# Regular Functions (Thursday, September 17)

> See chapter 3 in the notes.

We'll next want to attach certain rings of functions to a space.


:::{.example title="Some examples of regular functions"}
\envlist

- $X$ a manifold or an open set in $\RR^n$ has a ring of $C^\infty$ functions.
- $X \subset \CC$ has a ring of holomorphic functions.
- $X\subset \RR$ has a ring of real analytic functions

These all share a common feature: it suffices to check if a function is a member on an arbitrary open set about a point, i.e. they are **local**.
:::

## Defining Regular Functions

:::{.definition title="Regular Functions"}
Let $X$ be an affine variety and $U\subseteq X$ open.
A **regular function** on $U$ is a function $\phi: U\to k$ such that $\phi$ is "locally a fraction", i.e. a ratio of polynomial functions.
More formally, for all $p\in U$ there exists a $U_p$ with $p\in U_p \subseteq U$ such that $\phi(x) = g(x)/ f(x)$ for all $x\in U_p$ with $f, g\in A(X)$.
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:::{.example}
For $X$ an affine variety and $f\in A(X)$, consider the open set $U\da V(f)^c$.
Then ${1\over f}$ is a regular function on $U$, so for $p\in U$ we can take $U_p$ to be all of $U$.
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:::{.example}
For $X = \AA^1$, take $f=x-1$.
Then ${x\over x-1}$ is a regular function on $\AA^1 \sm\ts{1}$.
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:::{.example}
Let $X + V(x_1 x_4 - x_2 x_3)$ and 
\[  
U \da X\sm V(x_2, x_4) = \ts{\thevector{x_1, x_2, x_3, x_4} \st x_1 x_4 = x_2 x_3, x_2\neq 0 \text{ or } x_4\neq 0 }
.\]

Define
\[  
\phi: U &\to K \\
\thevector{x_1, x_2, x_3, x_4} &\mapsto
\begin{cases}
{x_1\over x_2} & \text{if } x_2 \neq 0 \\
{x_3\over x_4} & \text{if } x_4 \neq 0
\end{cases}
.\]

This is well-defined on $\ts{x_2\neq 0} \intersect \ts{x_4 \neq 0}$, since ${x_1\over x_2} = {x_3 \over x_4}$.
Note that this doesn't define an element of $k$ at $\thevector{0,0,0,1}\in U$.
So this is not globally a fraction.
:::

Notation: we'll let $\OO_X(U)$ is the ring of regular function on $U$.

:::{.proposition title="?"}
Let $U\subset X$ be an affine variety and $\phi \in \OO_X(U)$.
Then $V(\phi) \da \ts{x\in U \st \phi(x) = 0}$ is closed in the subspace topology on $U$.
:::

:::{.proof}
For all $a\in U$ there exists $U_a\subset U$ such that $\phi = g_a/f_a$ on $U_a$ with $f_a, g_a \in A(X)$ with $f_a \neq 0$ on $U_a$.
Then
\[  
\ts{x\in U_a \st \phi(x) \neq 0} = U_a \sm V(g_a)\intersect U_a
\]
is an open subset of $U_a$, so taking the union over $a$ again yields an open set.
But this is precisely $V(\phi)^c$.
:::

:::{.proposition}
Let $U\subset V$ be open in $X$ an *irreducible* affine variety.
If $\phi_1, \phi_2 \in \OO_X(V)$ agree on $U$, then they are equal.
:::

:::{.proof}
$V(\phi_1 - \phi_2)$ contains $U$ and is closed in $V$.
It contains $\bar U\intersect V$, by an earlier lemma, $X$ irreducible implies that $\bar U = X$ and so $V(\phi_1 - \phi_2) =V$.
:::

:::{.question}
Let $U\subset V \subset \RR^n$ be open.
If $\phi_1, \phi_2 \in C^\infty(V)$ such that $\phi_1, \phi_2$ are equal when restricted $U\subset V$.
Does this imply $\phi_1 = \phi_2$?
:::

:::{.answer}
For $\RR^n$, no, there exist smooth bump functions. 
You can make a bump function on $V\setminus U$ and extend by zero to $U$.
For $\CC$ and holomorphic functions, the answer is yes, by the uniqueness of analytic continuation.
:::


## Distinguished Open Sets

:::{.definition title="(Important) Distinguished Open Sets"}
A **distinguished open set** in an affine variety is one of the form 
\[  
D(f) \da X\sm V(f) = \ts{x\in X \st f(x) \neq 0}
.\]
:::

:::{.proposition}
The distinguished open sets form a base of the Zariski topology.
:::

:::{.proof}
Given $f, g\in A(X)$, we can check:

1. Closed under finite intersections: $D(f) \intersect D(g) = D(fg)$.
2. \[U = X\sm V(f_1, \cdots, f_k) = V\sm \bigcap V(f_i) = \bigcup D(f_i),\]
  and any open set is a *finite* union of distinguished opens by the Hilbert basis theorem.
:::

:::{.proposition title="?"}
The regular functions on $D(f)$ are given by
\[  
\OO_X(D(f)) = \ts{{ g \over f^n} \st g\in A(X), n\in \NN} = A(X)_{\gens{f}}
,\]
the localization of $A(X)$ at $\gens{f}$.
:::

Note that if $f=1$, then $\OO_X(X) = A(X)$.

:::{.proposition title="?"}
Note that ${g\over f^n} \in \OO_X(D(f))$ since $f^n\neq 0$ on $D(f)$.
Let $\phi: D(f) \to k$ be a regular function.
By definition, for all $a\in  D(f)$ there exists a local representation as a fraction $\phi = g_a/f_a$ on $U_a\ni a$.
Note that $U_a$ can be covered by distinguished opens, one of which contains $a$.
Shrink $U_a$ if necessary to assume it is a distinguished open set $U_a = D(h_a)$.
\

Now replace 
\[  
\phi = {g_a \over f_a} = {g_a h_a \over f_a h_a}
,\]
which makes sense because $h_a\neq 0$ on $U_a$.
We can assume wlog that $h_a = f_a$.
Why?
We have $\phi = {g_a \over f_a}$ on $D(f_a)$.
Since $f_a$ doesn't vanish on $U_a$, we have $V(f_a h_a) = V(h_a)$ since $V(f_a) \subset D(h_a)^c = V(h_a)$.
Consider $U_a = D(f_a)$ and $U_b = D(f_b)$, on which $\phi = {g_a\over f_a}$ and $\phi = {g_b \over f_b}$ respectively.
On $U_a\intersect U_b = D(f_a f_b)$, these are equal, i.e. $f_b g_a = f_a g_b$ in the coordinate ring $A(X)$.
\

Then $D(f) = \bigcup_a D(f_a)$, so take the component $V(f) = \intersect V(f_a)$ by the Nullstellensatz
$f\in I(V(f_a)) = I(V(g_a, a\in D_f)) = \sqrt{f_a \st a\in D_f}$.
Then there exists an expression $f^n = \sum k_a f_a$ as a finite sum, so set $g - \sum g_a k_a$.

:::{.claim}
$\phi = g/f^n$ on $D(f)$.
:::
This follows because on $D(f_b)$, we have $\phi = {g_b \over f_b}$, and so $gf_b = \sum k_a g_a f_b$.

> Finish next class

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