# Quadratic Fields (Lec. 3, Thursday, January 21)

:::{.remark}
Today: roughly corresponds to chapter 3 in the book.
Goal: do all of the big theorems in the setting of quadratic number fields, then redo everything for general number fields.
:::

## Quadratic Number Fields

:::{.remark}
Simplest case: $\QQ$, a degree 1 number field, so the next simplest case is degree 2.
:::

:::{.definition title="Quadratic Number Fields"}
A field $K$ is a **quadratic number field** if and only if $K$ is a number field and $[K: \QQ] = 2$.
:::

:::{.remark}
Some notation: if $d\in \RR\units$, then $\sqrt d$ means the *positive* square root of $d$ if $d \geq 0$, and if $d<0$ this denotes $i\sqrt{\abs{d}}$.
:::

:::{.proposition title="Quadratic fields are parameterized by squarefree integers"}
If $K$ is a quadratic number field, then $K = \QQ(\sqrt{d})$ for some squarefree 
[^note_squarefree]
integer $d\in \ZZ$.
Moreover, this $d$ is uniquely determined by $K$, so all quadratic number fields are parameterized by the set of squarefree integers.

[^note_squarefree]: 
*Squarefree* means not divisible by $n^2$ for any $n > 1\in \ZZ$, or equivalently not divisible by the square of any primes.

:::

:::{.proof title="of proposition, existence"}
**Existence**: 
Since $[K: \QQ] = 2$, we have $K\supsetneq \QQ$ so pick \( \alpha\in K\sm \QQ \) then $K = \QQ(\alpha)$.
Note that we could also furnish this \( \alpha \) from the primitive element theorem, although this is overkill here.
So \( \alpha \) is a root of some degree 2 $p\in \QQ[x]$, and by scaling coefficients we can replace this by $p\in \ZZ[x]$.
So write $p(x) = Ax^2 + Bx + C$, in which case we can always write 
\[
\alpha = {-B \pm \sqrt{B^2 - 4AC} \over 2A} 
\] 
where $A\neq 0$, since this would imply that \( \alpha\in\QQ \).
Writing \( \Delta\da B^2 - 4AC \), we have $K = \QQ(\alpha) = \QQ(\sqrt{\Delta})$.
This is close to what we want -- it's $\QQ$ adjoin some integer -- but we'd like that integer to be squarefree.

\

Now let $f\in \ZZ^{\geq 0}$ be chosen such that $f^2 \divides \Delta$ and $f$ is as large as possible, i.e. the largest square factor of \( \Delta \).
Writing \( \Delta = f^2 - d \) where $d$ is whatever remains.
Then $d$ must be squarefree, otherwise if $d$ had a square factor bigger than 1, say $d = r^2 d'$, in which case $f^2 r^2 > f^2$ would be a larger factor of \( \Delta \).
So $d$ is squarefree, and \( \Delta = f \sqrt d \) and thus \( \QQ(\Delta) = \QQ(\sqrt{d}) \).

**Uniqueness**: 
Well use some extra machinery.

:::

## Norm and Trace

:::{.definition title="Norm and Trace"}
Let $K$ be a number field with $K_{/\QQ}$ Galois.
For each \( \alpha\in K \) define
\[
N(\alpha) &\da \prod_{\sigma\in \Gal(K_{/\QQ})} \sigma(\alpha) && \text{the norm} \\
\Tr(\alpha) &\da \sum_{\sigma\in \Gal(K_{/\QQ})} \sigma(\alpha) && \text{the trace}
.\]
:::

:::{.remark}
Why use these kind of sum at all?
Applying any element in the Galois group just permutes the elements.
Note that $N( \alpha), \Tr( \alpha)$ are $G(K_{/\QQ})\dash$invariant, and thus rational numbers in $\QQ$.
The norm is multiplicative, and the trace is additive and in fact $\QQ\dash$linear: $\Tr(a \alpha + b \beta) = a \Tr( \alpha) + b \Tr( \beta)$ for all \( \alpha, \beta\in K \) and all \( a,b \in \QQ \).
:::
  
:::{.remark}
What do the norm and trace look like for a quadratic field?
We can write $K = \ts{a + b \sqrt d \st a,b \in \QQ}$ and there is a unique (non-identity) element \( g\in \Gal(K_{/\QQ}) \) with \( \sigma(a + b \sqrt d) = a - b \sqrt{d} \).
We'll refer to this automorphism as **conjugation**.
We can compute
\[
N(a + b \sqrt{d} ) &= a^2 - db^2 \\
\Tr(a + b \sqrt{d} ) &= 2a
.\]
:::

:::{.proof title="of proposition, uniqueness continued"}
Returning to the proof, suppose otherwise that $K = \QQ(\sqrt{d_1} ) = \QQ ( \sqrt{d_2} )$ with $d_1\neq d_2$ squarefree integers.
Note that they must have the same sign, otherwise one of these extensions would not be a subfield of $\RR$.
We know \( \sqrt{d_1} \in \QQ( \sqrt{d_2} ) \) and thus \( \sqrt{d_1} = a + b \sqrt{d_2} \) for some $a, b\in \QQ$.

\

Taking the trace of both sides, the LHS is zero and the RHS is $2a$ and we get $a=0$ and \( \sqrt{d_1} = b \sqrt{d_2} \).
Write $b = u/v$ with $u,v\in \QQ$.
Squaring both sides yields $v^2 d_1 = u^2 d_2$.
Let $p$ be a prime dividing $d_1$; then since $d_1$ is squarefree there is only one copy of $p$ occurring in its factorization.
Moreover there are an even number of copies of $p$ coming from $v^2$, thus forcing $d_2$ to have an odd power of $p$.
This forces $p\divides d_2$, and since this holds for every prime factor $p$ of $d_1$, we get $d_1 \divides d_2$ since $d_1$ is squarefree.
The same argument shows that $d_2 \divides d_1$, so they're the same up to sign: but the signs must match and we get $d_1 = d_2$.

:::

:::{.remark}
Note that this results holds for every squarefree number not equal to 1.
If $K = \QQ( \sqrt{d} )$, what is the ring of integers $\ZZ_K$?
Some more machinery will help here.
:::

## The Field Polynomial

:::{.definition title="The Field Polynomial of an Element"}
Assume $K_{/\QQ}$ is a Galois number field and for \( \alpha\in K \) define the **field polynomial of $\alpha$** as
\[
\varphi_{\alpha}(x) \da \prod_{ \sigma\in \Gal(K_{/\QQ})} \qty{ x - \sigma(\alpha)}
.\]
:::

:::{.remark}
For the same reasons mentioned for the norm/trace, we get \( \varphi_{\alpha} \in \QQ[x] \), and moreover \( \varphi_{ \alpha } (\alpha) = 0 \).
When is \( \alpha\in \ZZ_K \)? 
We have the following criterion:
:::

:::{.proposition title="The field polynomial detects integrality"}
\[
\alpha\in \ZZ_K \iff \varphi_{ \alpha } (x) \in \ZZ[x]
.\]

:::

:::{.proof title="of proposition"}
$\impliedby$:
This is easy, since if \( \varphi_\alpha \) is a monic polynomial with integer coefficients, meaning that \( \alpha \) is an algebraic integer and thus in $\ZZ_K$.

\

$\implies$:
If \( \alpha \in \ZZ_K \) then it's the root of some monic polynomial in $\ZZ[x]$, and the same is true for \( \sigma(\alpha) \) and thus each \( \sigma(\alpha) \in \bar\ZZ \).
So \( \varphi_{ \alpha}(x) \in \bar \ZZ[x] \).
We said \( \varphi_{ \alpha} \) has coefficients in $\QQ$ too, and thus in $\bar \ZZ \intersect \QQ = \ZZ$.
So the problem is reduced to finding out when \( \varphi_{\alpha}(x) \) has integer coefficients.

If $\deg(K_{/\QQ}) = n$, then 
\[
\varphi_{ \alpha}( \alpha) = \prod x- \sigma(\alpha) = x^n - \Tr(\alpha)x^{n-1} + \cdots + (-1)^n N( \alpha)
.\]
If $n=2$, these are the only terms, and so if $K$ is a quadratic number field then \( \alpha\in K \) is in $\ZZ_K$ if and only if $\Tr( \alpha), N(\alpha) \in \ZZ$.
:::

:::{.example title="of nonintuitive rings of integers"}
Let $K = \QQ( \sqrt{5} )$, then is it true that $\ZZ_K = \ZZ[\sqrt{5} ]$?
Since $1, \sqrt{5} \in \ZZ_K$, we have $\supseteq$ since $1, \sqrt{5}$ are algebraic.
The answer is **no**: take \( \alpha\da {1 + \sqrt{5} \over 2} \), then \( N( \alpha) -4/4 = -1 \) and \( \Tr( \alpha) = 1 \).
These are integers, so \( \alpha\in \ZZ_K \), and in fact \( \alpha \) is a root of $x^2 - x - 1 \in \ZZ[x]$.
:::

## Classification of $\ZZ_K$

:::{.theorem title="Classification of $\ZZ_K$ for quadratic fields"}
Let $K = \QQ( \sqrt{d} )$ be a quadratic number field.
Then 

- If $d = 2,3 \mod 4$, then $\ZZ_K = \ts { a + b \sqrt{d} \st a, b\in \ZZ}$.

- If $d=1 \mod 4$, then $\ZZ_K = \ts{ {1 + b \sqrt{d} \over 2} \st a,b\in \ZZ,\, a\equiv b \mod 2}$.
:::

:::{.remark}
For $d=1$, if $a, b$ are even then we just recover the $d=2,3$ case, so we're picking up extra elements from when $a,b$ both odd.
:::

:::{.proof title="?"}
Let \( \alpha\in K \) and write \( \alpha = A + B \sqrt{d} \) with $A, B\in \QQ$.


:::{.exercise title="?"}
Check that \( N( \alpha), \Tr( \alpha) \in \ZZ \) for both cases.
:::

Assuming now that $N( \alpha), \Tr( \alpha) \in \ZZ$, then $A^2 - dB^2 \in \ZZ$.
Multiply this by 2 to get $(2A)^2 - d(2B)^2 \in 4\ZZ$.
Recalling that \( \Tr( \alpha) = 2 A \), we have $(2A)^2 \in \ZZ$ and thus $d(2B)^2 \in \ZZ$ as well.
The claim now is that $2B \in \ZZ$: we know $2B\in \QQ$.
If $2B\not\in \ZZ$, then the denominator has some prime factor.
This prime factor appears twice in $(2B)^2$, and $d(2B)^2 \in \ZZ$ then means that two copies of $p$ appear in $d$ in order to cancel -- however, we assumed $d$ was squarefree.
We now know that $A, B \in {1\over 2}\ZZ$, so write $A = (1/2)a'$ and $B = (1/2)b'$.
Thus
\[ 
\alpha= (1/2)a' + (1/2)b' \sqrt{d} 
\implies
N( \alpha) = ((a')^2 - d(b')^2) / 4 \in \ZZ
.\]
So the numerator is a multiple of 4, which yields $(a')^2 \equiv d(b')^2 \mod 4$.
We proceed by cases.

\

**Case 1:**
$d = 2,3 \mod 4$.
If $b'$ is odd then $(b')^2 = 1\mod 4$, which holds for any odd number.
But then $(a')^2 = d(b')^2 = d \mod 4$, which is a problem -- squares modulo 4 can only be $0$ or $1$.
This is a contradiction, so $b'$ must be even.
Then $(b')^2 \mod 4 = 0$, which forces $a' \equiv 0 \mod 4$ and $a'$ must be even.
But if $a', b'$ are both even, $(1/2)a', (1/2)b'\in \ZZ$ and we obtain \( \alpha\in \ZZ + \sqrt{d} \ZZ \) .

\

**Case 2:**
If $d\equiv 1 \mod 4$, then $(a')^2 \equiv (b')^2 \mod 4$.
We can conclude that $a', b'$ are either both odd or both even, otherwise we'd get $0\equiv 1 \mod 4$, and thus we can write $a' \equiv b' \mod 2$.
But this was exactly the condition appearing in the theorem.
:::

:::{.remark}
Let $K$ be a quadratic number field.
Then we can reformulate the previous results as:

\[
\ZZ_K = 
\begin{cases}
\ZZ[ \sqrt{d} ] &  d \equiv 2,3 \mod 4
\\
\ZZ\left[{1 + \sqrt{d} \over 2}\right] & d \equiv 1 \mod 4.
\end{cases}
\]

We've also shown that $\ZZ_K$ is a free \(\ZZ\dash\)module of rank 2, with basis either \( \ts{ 1, \sqrt{d} } \) or \( \ts{ 1, {1 + \sqrt{d} \over 2 } } \).
:::

:::{.remark}
What is true for general number fields?
Important theorem: $\ZZ_K$ is always a free \(\ZZ\dash\)module, i.e. there always exists an *integral basis*.
Surprisingly, the it's not always true that $\ZZ_K = \ZZ[\ell]$ for $\ell$ a single element.
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