# Discriminants and Norms (Lec. 14, Saturday, March 13)

:::{.example title="of a discriminant"}
Suppose \( K = \QQ( \sqrt{d} ) \) where $d$ is squarefree.
What is its discriminant?
We need a $\ZZ\dash$basis of $\ZZ_K$, for $d=2,3 \mod 4$ we can take \( (1, \sqrt{d} ) \).
Then we construct a matrix whose columns are the different embeddings of each entry.
The embeddings here are the identity and complex conjugation, so we get
\[
\Delta_K = \Delta(1, \sqrt{ d} )
= 
\det 
\qty{
\begin{bmatrix}
1 & \sqrt{d} 
\\
 1 & -\sqrt{d} 
\end{bmatrix}
}^2
= (-2 \sqrt{d} )^2 = 4d
.\]
If $d = 1 \mod 4$, then we can take a basis $(1, {1 + \sqrt{d} \over 2})$,
and
\[
\Delta_{K}
=
\qty{
\begin{bmatrix}
1 & {1 + \sqrt{d} \over 2}
\\
1 & {1 - \sqrt{d} \over 2}
\end{bmatrix}
}^2
= (- \sqrt{d} )^2 = d
.\]
So we have
\[
\Delta_K = 
\begin{cases}
d & d = 1 \mod 4 
\\
4d & d = 2,3 \mod 4 .
\end{cases}
\]

:::

:::{.remark}
Note that $\Delta_\QQ = 1$ if you trace through the computation.
:::

## Norms of Ideals

:::{.definition title="Norm of an ideal"}
Let $I \normal \ZZ_K$ be a nonzero ideal, then define \( N(I) \da \# \ZZ_K/I \).
:::

:::{.remark}
Note that this was finite in the quadratic field case since nonzero ideals had a "standard basis".
For general number fields, the ideals can be more complicated, so we'll need another way to show finiteness.
:::

:::{.lemma title="Elements divide their norms"}
Let \( \alpha\in \ZZ_K \), then \( \alpha\divides N( \alpha) \) in \( \ZZ_K \).
:::

:::{.proof title="of lemma"}
Write down the obvious thing and see that it works!
\[
N \alpha 
&\da \prod_{ \sigma: K \embeds \CC} \sigma( \alpha) \\
&= \alpha \prod_{ \substack{ \sigma: K \embeds \CC \\ \sigma\neq \one_K } } \sigma( \alpha) \\
&\da \alpha C 
,\]
where we've used that one embedding is the identity and factored it out.
So it only remains to show that the *cofactor* $C$ (the product term) is actually in $\ZZ_K$.
It is \( \bar{ZZ} \), since \( \alpha \) was an algebraic integer, i.e. a root of some monic polynomial with integer coefficients.
But then under every embedding, \( \sigma( \alpha) \) is a root of the same monic polynomial, so each \( \sigma( \alpha) \in \bar{\ZZ} \), as is their product since it's a ring.
On the other hand, we can write \( C = N \alpha/ \alpha \).
Since \( N \alpha \) is a nonzero rational integer and \( \alpha\in K \), and since $K$ is a field, this quotient is in $K$.
But then \( C \in \bar{\ZZ} \intersect K = \ZZ_K \). 
:::

:::{.proposition title="Nonzero ideals have finite norms in rings of integers"}
For $I\normal \ZZ_K$ nonzero,
\[
N(I) < \infty 
.\]
:::

:::{.proof title="of proposition"}
We start with principal ideals.
Let $m\in \ZZ^+$, then \( \ZZ_K \gens{ m } \da \ZZ_K /m \ZZ_K \cong_{\zmod} \ZZ^n / m\ZZ^n \cong (\ZZ/m\ZZ)^n  \) where we've forgotten the ring structure and are just considering it as a \(\ZZ\dash\)module .
But this has size $m^n < \infty$.

Now let \( \alpha\in I \) be nonzero and let \( m \da \pm N \alpha \), choosing whichever sign makes $m>0$.
Since \( \alpha\divides N \alpha \), so \( N \alpha = \ell \alpha \) is a multiple of \( \alpha \).
But \( \alpha\in I \) and $I$ is an ideal, so \( N \alpha\in I \implies m \in I\).
Then (check!) the following map is surjective:
\[
\ZZ_K/ \gens{ m } &\surjects \ZZ_K/I \\
[ \alpha]_m &\mapsto [ \alpha]_I
,\]
where we've used \( m\in I \) for this to be well-defined.
So $\# \ZZ_K /I \leq \ZZ_K / \gens{ m } = m^n < \infty$. 
:::

:::{.theorem title="The norm is multiplicative"}
For every pair $I, J\normal \ZZ_K$ nonzero, 
\[
N( IJ) + N(I) N(J)
.\]
:::

:::{.proof title="that the norm is multiplicative"}
Deferred!
:::

:::{.theorem title="Formula for norm of principal ideals"}
For all \( \alpha\in \ZZ_K \) nonzero, 
\[
N( \gens{ \alpha }) = \abs{ N ( \alpha ) }
,\]
i.e. the norm of a principal ideal is the absolute value of the norm of the element-wise ideal.
:::

:::{.remark}
This will follow from the following proposition:
:::

:::{.proposition title="Index = Determinant"}
Let $M \in \zmod$ be free of rank $n$ and let \( H \leq M \).
Then $H$ is free of rank at most $n$, so suppose $\rank_\ZZ H = n$.
Suppose that \( \omega_1, \cdots, \omega_n \) is a $\ZZ\dash$basis for $M$ and \( \theta_1, \cdots, \theta_n \) a $\ZZ\dash$basis for $H$.
We can thus write \( \tv{ \theta_1, \cdots, \theta_n} = \tv{ \omega_1, \cdots, \omega_n } A \) for some $A \in \Mat(n\times n, \ZZ)$.
Then \( [M: H] = \#M/H = \abs{ \det A } \).
:::

:::{.proof title="Sketch"}
Idea: convert this problem about an arbitrary \( M \in \zmod \) to a problem about $\ZZ^n$.
We know \( M \cong \ZZ^n \), and if we send the \( \omega_i \) to the standard basis vectors, this identifies $H \cong A \ZZ^n$.
So \( M/H \cong \ZZ^n/A\ZZ^n \), and it's easy to see that \( \det A \neq 0 \): if not, there would be a linear dependence among the \( \theta_j \).
Using *Smith normal form*, we can choose $S, T \in \GL_n(\ZZ)$ with 
\[
SAT = \diag(a_1, \cdots, a_n) && a_i \in \ZZ
.\]
Since $\det A \neq 0$, we have \( \det S, \det T \neq 0 \), and so all of the \( a_i \) are nonzero.
We can write \( \ZZ^n/A\ZZ^n \cong \ZZ^n/SAT\ZZ^n \cong \bigoplus_{i=1}^n \ZZ/a_i \ZZ \), which has size \( \prod \abs{a_i} = \abs{ \prod a_i } = \abs{ \det (SAT) } = \abs{ \det(A) } \) since $S, T$ are invertible and thus have determinant $\pm 1$.
:::

:::{.proof title="of formula for norm of principal ideals"}
Let \( \omega_1, \cdots, \omega_n \) be a \( \ZZ\dash \)basis for $\ZZ_K$, then \( \alpha \omega_1, \cdots \alpha \omega_n \) is a $\ZZ\dash$basis for \( \alpha\ZZ_K = \gens{ \alpha }  \).
Now to compute $\# \ZZ_K/ \gens{ \alpha }$, we use the "index equals determinant" result: write
\[
\tv{ \alpha \omega_1, \cdots, \alpha \omega_n} = \tv{ \omega_1, \omega_n} A \implies \# \ZZ_K / \gens{ a } = \abs{ \det(A) } 
,\]
we now just need to show that this is equal to \( \abs{ N \alpha } \).
We'll proceed by taking discriminants of tuples, applied to the first equation above.
This yields 
\[
\discriminant( \alpha \omega_1, \cdots, \alpha \omega_n) 
&= 
\discriminant( \omega_1, \cdots, \omega_n) 
\det(A)^2 \\
\implies \det(A)^2 
&= 
{
\discriminant( \alpha \omega_1, \cdots, \alpha \omega_n) 
\over
\discriminant( \omega_1, \cdots, \omega_n) 
} \\
&=
{
\det(D_{\alpha \omega_1, \cdots, \alpha \omega_n })^2
\over
\det( D_{\omega_1, \cdots, \omega_n} )^2
}
=
\qty{
\det(D_{\alpha \omega_1, \cdots, \alpha \omega_n })
\over
\det( D_{\omega_1, \cdots, \omega_n} )
}^2
.\]
Recall that these matrices were formed by taking the $j$th tuple element for the $j$th column and letting the column entries be the images under all embeddings.
Just looking at the first rows in each, we'll have
\[
\tv{ \sigma_1( \alpha \omega_1), \cdots, \sigma_1( \alpha \omega_n) } && 
\tv{ \sigma_1( \omega_1), \cdots, \sigma_1( \omega_n) } 
.\]
In general, the $i$th row of the first matrix will be \( \sigma_i( \alpha) \) times the $i$th row of the second matrix.
But then this ratio of determinants will be \( \qty{ \prod_{i=1}^n \sigma_i( \alpha )}^2 \da (N \alpha)^2\).
So $\det(A)^2 = (N \alpha)^2$, and taking square roots yields the result.
:::

## Ch. 14: Integral Bases

:::{.question}
Given $K$ a number field, can you find an explicit $\ZZ\dash$basis for $\ZZ_K$?
:::

:::{.remark}
This depends on how one is given $K$, and in general this is hard!
This is a question in algorithmic number theory.
We'll focus on a specific sub-problem.
:::

:::{.question}
Let $K$ be a number field with \( [K : \QQ] = n \) and suppose \( \theta_1, \cdots, \theta_n \) in $\ZZ_K$ are a $\QQ\dash$basis for $K$.
Is there a simple condition for when they form a $\ZZ\dash$basis for $\ZZ_K$?
:::

:::{.remark}
We know there is *some* $\ZZ\dash$basis for $\ZZ_K$, so let \( \elts{ \omega}{n} \) be one.
Then express the \( \theta \) in terms of the \( \omega \):
\[
\tv{ \elts{ \theta}{n} } &= \tv{ \elts{ \omega}{n} } A
\\
\implies 
\discriminant(\elts{ \theta}{n} ) 
&= 
\discriminant(\elts{ \omega}{n} ) \det(A)^2
.\]
We can view \( \abs{ \det(A) } \) as the index of the subgroup generated by the \( \theta_i \)  in the group generated by the \( \omega_i \), so
\[
\abs{ \det(A) } = [ \ZZ_K : H], && H\da \spanof_\ZZ\ts{ \theta_i }
.\]
Thus 
\[
\discriminant(\elts{ \theta}{n} ) 
&= 
\discriminant(\elts{ \omega}{n} )
[ \ZZ_K: H]^2
.\]
We can thus form a simple condition for when $H = \ZZ_K$:
:::

:::{.corollary title="A sufficient condition"}
If \( \discriminant( \elts{ \theta}{n} \) is squarefree, then $\elts{\theta}{n}$ are a \( \ZZ\dash \)basis of $\ZZ_K$.
:::

:::{.remark}
Why?
If the left-hand side is squarefree, then use that $[\ZZ_K: H]^2$ divides the left-hand side to conclude it must be 1.
Note that this is *not* necessary! 
We saw that for $d = 2,3 \mod 4$ that $\discriminant_K = 4d$, which is not squarefree.
:::

:::{.example title="of finding bases"}
Let $K = \QQ( \theta)$ where \( \theta \) is a root of 
\[
f(x) = x^5 - 3x^2 + 1
,\]
which is irreducible over $\QQ$.
This yields a degree 5 number field.
We can look for an $n\dash$tuple of elements in $\ZZ_K$ which is a $\QQ\dash$basis for $\ZZ_K$with a squarefree discriminant.
A candidate would be \( \ts{ \theta^j \st 0\leq j \leq 4 } \), which are all in $\ZZ_K$ since $\theta\in \ZZ_K$ which is closed under multiplication.

:::{.claim}
\[
\discriminant( 1, \theta, \theta^2, \theta^3, \theta^4) \text{ is squarefree}
.\]
:::

We have
\[
\discriminant( 1, \theta, \theta^2, \theta^3, \theta^4)
&\da\det( \tv{ \sigma_i ( \theta^{j-1} ) } )^2 \\
&= \det( \tv{ \sigma_i ( \theta )^{j-1} } )^2 && \text{ since the $\sigma_i$ are embeddings } \\
&= \prod_{1\leq i < j \leq 5} ( \sigma_j( \theta) - \sigma_i( \theta ) )^2 &&\text{since this is a Vandermonde matrix}\\
&= \discriminant(f)
,\]
where this is the *polynomial* discriminant.
This can be computed in a computer algebra system, and 
in this case it equals $-23119 = (-61)(379)$ which is squarefree.
So this yields a $\ZZ\dash$basis for $\ZZ_K$, i.e. $\ZZ_K = \ZZ[ \theta]$.
Note that \( \discriminant_K = -23119 \) as well, since it's the discriminant of *any* integral basis.
:::

:::{.example title="of finding bases"}
Let $K = \QQ( \alpha)$ where \( \alpha \) is a root of 
\[
f(x) = x^3 + x^2 - 3x + 8
.\]
We can try \( 1, \alpha, \alpha^2 \), and check
\[
\discriminant(1, \alpha, \alpha^2) = \discriminant(f) = (-4)(503)
,\]
so we can't conclude this is a $\ZZ\dash$basis.
Going back to the proof, we *can* conclude that \( [\ZZ_K: H] ^2 \divides \discriminant(1, \alpha, \alpha^2) \) where \( H \da \ZZ + \ZZ \alpha + \ZZ \alpha^2 = \ZZ[ \alpha ] \).
This allows us to conclude that \( [\ZZ_K: H] = 1, 2 \), so this could still be an index 2 subgroup.
If this happens, $\# \ZZ_K/H = 2$ and every element is annihilated by 2, so \( 2\ZZ_K \subseteq H = \ZZ[ \alpha ] \).
This would mean
\[
\ZZ_K \subseteq {1\over 2} \ZZ[ \alpha ] 
= 
\ts{ {c_0 + c_1 \alpha + c_2 \alpha^2 \over 2} \st c_i \in \ZZ } 
.\]
So are there elements of $\ZZ_K$ of this form that are *not* in \( \ZZ[ \alpha ] \)?
If there's nothing of this form in \( \ZZ_K \sm \ZZ[ \alpha ]\) then we can conclude \( \ZZ_K = \ZZ[ \alpha ] \).
If there *is* something of this form in \( \ZZ_K \sm \ZZ [ \alpha ] \), then $\ZZ_K \supseteq \ZZ [\alpha]$.
One can check that \( {\alpha+ \alpha^2 \over 2} \in \ZZ_K \sm H \).
So the original candidate basis was wrong, but we can take \( 1, \alpha, {\alpha + \alpha^2 \over 2} \) instead, which is an integral basis.
:::

:::{.remark}
Why is this last part true?
These are 3 elements of $\ZZ_K$ that are still $\QQ\dash$linearly independent and contains the $\ZZ\dash$span of the previous 3 elements defining $H$.
But the index of $H$ was 2, so this forces it to be everything.
So $\ZZ_K \neq \ZZ[ \alpha]$, and in fact Dedekind showed that \( \ZZ_K \neq \ZZ[ \beta] \) for *any* choice of \( \beta\in \ZZ_K \).
So cubic number fields exhibit new behavior when compared to quadratic number fields!
:::

:::{.remark}
Next time: integral bases for cyclotomic fields.
:::