# Lecture 3: Last of Preliminaries

Today we'll be wrapping up the last of the preliminaries.
Upcoming: one-variable function fields and their valuation rings.

## Polynomials Defining Regular Function Fields

:::{.question}
Where's the curve in all of this?
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:::{.answer}
This will come from an equation like $f(x, y) = 0$.
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:::{.exercise}
Let $R_1, R_2$ be $k\dash$algebras that are also domains with fraction fields $K_i$.
Show $R_1 \tensor_k R_2$ is a domain $\iff$ $K_1 \tensor_k K_2$ is a domain.[^hint_for_exc1]

[^hint_for_exc1]: Hint: use a denominator clearing argument.

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## Geometric Irreducibility

:::{.definition title="Geometrically Irreducible Polynomial"}
A polynomial of positive degree $f\in k[t_1, \cdots, t_n]$ is **geometrically irreducible** if $f\in \bar k[t_1, \cdots, t_n]$ is irreducible as a polynomial.
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:::{.remark}
If $n=1$ then $f$ is geometrically irreducible $\iff$ $f$ is linear, i.e. of degree 1.
Let $f$ be irreducible, then since polynomial rings are UFDs then $\gens{f}$ is a prime ideal (irreducibles generate principal ideals) and $k[t_1, \cdots, t_n]/\gens{f}$ is a domain.
Let $K_f$ be the fraction field.
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:::{.exercise title="an easy one"}
\envlist
a. Above for $1\leq i \leq n$ let $x_i$ be the image of $t_i$ in $K_f$.
  Show that $K_f = k(x_1, \cdots, x_n)$.

b. Show that if $K/k$ is generated by $x_1, \cdots, x_n$, then it is the fraction field of $k[t_1, \cdots, t_n] /\mfp$ for some prime ideal $\mfp$ (equivalently, a height 1 ideal).
:::

:::{.proposition title="?"}
Suppose that $f$ is geometrically irreducible.

a. The function field $K/k$ is regular.

b. For all $\ell/k$, $f\in \ell[t_1, \cdots, t_n]$ is irreducible.

:::

:::{.definition title="Absolutely Irreducible Polynomial"}
In this case we say $f$ is **absolutely irreducible** as a synonym for geometrically irreducible.
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:::{.proof}
By definition of geometric irreducibility, $\bar k[t_1, \cdots, t_n]/\gens{f} = k[t_1, \cdots, t_n]/\gens{f} \tensor_k \bar k$ is a domain.
The exercise shows that $K_f \tensor_k k$ is a domain, so $K_f$ is regular.
It follows that for all $\ell/k$, $K_f \tensor_k \ell$ is a domain, so $\ell[t_1, \cdots, t_n]/\gens{f}$ is a domain.
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:::{.slogan}
Geometrically irreducible polynomials are good sources of regular function fields.
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:::{.exercise}
Let $k$ be a field, $d\in \ZZ^+$ such that $4\notdivides d$ and $p(x) \in k[x]$ be positive degree.
Factor $p(x) = \prod_{i=1}^r (x-a_i)^{\ell_i}$ in $\bar k[x]$.

a. Suppose that for some $i$, $d\notdivides \ell_i$. 
  Show that $f(x, y) \da y^d - p(x) \in k[x, y]$ is geometrically irreducible.
  Conclude that $K_f \da ff\qty{k[x, y] / \gens{y^d - p(x)}}$ is a regular one-variable function field over $k$, and thus elliptic curves yield regular function fields.[^hint_and_comment_on_exercise]

b. What happens when $4\divides d$?

[^hint_and_comment_on_exercise]: 
Referred to as *hyperelliptic* or *superelliptic* function fields.
Hint: use FT 9.21 or Lang's Algebra.

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:::{.exercise title="Nice, Recommended"}
Assume $k$ is a field, if necessary assuming $\ch(k) \neq 2$.

a. Let $f(x, y) = x^2 - y^2 -1$ and show $K_f$ is is rational: $K_f = k(z)$.

b. Let $f(x, y) = x^2 + y^2 - 1$.
  Show that $K_f$ is again rational.

c. Let $k = \CC$ and $f(x, y) = x^2 + y^2 + 1$, $K_f$ is rational.

d. Let $k= \RR$.
  For $f(x ,y) = x^2 + y^2 + 1$, is $K_f$ rational?[^example_genus_zero]

[^example_genus_zero]: 
This is an example of a non-rational genus zero function field.

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:::{.question}
Can we always construct regular function fields using geometrically irreducible polynomials?
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:::{.answer}
In several variables, no, since not every variety is birational to a hypersurface.
In one variable, yes, as the following theorem shows:
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## Our Function Fields are Geometrically Irreducible

:::{.theorem title="Regular Function Fields in One Variable are Geometrically Irreducible"}
Let $K/k$ be a one variable function field (finitely generated, transcendence degree one). 
Then

a. If $K/k$ is separable, then $K = k(x, y)$ for some $x, y\in K$.

b. If $K/k$ is regular (separable + constant subfield is $k$, so stronger) then $K \cong K_f$ for a geometrically irreducible $f\in k[x ,y]$.
:::

Recall separable implies there exists a separating transcendence basis.

:::{.proof title="of a"}
This means there exists a primitive element $x\in K$ such that $K/k(x)$ is finite and separable.
By the Primitive Element Corollary (FT 7.2), there exist a $y\in K$ such that $K = k(x, y)$.
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:::{.proof title="of b"}
Omitted for now, slightly technical.
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:::{.remark}
Importance of last result: a regular function field on one variable corresponds to a nice geometrically irreducible polynomial $f$.
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:::{.remark}
Note that the plane curve module may not be smooth, and in fact usually is not possible.
I.e. $k[x ,y]/\gens{f}$ is a one-dimensional noetherian domain, which need not be integrally closed.
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:::{.question}
Can every one variable function field be 2-generated?
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:::{.answer}
Yes, as long as the ground field is perfect.
In positive characteristic, the suspicion is no: there exists finite inseparable extensions $\ell/k$ that need arbitrarily many generators.
However, what if $K/k$ has constant field $k$ but is not separable?
Riemann-Roch may have something to say about this.
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:::{.example}
\hyperref[technical_example]{Example from earlier lecture:}
\[
ax^p + b - y^b
\]
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:::{.remark}
We can find examples of nice function fields by taking irreducible polynomials in two variables.
This will define a one-variable function field.
If the polynomial is geometrical reducible, this produces regular function fields.
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