# Lecture 1: Field Theory Preliminaries

The main theorems in this course, in order of importance:

- The Riemann-Roch Theorem
- The Riemann-Hurwitz Formula

## Finite Generation of Fields

See Chapter 11 of Field Theory notes.

### Notion 1

:::{.definition title="Finitely Generated Field Extension"}
A field extension $\ell/k$ is *finitely generated* if there exists a finite set $x_1, \cdots, x_n \in \ell$ such that $\ell = k(x_1, \cdots, x_n)$ and $\ell$ is the smallest field extension of $k$.

Concretely, every element of $\ell$ is a quotient of the form ${p(x_1, \cdots, x_n) \over q(x_1, \cdots, x_n)}$ with $p, q\in k[x_1, \cdots, x_n]$.
:::

There are three different notions of finite generation for fields, the above is the weakest.

### Notion 2

The second is being finitely generated as an algebra:

:::{.definition title="Finitely Generated Algebras"}
For $R\subset S$ finitely generated algebras, $S$ is finitely generated over $R$ if every element of $S$ is a polynomial in $x_1, \cdots, x_n$, with coefficients in $R$, i.e. $S = R[x_1, \cdots, x_n]$.
:::

Note that this implies the previous definition, since anything that is a polynomial is also a quotient of polynomials.

### Notion 3

The final notion: $\ell/k$ is finite (finite degree) if $\ell$ is finitely generated as a $k\dash$module, i.e. a finite-dimensional $k\dash$vector space.


:::{.definition title="Rational Function Field"}
A *rational function field* is $k(t_1, \cdots, t_n) \da ff \qty{ k[t_1, \cdots, t_n]}$.
:::

Note that we can make a similar definition for infinitely many generators by taking a direct limit (here: union), and in fact every element will only involve finitely many generators.

:::{.exercise}
\envlist
a. Show $k(t) / k$ is finitely generated by notion (3) but not by (2).

b. Show that $k[t]/k$ is (2) but not (1).[^not_a_field]


c. Show that it is not possible for a **field** extension to satisfy (2) but not (1).[^hint_zariskis_lemma_111]


d. Show that if $\ell/k$ is finitely generated by (3) and algebraic, then it satisfies (1).


[^hint_zariskis_lemma_111]: 
Hint: Zariski's lemma.

[^not_a_field]: 
Note $k[t]$ is not a field.

:::


:::{.theorem title="Field Theory Notes 11.19"}
If $L/K/F$ are field extensions, then $L/F$ is finitely generated $\iff$ $K/F$ and $L/K$ are finitely generated.[^see_artin_tate_lemma]

[^see_artin_tate_lemma]: 
See Artin-Tate Lemma, this doesn't necessarily hold for general rings.

:::


:::{.definition title="Algebraically Independent"}
For $\ell/k$, a subset $\ts{x_i}\subset \ell$ is *algebraically independent* over $k$ if no finite subset satisfies a nonzero polynomial with $k$ coefficients.

In this case, $k[\ts{x_i}] / k$ is *purely transcendental* as a rational function field.
:::

:::{.theorem title="Existence of transcendence bases"}
For $\ell/k$ a field extension,

a. There exists a subset $\ts{x_i}\subset \ell$ algebraically independent over $k$ such that $\ell/k(\ts{x_i})$ is algebraic. 

b. If $\ts{y_t}$ is another set of algebraically independent elements such that $\ell/k(\ts{y_t})$ is algebraic, then $\abs{\ts{x_i}} = \abs{\ts{y_t}}$.
:::

Thus every field extension is algebraic over a purely transcendental extension.
A subset as above is called a *transcendence basis*, and every 2 such bases have the same cardinality.

We have a notion of generation (similar to "spanning"), independence, and bases, so there are analogies to linear algebra (e.g. every vector space has a basis, any two have the same cardinality).[^note_on_matroids_common_gen]

[^note_on_matroids_common_gen]: 
There is a common generalization: matroids.

The following notion will be analagous to that of dimension in linear algebra:

:::{.definition title="Transcendence Degree"}
The *transcendence degree* of $\ell/k$ is the cardinality of any transcendence basis.
:::

:::{.theorem title="Transcendence Degree is Additive in Towers"}
If $L/K/F$ are fields then $\trdeg(L/F) = \trdeg(K/F) + \trdeg(L/K)$.
:::

:::{.theorem title="Bounds on Transcendence Degree"}
Let $K/k$ be finitely degenerated, so $K = k(x_1, \cdots, x_n)$.
Then $\trdeg(K/k) \leq n$, with equality iff $K/k$ is purely transcendental.
:::

:::{.proof}
Suppose $K$ is monogenic, i.e. generated by one element. Then $\trdeg(F(x)/F) = \indic{x/F\text{ is transcendental}}$.

So the degree increases when a transcendental element is added, and doesn't change when $x$ is algebraic.

By additivity in towers, we take $k \injects k(x_1) \injects k(x_1, x_2) \injects \cdots \injects k(x_1, \cdots, x)n) = K$ to obtain a chain of length $n$. 
The transcendence degree is thus the number of indices $i$ such that $x_i$ is transcendental over $k(x_1, \cdots, x_{i-1})$.[^checking_span]

[^checking_span]: 
This is similar to checking if a vector is in the span of a collection of previous vectors.

:::

:::{.definition title="Function fields in $d$ variables"}
For $d\in\ZZ^{\geq 0}$, an extension $K/k$ is *a function field in $d$ variables* (i.e. of dimension $d$) if $K/k$ is finitely generated of transcendence degree $d$.
:::

:::{.remark}
The study of such fields is birational geometry over the ground field $k$.
$k=\CC$ is of modern interest, things get more difficult in other fields.
The case of $d=1$ is much easier: the function field will itself be the geometric object and everything will built from that.
Our main tool will be **valuation theory**, where valuations will correspond to points on the curve.
:::


## Case Study: The Lüroth Problem.

:::{.question}
For which fields $k$ and $d\in \ZZ^{\geq 0}$ is it true that if $k \subset \ell \subset k(t_1, \cdots, t_d)$ with $k(t_1 ,\cdots, t_d)/\ell$ finite then $\ell$ is purely transcendental?
:::

:::{.answer}
It's complicated, and depends on $d$ and $k$.
We have the following partial results.
:::

:::{.theorem title="Lüroth"}
True for $d=1$: For any $k\subset \ell \subset k(t)$, $\ell = k(x)$.
:::

:::{.theorem title="Castelnuovo"}
Also true for $d=2, k=\CC$.
:::

:::{.theorem title="Zariski"}
No if $d= 2$, $k=\bar k$, and $k$ is positive characteristic.
Also no if $d=2, k\neq \bar k$ in characteristic zero.
:::

:::{.theorem title="Clemens-Griffiths"}
No if $d\geq 3$ and $k= \CC$.
:::

:::{.remark}
Note that unirational need not imply rational for varieties.
:::

:::{.exercise}
Let $k$ be a field, $G$ a finite group with $G\injects S_n$ the Cayley embedding.
Then $S_n$ acts by permutation of variables on $k(t_1, \cdots, t_n)$, thus so does $G$.
Set $\ell \da k(t_1, \cdots, t_n)^G$ the fixed field, then by Artin's observation in Galois theory: if you have a finite field acting effectively by automorphisms on a field then taking the fixed field yields a galois extension with automorphism group $G$.

So $\aut(k(t_1, \cdots, t_n)/ \ell) = G$.A

a. Suppose $k=\QQ$, and show that an affirmative answer to the Lüroth problem implies an affirmative answer to the inverse galois problem for $\QQ$.

  > Hint: works for any field for which Hilbert's Irreducibility Theorem holds.

b. $\ell /\QQ$ need not be a rational function field, explore the literature on this: first example due to Swan with $\abs{G} = 47$.

c. Can still give many positive examples using the Shepherd-Todd Theorem.
:::

\todo[inline]{What's a global field?}

## Integrals Closures and Constant Fields

:::{.definition title="Integral Closure and Field of Constants"}
For $K/k$ a field extension, set $\kappa(K)$ to be the algebraic closure of $k$ in $K$, i.e. special case of *integral closure*.
If $K/k$ is finitely generated, then $\kappa(K)/k$ is finite degree.

Here $\kappa(K)$ is called the *field of constants*, and $K$ is also a function field over $\kappa(K)$.
:::

:::{.remark}
In practice, we don't want $\kappa(K)$ to be a proper extension of $k$. 
If this isn't the case, we replace considering $K/k$ by $K/\kappa(K)$.
If $K/k$ is finitely generated, then

\begin{tikzcd}
k \arrow[rr, "\text{finite}", hook] &  & \kappa(K) \arrow[rr, "\text{finitely generated}", hook] &  & K
\end{tikzcd}

Where we use the fact that from above, $\kappa(K)/k$ is finitely generated and algebraic and thus finite, and by a previous theorem, if $K/k$ is transcendental then $K/\kappa(K)$ is as well, and thus finitely generated.
Thus if you have a function field over $k$, you can replace $k$ by $\kappa(K)$ and regard $K$ as a function field over $\kappa(K)$ instead.
:::