# Starting Over with General Number Fields (Lec. 13, Thursday, March 04)

## Recasting Old Definitions

:::{.remark}
This corresponds to chapter 13, "Starting Over".
Idea: we've phrased everything so far for quadratic fields, now we want to do everything for general number fields.
The basic objects and tools: norm and trace.
If $K/\QQ$ is a number field that's Galois, we define \( N \alpha \da \prod_{ \sigma \in G} \sigma( \alpha) \) and \( \Tr \alpha \da \sum_{ \sigma\in G } \sigma( \alpha) \).
We'll have to modify this for a general number field since there's not an immediate candidate for the Galois group.

Setup: let $K$ be a number field with \( [K: \QQ] = n \), then recall that there exist $n$ different embeddings \( \sigma: K \embeds \CC \).

:::

:::{.definition title="Field Polynomial"}
If \( \alpha\in K \) then define its **field polynomial**
\[
\varphi_{ \alpha}(x) \da \prod_{\sigma: K\embeds \CC} (x - \sigma( \alpha) )
.\]
:::

:::{.proposition title="The field polynomial is monic, has rational coefficients, and is a power of the minimal polynomial"}
This is a monic polynomial with $\CC\dash$coefficients, and in fact \( \varphi_{ \alpha} \in \QQ[x] \) and \( \varphi_{ \alpha}(x) = \min_{ \alpha} (x)^n\) (the minimal polynomial over $\QQ$) for some power $n$, and the correct choice turns out to be \( n \da [K: F[ \alpha] ] \).
:::

:::{.remark}
Note that the first claim follows from the second since \( \min_{ \alpha}(x) \in \QQ[x] \).
:::

:::{.lemma title="Number of embeddings of subfields"}
Let $K$ be a number field with $[K: \QQ] = n$ and $F\leq K$ a subfield with $[F: \QQ] = r$.
Note that $r\divides n$.
Then every embedding \( \tau: F\embeds \CC \) extends in $n/r$ ways to an embedding \( \sigma: K \injects \CC \).
:::

:::{.proof title="of lemma, sketch"}
Standard field theory exercise: by the primitive element theorem, write $K = F( \theta)$ where \( \deg( \theta) = [K: F] = n/r \) over $F$.
Since we're extending an embedding, it suffices to define what it does to \( \theta \).
If $m(x) = \min_{ \theta}(x)$ over $F$, then \( \sigma( \theta) \) must be a root of \( (\tau m)(x) \), where the latter polynomial is taking \( m \) and applying \( \tau \) to each of the coefficients.
Note that this preserves the degree, so \( \deg \tau m = n/r \), and there are \( n/r \) choices for \( \sigma( \theta) \).
Now the proof follows from checking that every single root is a possibility.
:::

:::{.proof title="of proposition"}
By definition, \( \varphi_{ \alpha}(x) \) is a product over embeddings \( \sigma:K \embeds \CC \), and each such \( \sigma \) restricts to an embedding $F[ \alpha] \injects \CC$, so applying the lemma to \( F[ \alpha] \leq K \) yields 
\[
\varphi_{ \alpha}(x) 
&= \prod_{ \sigma: K\embeds \CC} (x - \sigma( \alpha) ) \\
&= \prod_{ \tau: F[ \alpha] \embeds \CC} \,\,\, \prod_{ \sigma \text{ s.t. } \ro{ \sigma}{ F[ \alpha ] } = \tau } (x - \sigma( \alpha) ) \\
&= \prod_{ \tau: F[ \alpha ] \embeds \CC } (x - \tau( \alpha ) )^{n(\tau)} \\
&= \qty{ \prod_{ \tau: F[ \alpha] \embeds \CC} (x - \tau( \alpha ) ) }^{[K: F(\alpha)]} \\
&= \min_ \alpha(x) ^{[ K : F( \alpha ) ] }
.\]
where 

- We've first just reorganized the product by grouping, 
- Then we've used that all of the terms in the inner product must have the same value for \( \sigma( \alpha) \) since \( \alpha\in F[ \alpha] \) and this makes \( \sigma( \alpha) = \tau( \alpha) \),
- We note that the exponent should be the number of terms in the inner product, i.e. the number of \( \sigma \) extending \( \tau \), i.e. \( n(\tau) = [K: F[ \alpha] ] \) since \( r = [F[ \alpha] : \QQ] \) and \( n = [K: \QQ ] \),
- The last equality follows from remarks in chapter 1.
:::

:::{.remark}
The field polynomial gives us a way to determine whether an element of a number field is in its ring of integers:
:::

:::{.proposition title="Field polynomial has integer coefficients iff the element is an integer"}
\[
\alpha\in \ZZ_K \iff \varphi_{ \alpha}(x) \in \ZZ[x]
.\]
:::

:::{.proof title="of proposition"}
$\impliedby$:
This direction is easy, since having integer coefficients, being monic, and having \( \alpha \) as a root since \( x - \sigma( \alpha) = x - \alpha \) for some \( \sigma \).
But this puts \( \alpha \in \ZZ_K \) by definition.

$\implies$:
We proved that if \( \alpha\in \ZZ_K \) then \( \min_ \alpha(x) \in \ZZ[x] \), and \( \varphi_{ \alpha} \)  is just a power of \( \min_ \alpha(x)\).
:::

:::{.definition title="Norm and Trace"}
Write 
\[
\varphi_{ \alpha} (x) = x^n + \sum_{i=1}^n a_i x^i \in \QQ[x]
,\]
we then define the **norm** and **trace**[^norm_and_trace] respectively as
\[
N( \alpha) &\da (-1)^n a_0 \in \QQ\\
\Tr( \alpha) &\da -a_{n-1} \in \QQ
.\]
Note that if \( \alpha \in \ZZ_K \) then these are both in fact in $\ZZ$.

[^norm_and_trace]: 
These come from the trace and determinant of the map \( y \mapsto y\cdot x \) on $L/K$, viewed as a $K\dash$linear map on $L$.

:::

:::{.remark}
Note that $-(1)^n a_0 = \prod r_i$ is the product of the roots of \( \varphi_{ \alpha} (x)\) and \( -a_{n-1} = \sum r_i \), so equivalently we can think of these as 
\[
N( \alpha ) &= \prod_{ \sigma: K \embeds \CC } \sigma( \alpha) \\
\Tr( \alpha ) &= \sum_{\sigma: K \embeds \CC} \sigma( \alpha) 
.\]
It's also the case that $N(\wait)$ is multiplicative and $\Tr(\wait)$ is $\QQ \dash$linear.
:::

## Discriminants

:::{.remark}
Let $K$ be a number field and $[K : \QQ] = n$.
Pick an arbitrary ordering of embeddings $\sigma_1, \cdots, \sigma_n: K \embeds \CC$.
:::

:::{.definition title="Tuple Discriminant"}
For any $n\dash$tuple $(w_1, \cdots, w_n) \in K^n$ define the **tuple discriminant** as 
\[
\Delta(w_1, \cdots, w_n) \da \det( D_{w_1, \cdots, w_n} )^2 
\]
where
\[
D_{w_1, \cdots, w_n}
=
\begin{bmatrix}
\sigma_1(w_1) & \cdots  & \sigma_1(w_n) \\
\sigma_2(w_1) & \cdots  & \sigma_2(w_n) \\
\vdots & \cdots & \vdots \\
\sigma_n(w_1) & \cdots  & \sigma_n(w_n) 
\end{bmatrix}
.\]

:::

:::{.remark}
Why square this? 
Permuting two columns changes the sign of the determinant, which is just swapping the order of the embeddings.
So squaring keeps this invariant under relabeling the $\sigma_i$.
It turns out that this is a rational number, since we can write
\[
\Delta(w_1, \cdots, w_n) 
= \det(D) \det(D) 
\det(D^t D)
,\]
where $(D^t D)_{ij}  = \Tr(w_i w_j) \implies D^t D \in \Mat(n\times n, \QQ)$.
So taking the determinant yields a rational number, so $\Delta(w_1, \cdots, w_n) \in \QQ$. 
Moreover if you start with the $w_i \in \ZZ_K$, then $D^t D \in \Mat(n\times n, \ZZ)$ and thus $\Delta(w_1, \cdots, w_n) \in \ZZ$.

Why is this called the discriminant?
:::

:::{.theorem title="The discriminant detects $\QQ\dash$bases"}
Let $w_1, \cdots, w_n\in K$, then 
\[
\ts{ w_1, \cdots, w_n } \text{ form a $\QQ\dash$basis for $K$}
\iff
\Delta(w_1, \cdots, w_n) \neq 0
.\]
:::

:::{.remark}
So this *discriminates* between bases and non-bases.
:::

:::{.proof title="of theorem"}
$\impliedby$:
Suppose $\Delta(w_1, \cdots, w_n) \neq 0$.
Note that the $n$ elements $w_1, \cdots, w_n$ are $n$ elements in an $n\dash$dimensional $\QQ\dash$vector space, so the only way they could fail to be a basis would be if there were a linear dependence.
But then considering the matrix $D$ above, a $\QQ\dash$linear dependence between the $w_i$, this translates to a corresponding dependence between the columns of $D$, which would yield the contradiction $\det(D)^2 = 0$.

\

$\implies$:
This is the harder part.
Toward a contradiction suppose $w_1, \cdots, w_n$ are a $\QQ\dash$basis for $K$ but $\Delta(w_1, \cdots, w_n) = \det(D^t D) = 0$.
Then the columns of $D^t D$ are linearly dependent, so there are $c_i \in \QQ$ not all zero such that 
\[
\sum_{j=1}^n c_j \Tr( w_i w_j) = 0 && \forall i=1, \cdots, n
.\]
Introduce an element \( \beta \da \sum_{j=1}^n c_j w_j \in K\units \), which is not zero since not all of the $c_j$ are zero and the \( w_i \) are a basis.
Using linearity of the trace, we can write 
\[
\Tr(w_i \beta) = 0 && \forall i=1, \cdots, n
.\]
Again using linearity, we actually have \( \Tr( \alpha \beta) = 0 \) for all \( \alpha\in K \) since every \( \alpha \) is in the \( \QQ\dash \)span of the $w_i$, which are a basis.
It's then perfectly fine to take \( \alpha \da \beta\inv \), which forces \( \Tr(1) = 0 \).
But we can compute directly that \( \Tr(1) = n > 0 \) since every embedding \( \sigma \) must send 1 to 1.
:::

## Integral Bases 

:::{.theorem title="Integral Basis Theorem"}
For $K$ any number field of degree $n$, $\ZZ_K \in \mods{\ZZ}$ is free of rank $n$.
:::

:::{.observation}
Suppose $\vector (w_1, \cdots, w_n), (\theta_1, \cdots, \theta_n) \in K$ where \( \tv{w_1, \cdots, w_n} = \tv{ \theta_1, \cdots, \theta_n } M \) for some matrix $M\in \Mat(n\times n, \QQ)$.
Then 
\[
\Delta(\elts{w}{n}) = \Delta(\elts{\theta}{n}) \det(M)^2
.\]

:::

:::{.proof title="of observation"}
Applying the embeddings $\sigma_i$ yields an equality $D_{ \elts w n }= D_{\elts \theta n } M$.
Now taking determinants and squaring yields the result.
:::

:::{.proof title="of integral basis theorem"}
Choose \( \elts w n \in \ZZ_K \) such that

1. \( \discriminant(\elts w n ) \neq 0 \) 
2. \( \abs{ \discriminant(\elts w n ) } \) is minimal among those satisfying (1).

Does this make sense?
The claim is that if (1) is possible, then (1) and (2) is also possible.
This is because $\discriminant(\elts w n ) \in \ZZ$, taking absolute values makes it positive, and then we can minimize among the positive integers occurring using the well-ordering principle.
But we can choose tuples satisfying (1): we can always choose a $\QQ\dash$basis, and to get them down to $\ZZ_K$ instead of $K$, they can just be scaled by a rational integer without changing that they form a basis.


:::{.claim}
Any tuple \( \elts w n \) satisfying (1) and (2) will be a \( \ZZ\dash \)basis for $\ZZ_K$.
:::

How could this fail?
No elements could have multiple representations as a $\ZZ\dash$basis, since they don't admit any in the $\QQ\dash$basis.
So it suffices to show \( \spanof_\ZZ \ts{ \elts w n } = \ZZ_K \).
If not, choose \( \alpha\in \ZZ_K \) not in their $\ZZ\dash$span -- it must still be in the $\QQ\dash$span, so we can write \( \alpha= \sum c_i w_i \) where the $c_i\in \QQ$.
We can assume that $c_1\not \in \ZZ$ by renumbering.
Now write 
\[
\beta \da \alpha - \integerpart{c_1} w_1 \in \ZZ_K
,\]
where \( \integerpart{\wait} \) denotes taking the integer part.
We can write \( \beta = \fractionalpart{c_1}w_1 + c_2 w_2 + \cdots + c_n w_n \).
Observe that the tuple 
\[
\tv{ \beta, w_2, \cdots, w_n} 
= \tv{ w_1 , w_2, \cdots, w_n} 
M,
&& 
M \da 
\begin{bmatrix}
\fractionalpart{c_1} & 0  & \cdots & \vdots \\
c_2 &  1 & \ddots & \vdots \\
c_n &  0 & 0 & 1
\end{bmatrix}
.\]
noting that the first column describes how to write \( \beta \) as a linear combination of the \( w_i \).
Taking discriminants yields
\[
\discriminant(\beta, w_2, \cdots, w_n) 
= \discriminant( \elts w n )\det(M)^2
= \discriminant( \elts w n )\fractionalpart{c_1}^2
,\]
where we've computed the determinant using the fact that it is lower triangular.
Since $c_1\not\in\ZZ$, we have \( \fractionalpart{c_1} \) nonzero, real, between 0 and 1.
Since the discriminant was nonzero, the right-hand side is nonzero and thus neither is the left-hand side.
We would then have
\[
0 < 
\abs{ \discriminant( \beta, w_2, \cdots, w_n) }
<
\abs{ \discriminant( \elts w n ) }
,\]
which contradicts the minimality of the $w_i$.
$\contradiction$
:::

:::{.remark}
Note that this is non-constructive, finding a basis is another question!
:::

## Discriminant of Number Fields

:::{.definition title="Discriminant of a Number Field"}
Let $K$ be a number field, then the **discriminant** of $K$ is defined as 
\[
\discriminant_K \da \discriminant(\elts w n) 
,\]
where \( \ts{ w_i } \) is any $\ZZ\dash$basis for $\ZZ_K$.
:::

:::{.remark}
Is this actually an invariant of $K$, since we made a choice of basis?
Given two $\ZZ\dash$bases for $\ZZ_K$, say \( \elts w n, \elts \theta n \), then
\[
\tv{ \elts w n} = \tv{ \elts \theta n } M && M \in \Gl(n, \ZZ)
.\]
Hence 
\[
\discriminant(\elts w n) 
= \discriminant( \elts \theta n ) \det(M)^2
= \discriminant( \elts \theta n )
.\]
using that invertible matrices have unit determinants, which in $\ZZ$ are just \( \pm 1 \).
:::

:::{.remark}
Why do we care?
The discriminant measures the complexity of the number field and carries arithmetic information:

:::{.theorem title="Hermite"}
For every $X>0$, there are only finitely many number fields such that $\abs{ \discriminant_K } \leq X$.
:::

:::{.remark}
Interesting question: how many are there as a function of $X$?
This is studied today by fixing a degree $n$, and we have good answers for $n=2,3,4,5$, but it's still open to get an asymptotic formula for $n>5$.
Note that our new faculty hire this year is an expert on these kinds of questions!
:::

:::{.theorem title="Dedekind"}
Taking a prime $p\in \ZZ$, we have
\[
p \text{ ramifies in } \ZZ_K \iff p \divides \discriminant_K
,\]
where **ramification** occurs if when \( \gens{ p }\normal \ZZ_K  \) factors into prime ideals with a repeated prime factor.
In particular, \( \discriminant(\wait) < \infty \), and so only finitely many such primes can occur.
:::

:::