## Functoriality of the Artin Map

Instead of restating the functoriality of the Artin map, let’s just review the statement through an example. We’ll re-use our example from last time. Let ${L}$ be the splitting field of ${x^3-x-1}$ over ${\mathbb{Q}}$. We get a non-abelian Galois group ${H\simeq S_3}$ (to keep notation the same, we called this ${H}$ last time). Take the quadratic subextension ${K=\mathbb{Q}(\sqrt{-23})}$. We have an abelian Galois group ${G\simeq \mathbb{Z}/3}$. We need the abelianization ${H^{ab}\simeq \mathbb{Z}/2}$.

By Galois theory we know ${H^{ab}}$ gives us a field extension ${L'}$ sitting between ${L}$ and ${\mathbb{Q}}$. Class field theory tells us that the conductor ${\frak{m}_{L'}=\{(23), w_1\}}$ because we must pick up all ramification from ${\mathbb{Q}(\sqrt{-23})/\mathbb{Q}}$. The general argument we gave a few posts ago shows us that ${Cl_{\frak{m}_{L'}}(\mathbb{Q})\simeq (\mathbb{Z}/23)^*}$. Now ${K}$ is the Hilbert class field, so ${Cl(\mathcal{O}_K)\stackrel{\sim}{\rightarrow} G}$ via the Artin map. We take ${\frak{m}=(\sqrt{-23})\mathcal{O}_K}$ with no embeddings specified.

This gives us the diagram:

$\displaystyle \begin{matrix} Cl_\frak{m}(K) & \twoheadrightarrow & G & \stackrel{\Psi_{L/K}}{\rightarrow} & \mathbb{Z}/3 & \rightarrow & 1 \\ id \downarrow & & & & \downarrow & & \\ Cl_\frak{m}(K) & \stackrel{N_{K/F}}{\rightarrow} & (\mathbb{Z}/23)^* & \stackrel{\Psi_{L'/F}}{\rightarrow} & \mathbb{Z}/2 & \rightarrow & 1 \end{matrix}$

First, the right vertical arrow is clearly the zero map. The other important part of the diagram is that the norm map is taking a fractional ideal (class) that is relatively prime to ${(\sqrt{-23})\mathcal{O}_K}$ and taking the norm of it which lands you in the units ${(\mathbb{Z}/23)^*\simeq \mathbb{Z}/22}$. Moreover, the map ${\mathbb{Z}/22\rightarrow \mathbb{Z}/2}$ is the unique surjective one and the image of the norm map must land in the kernel of this by exactness. Interestingly, this tells us that the positive generator of ${N_{K/F}(\frak{b})}$ for any ${\frak{b}}$ prime to ${(\sqrt{-23})\mathcal{O}_K}$ is a square mod 23.

Let’s wrap up today by stating another functoriality result. Given the same setup of ${L/K/F}$ where ${G=Gal(L/K)}$ is abelian and ${H=Gal(L/F)}$ is finite possibly non-abelian. Suppose now that ${G\triangleleft H}$ and ${K/F}$ Galois with ${T=Gal(K/F)}$. Now ${T}$ acts on ${G}$ as follows. Let ${t\in T}$. Choose a lift ${h_t\in H}$. The action is given by ${t\cdot g=h_t g h_t^{-1}}$. Call this action ${\sigma_t}$.

We can transfer this Galois action to the ray class group as follows:

$\displaystyle \begin{matrix} Cl_\frak{m}(K) & \stackrel{\Psi}{\rightarrow} & G & \rightarrow & 1 \\ \downarrow & & \downarrow \sigma_t & & \\ Cl_\frak{m}(K) & \stackrel{\Psi}{\rightarrow} & G & \rightarrow & 1 \end{matrix}$

where the vertical arrow is just the natural map on ideals. Commutativity of the diagram just comes from the standard fact that ${hFrob_p(Q)h^{-1}=Frob_p(hQ)}$. There is another functoriality we could do, but it doesn’t seem worth it at this point because it is overly complicated and there isn’t a plan to use it anytime soon.

## Class Field Theory 2

Today we’ll start by sketching an example of a Hilbert class field. Let ${L}$ be the splitting field of ${x^3-x-1}$ over ${\mathbb{Q}}$. Thus ${L=\mathbb{Q}(\alpha_1, \alpha_2, \alpha_3)}$ where ${\alpha_j}$ are the roots of the polynomial. Standard Galois theory shows us that ${Gal(L/\mathbb{Q})\simeq S_3}$. I know, we assumed abelian extensions last time, but we aren’t interested in this extension. It turns out that one can also argue that we have a subextension ${K=\mathbb{Q}(\sqrt{-23})\subset L}$.

To see how useful this class field theory machinery is, we can check with some basic algebraic number theory (involves the Minkowski bound) that ${Cl(\mathcal{O}_K)\simeq \mathbb{Z}/3}$. But above we said that ${Gal(L/K)\simeq \mathbb{Z}/3}$. Thus by the main theorem last time we see that ${L}$ must be the Hilbert class field of ${K}$, and hence ${L}$ is the maximal unramified abelian extension of ${\mathbb{Q}(\sqrt{-23})}$. That’s pretty interesting that it is so small.

This gives us an idea of what these Hilbert class fields look like. Note also that class field theory tells us that if the class number is ${1}$, then there are no unramified abelian extensions, and hence fields like ${\mathbb{Q}(i)}$ or ${\mathbb{Q}(\sqrt{-11})}$ have Hilbert class field equal to themselves. The word “maximal” makes these things sound big, but being unramified is pretty strict and we see that in order to have large Hilbert class fields the class group needs to be large.

Let’s try to go the other direction now. Recall a few posts ago we calculated that ${Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m)^*}$ when we include the embedding in the conductor. We have an isomorphism ${Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m)^*\stackrel{\sim}{\rightarrow} Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})}$, where ${\zeta_m}$ is a primitive ${m}$-th root of unity. We’ll suggestively call the isomorphism ${\Psi([a])=\sigma_a}$ where ${\sigma_a(\zeta_m)=\zeta_m^a}$. This comes from standard theory of cyclotomic fields.

We have that the ring of integers is ${\mathbb{Z}[\zeta_m]}$. But now on the primes, we see that ${\Psi(p\mathbb{Z})}$ is just the Frobenius map of raising to the ${p}$. This shows us that our standard isomorphism is actually the Artin map ${\Psi_{\mathbb{Q}(\zeta_m)/\mathbb{Q}}}$ and hence class field theory tells us that the ray class field of ${\mathbb{Q}}$ of conductor ${\frak{m}=\{(m), \mathbb{Q}\hookrightarrow \mathbb{R}\}}$ is just the cyclotomic extension ${\mathbb{Q}(\zeta_m)}$.

This is an incredibly important example, because now let’s apply the Galois correspondence. Let ${L/\mathbb{Q}}$ be any abelian extension at all. We know that ${L}$ corresponds to some conductor ${\frak{m}_L}$. This must be of the form ${\{\mathbb{Z}m\}}$ or ${\{\mathbb{Z}m, w_1\}}$. Since the first one divides the second, we get two surjections ${Cl_{\{(m), w_1\}}(\mathbb{Q})\twoheadrightarrow Cl_\frak{m}(\mathbb{Q})\twoheadrightarrow Gal(L/\mathbb{Q})}$. But the first term of this by the above is that ${Cl_{\{(m), w_1\}}(\mathbb{Q})\simeq Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})}$. Therefore, ${L\subset \mathbb{Q}(\zeta_m)}$. Out of class field theory we get the classical result called the Kronecker-Weber theorem that any abelian extension of ${\mathbb{Q}}$ must be contained inside a cyclotomic field.

To finish today let’s talk about one more topic. Suppose we have our abelian extension ${L/K}$ of number fields with Galois group ${G}$. If we have a further subextension ${K/F}$, then we get ${Gal(L/F)=H}$ is possibly non-abelian. But we can take the abelianization ${H^{ab}=H/[H,H]}$ and this corresponds via Galois theory to a maximal abelian subextension say ${L'}$ between ${L}$ and ${F}$, i.e. ${Gal(L'/F)\simeq H^{ab}}$. By the universal property of the abelianization, we have a map ${G\rightarrow H^{ab}}$ since ${G\subset H}$.

The point is that ${G}$ corresponds to an abelian extension ${L/K}$ and so class field theory over ${K}$ tells us something about it, and ${H^{ab}}$ corresponds to an abelian extension ${L'/F}$, so class field theory over ${F}$ tells us something about it. We just produced a natural map ${G\rightarrow H^{ab}}$, and hence there should be a corresponding statement in class field theory.

Here’s the theorem. Let ${\frak{m}=\frak{m}_{K/F}}$ so that ${\frak{m}_L=\frak{m}_{L/K}}$ divides ${\frak{m}}$ and ${L/F}$ is unramified outside the places of ${F}$ under ${\frak{m}}$. Then we get a commutative diagram:

$\displaystyle \begin{matrix} Cl_\frak{m}(K) & \twoheadrightarrow & Cl_{\frak{m}_L}(K) & \stackrel{\Psi_{L/K}}{\rightarrow} & G & \rightarrow & 1 \\ id \downarrow & & & & \downarrow & & \\ Cl_\frak{m}(K) & \stackrel{N_{K/F}}{\rightarrow} & Cl_{\frak{m}_{L'}}(F) & \stackrel{\Psi_{L'/F}}{\rightarrow} & H^{ab} & \rightarrow & 1 \end{matrix}$

We could call this a functoriality property of the Artin map. The loose description of this is that inclusions of Galois groups go to the corresponding norm map. We’ll pick up here next time.

## Class Field Theory 1

Today we’ll to get to stating class field theory in terms of ray class groups. Let ${L/K}$ be an extension of number fields. Recall that if ${G=Gal(L/K)}$ is abelian and ${\frak{p}\subset \mathcal{O}_K}$ is a prime unramified in ${\mathcal{O}_L}$, then the Frobenius element, ${\Phi(\frak{q}/\frak{p})=Frob_{L/K}(\frak{q})}$, is really an element and is independent of the ${\frak{q}}$ over ${\frak{p}}$. The independence is just because we are assuming an abelian Galois group, and all the Frobenii are conjugate.

Let ${S}$ be the set of places that ramify in ${L}$. Recall that ${I_S(K)}$ are the fractional ideals relatively prime to ${S}$. Thus we can define a group homomorphism called the Artin map ${\Psi_{L/K}: I_S(K)\rightarrow G}$ by sending a prime to its Frobenius element and then extending.

To have something concrete in your head take the example of ${L=\mathbb{Q}(\sqrt{3})/\mathbb{Q}}$. We have that ${\mathcal{O}_L=\mathbb{Z}[\sqrt{3}]}$, and the discriminant is ${12}$. Take a prime ideal downstairs, ${p\mathbb{Z}}$. There are three possibilities, it ramifies so ${p\mathcal{O}_L=Q^2}$, it splits so ${p\mathcal{O}_L=Q_1Q_2}$, or it is inert in which case ${p\mathcal{O}_L=Q}$.

From the discriminant we know the only primes that ramify are ${p=2, 3}$. To determine whether the other primes split or are inert depends on whether or not ${x^2-3}$ has a root mod ${p}$ by Hensel’s lemma. It turns out ${p}$ splits if ${3}$ is a square mod ${p}$ and is inert if ${3}$ is not a square mod ${p}$. Thus we can read off these two cases from the Legendre symbol.

In the case that it splits, ${p\mathcal{O}_L=Q_1Q_2}$. The decomposition group ${D_{Q_1}=\{id\}}$ because the Galois group acts transitively on ${Q_1}$ and ${Q_2}$, but there is only one non-trivial element which must switch the two primes. Thus only the identity fixes ${Q_1}$ which is the definition ${D_{Q_1}}$. So we get ${\Psi_{L/\mathbb{Q}}(p)=1}$ when ${p}$ splits (i.e. 3 a square mod ${p}$). Likewise, in the inert case we get ${D_{Q}=G}$. Thus the Frobenius must be the non-trivial element of ${G}$ since it generates ${G}$. Thus ${\Psi_{L/\mathbb{Q}}(p)=\sigma}$ (the generator, i.e. raising to the ${p}$ power). This should make us think that quadratic reciprocity will turn out to be a special case of where we’re going.

Now let’s state the main theorem of class field theory (in our special case above). Artin actually proved this version of it. There is a conductor ${\frak{m}=(\frak{m}_f, w_1, \ldots, w_i)}$ depending on ${L/K}$ such that:

a) ${\text{Supp}(\frak{m})=S}$ where the support is the set of places that divide ${\frak{m}_f}$ together with real embeddings ${K\hookrightarrow \mathbb{R}}$.

b) The Artin map ${\Psi_{L/K}: I_S(K)\rightarrow G}$ factors through the ray class group, i.e. ${\Psi_{L/K}: I_S(K)\rightarrow Cl_{\frak{m}}(K)\rightarrow G}$.

c) ${\Psi_{L/K}}$ is surjective.

There is a smallest such ${\frak{m}}$ (under the partial order by dividing on the ideals and inclusion on the embeddings) called the conductor of the extension. There is an existence part of the theorem that gives a Galois correspondence. It says that for each conductor ${\frak{m}}$ and subgroup ${H}$ of ${Cl_\frak{m}(K)}$, there is an abelian extension ${E/K}$ such that ${\frak{m}_{E/K}}$ (the conductor coming from the earlier statement) divides ${\frak{m}}$ and the composition ${Cl_\frak{m}(K)\rightarrow Cl_{\frak{m}_{E/K}}(K)\stackrel{\Psi}{\rightarrow} Gal(E/K)}$ is surjective and has kernel ${H}$.

In particular, you can take ${H=\{e\}}$ and force the maps to be isomorphisms. In that case ${E}$ is called the ray class field of ${K}$ of conductor ${\frak{m}_{E/K}}$. One caution is that even in this nice situation you can’t force ${\frak{m}_{E/K}=\frak{m}}$. For example, take ${\frak{m}}$ to be ${\mathcal{O}_K}$ with no embeddings. The ray class field in this situation is called the Hilbert class field and the ray class group is just the ideal class group.

Thus we get ${Cl(\mathcal{O}_K)\stackrel{\sim}{\rightarrow} Gal(E/K)}$ via the Artin map for the Hilbert class field. Now take ${\frak{m}=\{\mathcal{O}_K, w_1\}}$. Just throw in an embedding. The ideal class group ${Cl(\mathcal{O}_K)}$ is the group of fractional ideals mod the principal ideals. This ${Cl_\frak{m}(K)}$ is the group of fractional ideals mod principal ideals generated by something that is positive under ${w_1}$. These are isomorphic because you can always multiply the generator by ${-1}$. Thus starting with this ${\frak{m}}$ and ${H=\{e\}}$ we get the Hilbert class field again, but ${\frak{m}_{E/K}=\{\mathcal{O}_K\}}$ without the embedding and hence in general we can’t force ${\frak{m}=\frak{m}_{E/K}}$ in general.

To end today let’s just say a few more things about the Hilbert class field. It is the maximal abelian unramified extension of ${K}$, say ${L_0}$. The proof that it is unramified is just that ${\text{Supp}(\mathcal{O}_K)=\emptyset}$. Let ${L/K}$ be any unramified abelian extension. Suppose ${L}$ is not contained in ${L_0}$. Then ${LL_0}$ is an unramified extension strictly containing ${L_0}$. Thus by the main theorem of class field theory ${LL_0}$ has a conductor dividing ${\frak{m}}$ and hence equal to ${\mathcal{O}_K}$. Thus ${Cl(\mathcal{O}_K)\simeq Gal(L/K)\rightarrow Gal(LL_0/K)}$ is surjective, a contradiction. This tells us we can take as our definition of the Hilbert class field to be the ray class field of conductor ${\mathcal{O}_K}$ and get the standard definition that it is the maximal unramified abelian extension of ${K}$.

## Ray Class Groups 2

Today let’s relate the ray class groups to the ideal class group. Fix ${K}$ a number field and choose a conductor ${\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}$. There is certainly always a map ${Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}$. This is just because given some fractional ideal ${\frak{a}}$ that is relatively prime to ${\frak{m}}$ it is a fractional ideal, so it maps to its class in the ideal class group (well-definedness is just because we mod out by less principal ideals, but they’re all principal).

It turns out that under this map we get an exact sequence

$\displaystyle 1\rightarrow \left(\frac{(\mathcal{O}_K/\frak{m}_f)^*\times \prod_{j=1}^i\{\pm 1\}}{\text{im} \Delta}\right)\rightarrow Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)\rightarrow 1,$

where the diagonal map ${\Delta: \mathcal{O}_K^* \rightarrow (\mathcal{O}_K/\frak{m}_f)^*\times\prod_{j=1}^i\{\pm 1\}}$ is given by ${\Delta(u)=([u], \text{sgn}(w_i(u)))}$.

This might seem scary, but let’s go back to our example of ${Cl_m(\mathbb{Q})}$. Something fishy was going on with having to pick the positive generator, and in general when we have lots more units floating around the above tells us exactly how that is affecting everything. For one thing we have ${Cl(\mathbb{Z})=1}$, and so this theorem tells us we have an isomorphism from the big quotient to ${Cl_m(\mathbb{Q})}$.

Our units are just ${\{\pm 1\}}$, and our embedding is ${w_1:\mathbb{Q}\hookrightarrow \mathbb{R}}$. Thus our diagonal map is ${1\mapsto ([1], 1)}$ and ${-1\mapsto ([-1], -1)}$. Thus ${\displaystyle Cl_m(\mathbb{Q})\simeq \frac{(\mathbb{Z}/m\mathbb{Z})^*\times\{\pm 1\}}{\text{im}(\Delta)}\simeq (\mathbb{Z}/m\mathbb{Z})^*}$ which is what we got last time.

What this shows is that if ${\frak{m}}$ doesn’t involve the real place, i.e. no embedding is part of the conductor, the map given last time is not well-defined because the two generators of the same fractional ideal ${\frac{a}{b}}$ and ${-\frac{a}{b}}$ must get mapped to the same place. But this exact sequence tells us exactly what gets changed, and we have to mod out by ${\{\pm 1\}}$ (the image of the diagonal) so that they really do go to the same place and hence we get a smaller ray class group ${Cl_\frak{m}(\mathbb{Q})\simeq \frac{(\mathbb{Z}/m\mathbb{Z})^*}{\{\pm 1\}}}$.

Now let’s prove exactness of that sequence in general. We’ll start with surjectivity of the right map. Let ${[\frak{b}]\in Cl(\mathcal{O}_K)}$. We want to find some fractional ideal ${\frak{a}}$ relatively prime to ${\frak{m}}$ such that ${[\frak{a}]=[\frak{b}]}$. We can use the same approximation theorem trick as in this post (see it for more details). We have ${\frak{b}=\prod \frak{p}_i^{e_i}}$, but for any prime appearing that also appears in ${\frak{m}}$ choose some ${a_i}$ with ${ord_{p_i}(a_i)=-e_i}$. Multiplying these together we get some element ${a}$ and ${\frak{a}=\frak{b}\cdot (a)}$ is altered by a principal and hence in the same ideal class, but is by construction relatively prime to ${\frak{m}}$.

Just by definition the kernel of that map is exactly the principal ideals that are relatively prime to ${\frak{m}}$ modulo the principal ideals with the conductor condition, i.e. ${\displaystyle \frac{Prin(\mathcal{O}_K)\cap I_\frak{m}(K)}{Prin_\frak{m}(K)}}$. We’ll establish an isomorphism between this and the first term of the sequence.

Suppose ${\mathcal{O}_K\cdot \beta}$ is a class in this group, so that ${\beta}$ is relatively prime to ${\frak{m}}$. Thus our map will be ${\beta\mapsto \overline{\beta}:=((\beta \mod \frak{p}^{ord_{\frak{p}}(\frak{m}_f)}\mathcal{O}_K), \text{sgn}(w_j(\beta)))}$ in

$\displaystyle \prod_{\frak{p}|\frak{m}_f} \mathcal{O}_{K, \frak{p}}^*/(1+\frak{p}^{ord_{\frak{p}}(\frak{m}_f)}\mathcal{O}_K)^*\times\prod_{j=1}^i\{\pm 1\}$

Now it is not clear this is well-defined because there is ambiguity in the choice of generator ${\beta}$. Thus we must check that ${\mathcal{O}_K \beta=\mathcal{O}_K \beta'}$ (as classes mod the principal ideals generated by elements congruent to ${1\mod \frak{m}}$!) if and only if ${\beta=u\beta'}$ for some ${u\in \mathcal{O}_K^*}$.

Claim: The ideal ${\mathcal{O}_K\beta}$ is in ${Prin_{\frak{m}}(K)}$ if and only if ${\overline{\beta}\in \text{im}(\Delta)}$. If ${\overline{\beta}\in \text{im}(\Delta)}$, then let ${u}$ a unit such that ${\Delta(u)=\overline{\beta}}$. Take ${\beta'=u\beta}$, and you get that ${\beta'\equiv 1\mod \frak{m}}$, thus ${\mathcal{O}_K\beta=\mathcal{O}_K\beta'}$ which is in ${Prin_{\frak{m}}(K)}$.

Conversely, if ${\mathcal{O}_K\beta}$ has the property that ${\mathcal{O}_K\beta=\mathcal{O}_K\beta'}$ where ${\beta'\equiv 1\mod \frak{m}}$, then ${\beta=u\beta'}$ for some unit. But now by definition ${\overline{\beta'}}$ is just ${1}$. Thus ${\overline{\beta}=\Delta(u)}$. This proves the claim.

The proof of this claim actually shows that the map descends to an isomorphism and hence the sequence is exact by identifying the appropriate groups using the Chinese remainder theorem. That seems like enough for today.

## Ray Class Group 1

There are so many ways I was tempted to branch out from where I left off. Eventually I want to come back and talk some more about complex multiplication which is a fascinating topic, but before doing that it would be better to cover some other ground first. So we’re moving on to a new topic. Instead of useful algebra that could be taught in a first year course, we’re going to jump all the way to assuming knowledge of a one quarter (10 week) class in algebraic number theory.

This means we’re not assuming that much, but basic facts about number fields, discriminants, rings of integers, Dedekind domains, finiteness of class group, valuations/Hensel’s lemma, and basic elliptic curve theory will be assumed (but maybe not used).

It might seem a bit of a jump, but let’s do some class field theory. The way I saw this was via ray class groups, but it seems maybe this isn’t the most popular current viewpoint. Let’s fix a number field ${K}$ with ring of integers ${\mathcal{O}_K}$. A conductor for ${K}$ is a tuple ${\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}$ where ${\frak{m}_f}$ is an ideal of ${\mathcal{O}_K}$ and ${w_j: K\hookrightarrow \mathbb{R}}$ are embeddings.

A non-zero ${\alpha}$ in ${K}$ is defined to be ${\alpha\equiv 1 \mod \frak{m}}$ if it satisfies ${ord_\frak{p}(\alpha -1)\geq ord_\frak{p}(\frak{m}_f)}$ for all primes and the images under all the embeddings ${w_j(\alpha)>0}$ are positive. In the simplest case, we could take ${\frak{m}=\{\frak{m}_f\}}$ where we don’t pick any embeddings. This makes the second condition empty. Let’s unravel the first condition a little more.

To think about the first condition recall that ${ord_\frak{p}(\alpha-1)}$ can be interpreted as the first ${r}$ such that ${\alpha-1\in \frak{p}^r}$ but ${\alpha-1\notin \frak{p}^{r+1}}$ and ${ord_\frak{p}(\frak{m}_f)}$ is the power of ${\frak{p}}$ occuring when decomposing ${\frak{m}_f=\prod \frak{p}_i^{e_i}}$. Thus the inequality tells us that ${\alpha-1}$ is in all those powers occuring in ${\frak{m}}$, i.e. ${[\alpha-1]=[0]\in \mathcal{O}_K/\frak{m}_f}$. Thus ${\alpha\equiv 1\mod \frak{m}_f}$, so the notation makes sense. Another way to grapple with this definition is to set ${S=\{\frak{p}: ord_\frak{p}(\frak{m}_f)>0\}}$. Then any ${\alpha}$ such that ${\alpha\equiv 1\mod \frak{m}}$ is an ${S}$-unit in ${\mathcal{O}_K}$.

Define ${I_\frak{m}}$ to be the group of fractional ideals ${Q}$ such that ${ord_\frak{p}(Q)=0}$ if ${ord_\frak{p}(\frak{m}_f)>0}$. Unravelling this a little tells us that if we decompose ${Q=\prod \frak{p}_j^{e_j}}$, then since ${\frak{m}_f}$ is assumed an honest ideal, the decomposition involves only positive powers and hence ${ord_\frak{p} (\frak{m}_f)>0}$ just says ${\frak{p}}$ appears. So ${I_\frak{m}}$ is the group of fractional ideals such that no power (positive or negative) appears in the decomposition of the ideal if the prime appears in ${\frak{m}_f}$. Or more colloquially, it is the group of fractional ideals relatively prime to ${\frak{m}_f}$.

Now in the same way we form the ideal class group, we can form the group of principal ideals: ${Prin(\frak{m})=\{\mathcal{O}_K\cdot \alpha : \alpha \equiv 1 \mod \frak{m}\}}$. This is a subgroup of ${I_\frak{m}}$ (this is an exercise, but shouldn’t be immediately obvious that multiplication preserves congruence to ${1 \mod \frak{m}}$).

Now we can define ${Cl_\frak{m}(K)=I_\frak{m}/Prin(\frak{m})}$ to be the ray class group of ${K}$ of conductor ${\frak{m}}$. We easily recover the full ideal class group by taking ${\frak{m}}$ to be the “empty” conductor, or in other words no embeddings together with the ideal ${\frak{m}_f=\mathcal{O}_K}$. Since ${ord_p(\mathcal{O}_K)=0}$ for all primes, we get that ${I_\frak{m}}$ contains every fractional ideal and trivially all elements ${\alpha\equiv 1 \mod \mathcal{O}_K}$, so ${Prin(\mathcal{O}_K)}$ is the full group of principal fractional ideals. So ray class groups really are just generalizations of the ideal class group.

Another basic example is to take ${K=\mathbb{Q}}$ and ${\frak{m}=\{\mathbb{Z}\cdot m, i:\mathbb{Q}\hookrightarrow \mathbb{R}\}}$. The fractional ideals are principal, so ${I_\frak{m}=\{\mathbb{Z}\cdot\frac{a}{b} : (a,m)=1 \ \text{and} \ (b,m)=1\}}$ (of course WLOG assume ${(a,b)=1}$). Now for the principal ideals we finally have an embedding condition. But it turns out to be vacuous because all we need is to find some generator that maps to a positive under the embedding, and we have ${\mathbb{Z}\cdot n=\mathbb{Z}\cdot (-n)}$ so one or the other will map to a positive and we’ll assume we’ve chosen that one.

Now the other condition on principals is that ${ord_p(\frac{a}{b}-1)\geq ord_p(m)}$, and since all of these things are integers we convert this to ${m|(a-b)}$, i.e. ${a-b\equiv 0 \mod m}$. Thus we get a map ${I_\frak{m}\rightarrow \mathbb{Z}}$ by taking the positive generator ${\frac{a}{b}\mapsto a-b}$. The condition we just wrote down shows that any fractional ideal in ${Prin(\mathbb{Z}\cdot m)}$ will map to a multiple of ${m}$ and hence the map passes to the quotient ${Cl_m(\mathbb{Q})\rightarrow \mathbb{Z}/m\mathbb{Z}}$. By the earlier statement about being relatively prime to ${m}$ we see that the image of the map lands inside the units ${(\mathbb{Z}/m\mathbb{Z})^\times}$. This gives us a plausibility argument that ${Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m\mathbb{Z})^\times}$, but to get an actual isomorphism the standard argument goes through a bit of trouble involving completions and approximation theorems which we’ll skip.