a subgroup of $\Gal(L/K)$, and is called the decomposition group $D_{\mf{q}/\mf{p}}$. In fact, $\Gal(L/K)$ acts transitively on the primes of $\oO_L$ lying above $\mf{p}$, and thus if $\sigma\in\Gal(L/K)$ sends $\mf{q}\mapsto\mf{q}'$, then $D_{\mf{q'}/\mf{p}} = \sigma(D_{\mf{q}/\mf{p}})\sigma^{-1}$, so the decomposition groups for primes lying above $\mf{p}$ are all conjugate (but still in general they are different subgroups).

Now, since $D_{\mf{q}/\mf{p}}$ fixes $\mf{q}$, it determines a well-defined action on $\oO_L/\mf{q}$, which is naturally seen to be a finite field extension of $\oO_K/\mf{p}$. In general, if you have a surjection of groups or rings $A\twoheadrightarrow B$, then an automorphism of $A$ descends to an automorphism of $B$ if and only if it fixes the kernel.

The elements of $D_{\mf{q}/\mf{p}}$ which induce the identity on $(\oO_L/\mf{q})/(\oO_K/\mf{p})$ are precisely the automorphisms $\sigma\in D_{\mf{q}/\mf{p}}$ such that $\sigma(x)\equiv x\mod \mf{q}$ for all $x\in\oO_L$. These automorphisms form a subgroup called the inertia group $I_{\mf{q}/\mf{p}}$.

Thus, there are inclusions $\{1\}\subset I_{\mf{q}/\mf{p}}\subset D_{\mf{q}/\mf{p}}\subset\Gal(L/K)$, which by galois theory corresponds to extensions of fields $L$ over $L^{I_{\mf{q}/\mf{p}}}$ over $L^{D_{\mf{q}/\mf{p}}}$ over $K$. Going right to left, the prime $\mf{p}$ of $K$ decomposes completely in $L^{D_{\mf{q}/\mf{p}}}$, remains inert until $L^{I_{\mf{q}/\mf{p}}}$, and ramifies from there until $L$.

For (3), imagine the horizontal parabola given by, say, $x = y^2$, projecting onto the $x$-axis. At $x = 1$, you can solve the equation to find that $(1,1)$ and $(1,-1)$ are both points on the parabola which map to 1. Indeed, in general for $x = a$, the preimage of $a$ is $\{(a,\sqrt{a}),(a,-\sqrt{a})\}$. Thus, the preimage almost always has size 2, with one exception, namely when $a = 0$ (if $a < 0$ then the picture becomes deficient, but if you work over $\mathbb{C}$, you still have two (imaginary) solutions). Thus, this is an example of a finite map (ie, map with finite fibers, in this case generically of size 2, so you call the map a degree 2 map), which is ramified above $x = 0$, in the sense that above $x = 0$, there are fewer preimages than normal (in this case 1). In general this is what "ramification" means in geometry. An unramified map is one where the preimage (ie, fiber) above any point are all finite and have the same size.

What does this have to do with primes? Well, one of the great insights of Grothendieck is that prime ideals are kind of like points! For example, if you consider the ring $\CC[x]$, what are its prime ideals? Well, there is the zero ideal $(0)$, and then by the fundamental theorem of algebra, the other primes are all of the form $(x - a)$, where $a\in\CC$. Thus, we get a bijection between the maximal prime ideals of $\CC[x]$ and the points of $\CC$ by sending $(x-a)\mapsto a$. The $(0)$ ideal is kind of special - it's called the generic point, and in a sense represents the entirety of $\CC$.

Note that the rings $\CC[x]$, $\oO_K$ are very similar! They're both Dedekind domains, and in fact $\CC[x]$ is even a PID (class number 1). The fraction field of $\CC[x]$ is $\CC(x)$ and is analogous to $K$. A galois extension of $\CC(x)$ looks like $\CC(x)[y]/f(y)$ and is analogous to the extension $L/K$. Then, $\oO_L$ is analogous to the integral closure of $\CC[x]$ in $\CC(x)[y]/f(y)$.

If you take our example with the parabola, you could set $f(y) = y^2 - x$, then $L' := \CC(x)[y]/(y^2-x)$ is a quadratic extension of $K' := \CC(x)$. It is galois, with galois group of order 2 (hence isomorphic to $\mathbb{Z}/2\mathbb{Z}$, and is generated by the automorphism which fixes $\CC(x)$ and sends $y\mapsto -y$. The integral closure of $\oO_{K'} := \CC[x]$ inside $L'$ in this case is just $\oO_{L'} := \CC[x,y]/(y^2-x)$.

For the prime $\mf{p}_a := (x-a)$ of $\CC[x]$, then there are two primes of $\oO_{L'}$ lying above $\mf{p}_a$, namely $(y-\sqrt{a}),(y+\sqrt{a})$, which are different primes, unless $a = 0$, in which case they are the same.

Exercise: Compute the inertia and decomposition groups for primes of $\oO_{L'}$ lying above $(x - 1),(x-2)$ and $(x-0) = (x)$. You'll find that in the first two cases the decomposition group and inertia group are both trivial. In the second case, the decomposition group and inertia group are both the whole galois group. Note that in this situation primes never remain inert, simply because all the residue fields of $\CC[x]$ are just $\CC$, which is algebraically closed. Note that in each case the galois group acts transitively on the primes of $\oO_{L'}$ lying above a prime of $\oO_{K'}$.

Exercise: Play the same game above with $\CC[x]$ replaced by $\QQ[x]$. Note that $\QQ[x]$ is still a Dedekind domain (in fact a PID), so all the definitions of decomposition/inertia still make sense. You should find that $(x-1)$ splits, $(x-2)$ remains inert, and $(x-0) = (x)$ is ramified.

Remark: In the language of schemes, the topological space $\CC$, viewed as a scheme, is Spec $\CC[x]$, where Spec $R$ for a ring $R$ is a topological space whose underlying set is the set of prime ideals of $R$. The topology however is called the Zariski topology and in the case of Spec $\CC[x]$, does not agree with the complex topology on $\CC$. However, the cool thing here is that the traditionally geometric notions of dimension, smoothness, connectedness, compactness, fundamental groups, picard groups, tangent spaces, differential forms,...etc can all be defined purely algebraically for RINGS. From this viewpoint, the ring of integers of an algebraic number field is a smooth connected non-compact curve. $\ZZ$ has trivial fundamental group, since all rings of integers are ramified over $\ZZ$ (this follows I believe from the fact that the discriminant of a number field is never a unit, and comes from the geometry of numbers (NOT algebraic geometry)).

• Thanks a lot for all this effort, I'll read it carefully and work out your examples and exercises. Thanks again! – Shoutre Oct 7 '15 at 19:47
• In line 4, you mean a prime $\mathfrak{q}$ above $\mathfrak{p}$, right? Besides that, I have a question about what you said here: "...Going right to left, the prime $\mathfrak{p}$ of $K$ decomposes completely in $L^{D_q/p}$, remains inert until $L^{I_q/p}$, and ramifies from there until $L$." You are saying that a prime $\mathfrak{p}$ decomposes "all it can decompose" at most at the extension $L^{D_q/p}$? Like, I can guarantee that it will not split more when we go from $L^{D_q/p}$ up to $L$? – Shoutre Oct 7 '15 at 20:31
• @Shoutre Yes, and Yes. In the first stage, the prime decomposed as much as it can, but the residue fields remain the same and it remains unramified. In the next stage it stops decomposing, and the effect of the further extension can be completely measured by the degree of extension of the residue field, and again it remains unramified. In the last stage, the residue fields remain the same, it doesn't decompose, and the effect of the extension is completely measured by the ramification index. It's from this that you get the formula: $[L:K] = nef$ – oxeimon Oct 7 '15 at 20:37
• How amazing!! I've never realized that. Thanks a lot! – Shoutre Oct 7 '15 at 20:40
• @oxeimon Wow man, your answer is incredible, clear and detailed, very nice job! – PITTALUGA Oct 9 '15 at 18:53