# Class Groups (Lec. 10, Thursday, February 18)

## Computing Class Groups

:::{.remark}
Last time: we defined an equivalence relation on nonzero ideals of $\ZZ_K$, namely \( I \approx J \iff I = \alpha J \) for some \( \alpha \in K\units \).
We then defined the **class group**
\[
\Cl(\ZZ_K) \da \Id(\ZZ_K) / \approx
.\]
We saw that ideal multiplication descends to a well-defined group structure on ideal classes. 
Since ideal multiplication is commutative, this is an abelian group, and moreover it is finite.
:::

:::{.example title="Computing the Class Group"}
Let $K = \QQ(d)$ where \( d \da \sqrt{ -5 } \).
We saw that every ideal class is represented by an element with bounded norm.
Applying it to this specific value of $d$, every element is represented by $[I]$ where \( N(I) \leq 6 \).
If we have such an ideal, it will factor into primes, and thus the class will factor into prime classes. 
Thus the group is actually *generated* by prime ideals of norm at most 6.
Any such ideal will lie above a prime $p \leq 6$, so $p=2,3,5$.
We saw
\[
\gens{ 2 } = P_1^2 && P_1 \da \gens{ 2, 1 + \sqrt{-5} } \\
\gens{ 3 } = P_2 P_3 && P_2 \da \gens{ 3, 1 + \sqrt{-5} },\, P_3 \da \gens{ 3, 1 - \sqrt{-5} }    \\
\gens{ 5 } = P_4^2 && P_4 \da \gens{ \sqrt{-5} }.  
.\]
We conclude that \( \Cl(\ZZ_K) = \gens{ P_1, \cdots, P_4 }  \).
What are the relations?
Consider $P_4$, and note that $\gens{ \sqrt{-5} } \approx  \gens{ 1 }$ since \( P_4 = \sqrt{-5} \gens{ 1 }  \). 
A similar argument works for any principal ideal, and we can throw out $P_4$.
Consider $P_2$ and $P_3$.
Since \( \gens{ 3 } \approx \gens{ 1 }   \), we have $P_2 = P_3 \inv$, so we can also throw out $P_2$, since we don't need to include the inverse of a generator.
Recall that there is a factorization
\[
\gens{ 1 - \sqrt{-5} } 
=
\gens{ 2, 1 - \sqrt{-5} } 
\gens{ 3, 1 - \sqrt{-5} }  
= P_1 P_3
\]
and so these are inverses and we can get rid of $P_3$.
So \( \Cl(\ZZ_K) = \gens{ [P_1] } \), which is a cyclic group.
The generator has to have order 1 or 2, since $P_1^2 = \gens{ 1 }$. 
The claim is that the order is 2: otherwise, it would be trivial, making the class group trivial, which would imply that $\ZZ_K$ is a PID.
Why?
This implies that every $I \in \Id(\ZZ_K)$ is dilation equivalent to the unit ideal, so $I = \alpha \gens{ 1 }$ for some \( \alpha\in K\units \). 
But since $I$ is an ideal in $\ZZ_K$, this forces \( \alpha\in \ZZ_K \) and $I = \gens{ \alpha }$. 
This is a contradiction, since every PID is a UFD, and \( \QQ( \sqrt{-5} ) \) has non-unique factorization.
So we can write \( \Cl(\ZZ_K) \cong \GG_a(\ZZ/2\ZZ) \).
:::

:::{.remark}
What is the class group useful for?
We'll tie this into Diophantine equations.
:::

:::{.example title="of using the class group to solve Diophantine problems"}
Solve the following equation in $\ZZ$:
\[
y^2 + 5 = x^3
.\]
Recall that we originally tried to do this by factoring the left-hand side and appealing to unique factorization in a number field to deduce that various factors were powers.
However, we don't have unique factorization.
Although we can write \( ( y + \sqrt{-5} ) (y - \sqrt{-5} ) = x^3 \), it's not clear that this is helpful.
The fix will be to go to $\Id(\ZZ_K)$, which does have unique factorization, where we'll also be able to use facts about the class group.
We can turn this into an equation in ideals:
\[
\gens{ y + \sqrt{-5} }  
\gens{ y - \sqrt{-5} }  
= 
\gens{ x }^3 && \in \Id(\ZZ[\sqrt{-5} ]) 
.\]
The original strategy was to show the left-hand factors were coprime in order to deduce they were both cubes.
We'll try to show these ideals are coprime in the monoid sense, then since their product is a cube they'll have to be cubes.
This uses the fact that this is a reduced unique factorization monoid, so being a cube up to a unit is not something we have to worry about here.

:::{.claim}
There is no common prime ideal that divides both factors, using unique factorization.
:::

:::{.proof title="?"}
Suppose toward a contradiction that $P$ is prime and divides both.
Using that ideal norms are multiplicative, $N(P) \divides N( \gens{ y + \sqrt{-5} }) = y^2 + 5$. 
We also know $P$ contains both factors, so it contains $(y + \sqrt{-5} ) - (y - \sqrt{-5} ) = 2 \sqrt{-5}$, so $N(P) \divides N( \gens{ 2 \sqrt{-5} } ) = 20$. 
Thus $N(P) \divides \gcd(y^2 + 5, 20)$ in $\ZZ$.
This is impossible!

- $y$ is necessarily even for the original equation to be true.
  If $y$ is odd, take the equation $\mod 8$: an odd squared is $1\mod 8$, so $y^2 +5 \equiv 6 \mod 8$, which is not a cube in $\ZZ/8$ since any cube is $0\mod 8$.

- $5$ can not divide $y$.
  If so, $5$ would divide the left-hand side and thus the right-hand side, which forces $5\divides x$ since $5$ is prime.
  Then $5^3 \divides x^3$, meaning $5^3 \divides y^2 + 5$.
  In this case, $5^2 \divides y^2 + 5$, and if $5\divides y$ then $5^2 \divides y^2$, so we'd need $5^2 \divides 5$.

These together imply that $\gcd(y^2 + 5, 25) = 1$.
This $N(P) \divides 1$, forcing $P = \gens{ 1 }$, a contradiction.

:::

Thus we can write
\[
\gens{ y + \sqrt{-5} }  &= I^3 \\
\gens{ y - \sqrt{-5} }  &= J^3
.\]

In the previous argument, we wrote out $(a + b \sqrt{-5} )^3$, expanded, and compared coefficients.
Here we have an equation in ideals, and we can't do something similar unless $I, J$ are principal.
This is in fact the case: we'll restrict our attention to the class group.
The left-hand side is the unit ideal, since it is principal.
So we can write $[I]^3 = [J]^3 = e$, but we also know $\Cl(\ZZ_K) \cong \GG_a(\ZZ/2\ZZ)$, so this can only happen if $[I] = [J] = e$ and $I, J$ must be principal.
So we can write \( I = \gens{ a + b \sqrt{-5}}  \) for some \( a, b \in \ZZ \).
Thus 
\[
\gens{ y + \sqrt{-5} } = \gens{ (a + b \sqrt{-5} )^2 } \implies y + \sqrt{-5} = \pm 1 \qty { a + b \sqrt{-5} }^3  
,\]
using the fact that they differ by a unit but the only units in $\ZZ[ \sqrt{-5} ]$ are $\pm 1$.
The original proof now goes through, comparing coefficients of \( \sqrt{-5} \).
This will force $b = \pm 1$, then plug things back in to find $a$, then $y$, then $x$.
The conclusion is that there are no solutions.
:::

:::{.remark}
The critical takeaway: unique factorization failed, but the structure of the class group saved us! 
We crucially used that it had no elements of order 3.
See the book for a general theorem about equations $y^2 + d = x^3$.
Ideal theory gives us a way to study Diophantine equations.
:::


## The Class Group as a Measure of Non-unique Factorization

:::{.remark}
This is chapter 10.
This statement shows up in talks: it's more of a vague sentiment than an actual theorem, but we'll discuss a way to make it precise.
:::

:::{.theorem title="Class number 1 iff UFD"}
Recall that the class number is defined as $h_K \da \# \Cl(\ZZ_K)$.
Then
\[
h_K = 1 \iff \ZZ_K \text{ is a UFD}
.\]
:::

:::{.proof title="of theorem"}
$\implies$:
Every ideal is equivalent to the unit ideal, so every ideal is principal and PID implies UFD.

\

$\impliedby$:
Note that this is subtle: this is the claim that $\ZZ_K$ is a UFD $\implies \ZZ_K$ is a PID, which isn't true for general rings (e.g. $\ZZ[x]$).
Suppose $\ZZ_K$ is a UFD, then it's enough to show that every prime ideal is principal.
Let $P$ be prime, then $P$ lies above some ordinary prime $p$, so $P \divides \gens{ p }$.
We can factor \( \gens{ p }= \gens{ \prod_{i=1}^k \pi_i  } = \prod_{i=1}^k \gens{ \pi_i } \) for some $\pi_i$ irreducible.
A prime ideal dividing a product, by unique factorization, must divide a factor, so $P \divides \gens{ \pi_i }$ for some $i$. 
In a UFD, irreducibles are prime, so \( \gens{ \pi_i }  \) is a prime ideal, so we have a prime ideal dividing a prime ideal.
By unique factorization, this forces $P = \gens{ \pi_i }$, make $P$ principal.
:::

:::{.question}
Can anything be said if $h_K = 2$, even though we know $\ZZ_K$ is not a UFD?
:::

:::{.theorem title="Carlitz"}
$h_K = 2 \iff$ in $\ZZ_K$, any two factorizations of nonzero nonunit $\alpha$ into irreducibles have the same number of terms.
:::

:::{.remark}
For example, in $\ZZ[ \sqrt{-5} ]$ we have $6 = (2)(3) = (1 + \sqrt{-5} )(1 - \sqrt{-5} )$, which have the same number of factors.
To prove this theorem, we'll first need a lemma:
:::

:::{.lemma title="Class Number 2 implies 2 factors"}
Suppose $h_K = 2$, and suppose \( \pi \in \ZZ_K \) is an irreducible that is not prime (which is possible in a non-UFD).
Then factoring \( \gens{ \pi } = P_1 P_2  \) involves exactly two prime ideals $P_1, P_2$ in $\ZZ_K$.
:::

:::{.proof title="of lemma"}
Write \( \gens{ \pi }= \prod_{i=1}^g P_i  \), we then want to show $g=2$.
We have $g\geq 2$, since otherwise this would be a prime ideal, which would make $\pi$ a prime element.
The claim is that none of the $P_i$ can be principal.
Suppose toward a contradiction $P_1 = \gens{ \rho }$. 
Note that multiplying ideals yields smaller sets, so the right-hand side is a subset of \( \gens{ \rho }  \), as is the left-hand side, and so \( \rho \divides \pi \).
Since the $P_i$ were principal prime ideals, $\rho$ is prime and thus irreducible (since prime $\implies$ irreducible for any domain), so $\rho = u \pi$ for some unit.
Thus they generate the same ideal, and \( P_1 = \gens{ \rho } = \gens{ \pi }   \).
But then $\pi$ generates a prime ideal, make $\pi$ prime, a contradiction.

So none of the $P_i$ are principal.
Look at this equation in the class group.
The left-hand side is the identity, and the right-hand side are all non-identity elements a group of order 2.
So $[P_1][P_2] = e$, making $P_1 P_2 = \gens{ \omega }$ principal.
Then \( \gens{ \pi }\subseteq \gens{ \omega }   \) and so \( \omega \divides \pi \).
Moreover, \( \omega \) is not a unit since the product of two prime ideals is not the unit ideal.
Since \( \pi \) is irreducible, this makes \( \omega= u \pi \) and thus \( \gens{ \omega } = \gens{ \pi }   \).
If this were the case, we could cancel in the original equation:
\[
\gens{ \pi } &= (P_1 P_2)P_3 \cdots P_g = \gens{ \pi } P_3 \cdots P_g   \\
\implies \gens{ 1 } = P_3 \cdots P_g 
,\]
but this is a product of prime ideals resulting in the unit ideal.
This can only happen if there are no terms in this product, so $g=2$.
:::

:::{.proof title="of theorem ($\implies$)"}
Suppose $h_K = 2$.
We already know $\ZZ_K$ is not a UFD by the previous theorem.
The nontrivial part is showing factorization into nonzero nonunits of the same number of terms.
Instead of working with factorization of elements, we'll work with factorization of their principal ideals into principal ideals generated by irreducibles, which will obviate the need to worry about units.
We thus want to show that any principal ideal $P \neq \gens{ 0 }, \gens{ 1 }$ has all of its factorizations into principal ideals generated by irreducibles the same length.

:::{.example title="?"}
Much like the previous example, we have
\[
\gens{ 6 } = \gens{ 1 + \sqrt{-5} } \gens{ 1 - \sqrt{-5} } = \gens{ 2 } \gens{ 3 }     
.\]
:::

Suppose that $P$ factors as 
\[
P = 
\prod_{i=1}^k \gens{ \pi_i }
=
\prod_{j=1}^\ell \gens{ \rho_j }
&&
\pi_i, \rho_j \text{ irreducible}
,\]
we'd then like to show that $k=\ell$.

:::{.observation}
If $\pi_1$ is prime, we can use that \( \gens{ \pi_1 }\divides \prod \gens{ \rho_j }   \), and would thus have to divide (say) \( \gens{ \rho_1 }  \) up to relabeling.
Since everything is irreducible, if $\pi_1 \divides \rho_1$ then \( \gens{ \pi_1 }= \gens{ \rho_1 }   \), meaning we can cancel.
:::

So after cancellation, we can suppose that all the $\pi_i, \rho_j$ and none are prime.
Consider the number of prime ideals that show up after factoring all of the principal ideals on either side.
By the lemma, any irreducible that's not prime factors into two primes, so we get $2k$ primes on the left-hand side (not necessarily distinct) and $2\ell$ on the right-hand side.
But the factorization into primes *is* unique, so $2k=2\ell$ and $k=\ell$.
:::

:::{.remark}
See the book for the other direction!
:::

:::{.theorem title="Landau"}
Every ideal class contains infinitely many prime ideals.
:::

:::{.remark}
This is an analytic theorem!
The proof is similar to how Dirichlet proved the infinitude of primes in arithmetic progressions, which involves $L\dash$functions.
:::

:::{.remark}
What about $h_K \geq 2$? 
We'll introduce a way of measuring how bad unique factorization fails in a ring, the notion of *elasticity*.
:::


## Elasticity

:::{.definition title="Elasticity of a Ring"}
Let \( \alpha\in \ZZ_K \) where \( \alpha\neq 0 \) and is not a unit.
Define
\[
\rho(\alpha) \da { L( \alpha) \over S( \alpha) }
,\]
where $L( \alpha)$ is the number of terms in the longest[^well_def_max]
factorization of \( \alpha \) and \( S( \alpha ) \) is the shortest number of terms.
This measures how far away from unique the factorization of \( \alpha \) is.
Now define the **elasticity** of $\ZZ_K$ as
\[
\rho(K) \da \sup_{\alpha} \rho( \alpha)
.\]

[^well_def_max]: 
There is a way to factor that maximizes the number of irreducibles appearing, and there are not arbitrarily long factorizations.

:::

:::{.remark}
Note that $h_K = 1, 2\iff \rho(K) = 1$, and $h_K > 2 \implies \rho(K) > 1$.
:::

:::{.theorem title="Elasticity in terms of the Davenport constant"}
For $h_K \geq 2$, 
\[
\rho(K) = {1\over 2} D( \Cl( \ZZ_K))
,\]
where $D(G)$ is the **Davenport constant** of the finite abelian group $G$: the smallest number $D$ such that every sequence of $D$ elements of $G$ contains a nonempty subsequence whose product is the identity.
This is a function from combinatorial group theory.
:::

:::{.exercise title="bounding the Davenport constant"}
Show that $D(G) \leq \abs{G}$.
:::

:::{.fact}
$D(G) \to \infty$ as $\abs{G} \to \infty$.
:::

:::{.corollary title="?"}
If $h_K \to \infty$ for a sequence of number fields, then $\rho(K) \to \infty$.
:::

:::{.remark}
This just follows from the above facts, since $h_K \to \infty$ means the size of the group $G \da \Cl( \ZZ_K)$ goes to infinity, which is a constant times $\rho(K)$.
So as the class group gets larger, factorization gets worse.
:::