First, while you are reading, please fill in proofs of theorems or lemmas below. Some of them are left as exercises. For the discussion in this afternoon about representatives of equivalent classes, it is necessary to mention a little about regular functions to understand things more clearly.
Recall about the Zarisky topology on $\overline{k}$, every single point is closed $\Leftrightarrow$ a closed set is finite. Note that if we endow $\overline{k}\times \overline{k}$ with the product topology then $\overline{k}$ is not a topological field. Recall that $F$ is said to be a topological field if the addition and multiplication are continuous.
Consider the product topology on $\overline{k}\times \overline{k}$, then a closed subset in this space is either finite, or vertical lines. Then the multiplication map $(+): \overline{k}\times\overline{k}\rightarrow \overline{k}$ that defines $(a,b)\mapsto a+b$. Then for $c\in\overline{k}, (+)^{-1}=\{(a,b)|a+b=c\}$. It is actually diagonal. Then $(+)$ is not a continuous map. Hence, $\overline{k}$ is not a topological field.
The Zarisky topology on $\overline{k}^n$ is a closed set on $\overline{k}^n$ is an algebraic set, that is $\cap Z_f(\overline{k})$. Then the Zarisky topology on $Z_f(\overline{k})$ is the induced topology. That is, the closed subset in $Z_f(\overline{k})$ is just the intersection of $f$ and other curves.
Exercise. Given $U,V$ are non-empty open sets in $\overline{k}^n$ then $U\cap V\ne 0$.
We now check that the multiplication $(.)$ and addition $(+)$ map are closed. For any $c\in\overline{k}$, we have $(+)^{-1}=\{(x,y)|x+y=c\}$. By our definition, it is closed in $\overline{k}\times \overline{k}$. The same method works for $(.)$. Hence, every polynomial function is continuous.
We begin with an important
Proposition 2.2. Let $f\in\overline{k}[x,y]$ be an irreducible polynomial. A non empty set $C\subset Z_f(\overline{k})$ is closed if and only if it is finite or $Z_f(\overline{k})$ itself.
Proof. $(\Leftarrow)$ easy to see.
$\Rightarrow$ Let $C$ be a non-empty proper subset of $Z_f(\overline{k})$. To prove $C$ is closed, it is sufficient to prove that $g^{-1}(c)$ is finite or equal $Z_f(\overline{k})$. Or it is equivalent to prove that $Z_f(\overline{k})\cap Z_g(\overline{k})$ is either finite or equal to $Z_f(\overline{k})$. If $f|g$ then it is obvious to see that the intersection is actually $Z_f(\overline{k})$. It is remaining to show
Theorem 2.3. Let $f\in\overline{k}[x,y]$ be an irreducible polynomial. Let $g\in \overline{k}[x,y]$ be a polynomial not divisible by $f$. Then $Z_f(\overline{k})\cap Z_g(\overline{k})$ is finite.
Proof. We consider some cases.
Case 1. If $f(x,y)=c(x-a)$, then $Z_f(\overline{k})$ = {(a,y)|y\in\overline{k}}. Thus $Z_f(\overline{k})\cap Z_f(\overline{k})=\{(a,y)|g(a,y)=0\}$. Consider the polynomial $g(a,y)\in\overline{k}[y]$. Then $C$ is finite iff $g(a,y\ne 0)$ in $\overline{k}[y]$.
Lemma 2.4. Let $f(x)\in R[x]$ is a polynomial, $R$ is UFD, then $f(a)=0$ iff $(x-a)|f(x)$.
Proof. Using Taylor's expansion.
Hence, applying the lemma, $g(a,y)=0\Leftrightarrow (x-a)|g(x,y)$. Hence, $f|g$, that leads the contradiction.
Case 2. If $f(x,y)\ne c(x-a)$. Consider the map $X:Z_f(\overline{k})\rightarrow \overline{k}$ that projects $(x,y)\mapsto x$. Applying Case 1, then $Z_f(\overline{k})\cap Z_X(\overline{k})$ is finite, or equivalently, $Z_f(\overline{k})\cap X^{-1}(0)$ is finite. By previous argument $Z_f(\overline{k})\cap X^{-1}(a)$ is finite. Hence, to prove $C$ is finite, it is sufficient to prove $X(C)$ is finite, because $C\subset X^{-1}(X(C))$. And we are now prove that $X(C)\subset Z_h(\overline{k}), h\in \overline{k}[x], h\ne 0$.
Definition 2.5. Let $A$ be any ring, and $f(y)=a_ny^n+...+a_0, g(y)=b_my^m+...+b_0$ be two polynomials in $A[y]$ with $a_nb_m\ne 0$. The resultant of $f$ and $g$ is the determinant
...
Lemma 2.6. Let $A$ be a factorial domain. Then $f$ and $g$ have common non-constant divisor if and only if $Res(f,g)=0$.
Claim. The polynomial $f,g$ have a common factor $A[y]$ if and only if there exists two non-constant polynomials $u,v\in A[y]$ such that
(i) $\deg u<\deg f,\deg v<\deg g$.
(ii) $vf-ug=0$.
Proof. Please fill in as an exercise.
Let $u(y)=c_{n-1}y^{n-1}+...+c_0$, $v(y) = d_{m-1}y^{m-1}+...+d_0$. Assume that $vf-ug = 0$, then the coefficient of $y^k$ is $\sum_{i+j=k}a_id_j-\sum_{i+j=k}b_ic_j=0$, for all $k$. We can write this as $Res(f,g) (d_j....-c_j...)^T=0$. Then the system of equation has non-trivial solution iff $Res(f,g)=0$, or it is equivalent to say that $f,g$ have a common factor.
Return to Theorem 2.3, let $f(x,y)=\sum_{i=0}a_i(x)y^i(a_n(x)\ne 0), g(x,y)=\sum_{j=0}b_j(x)y^j (b_m(x)\ne0)$. Consider $Res(f,g)$ as a polynomial in one variable $x$, which denotes $R(x)$. Because $f,g$ have no common factor, $Res(f,g)\ne 0$. Let $a\in \overline{k}$, if $a_n(a)b_n(a)\ne 0$, then $Res(f(a,y),g(a,y))=R(a)$.
Claim 2.7. $X(C)\subset \{a\in \overline{k}|a_n(a)b_m(a)R(a)=0\}$.
Let $a\in\overline{k}$ such that $a_n(a)b_m(a)\ne 0$, if $a\in X(C)$ then $f(a,y), g(a,y)$ have a common root in $\overline{k}$. By Lemma 2.6, $R(a)=0$.
Then $a\in \{a\in\overline{k}|a_n(a)b_m(a)R(a)=0\}$. Hence $X(C)$ is finite.
Corollary 2.8. Let $f,g$ be coprime polynomials in $\overline{k}[x,y]$. Then $Z_f(\overline{k})\cap Z_f(\overline{k})$ is finite.
Proof. Left as an exercise.
Exercise. Let $f$ and $g$ be coprime polynomials in $\overline{k}[x,y]$. Assume that $f(x,y)$ is monic in $y$.Then $a\in X(C)$ iff $Res_y(f,g)(a)=0$.
Remark 2.10. Given a closed subset $C$ on an affine curve $Z_f(\overline{k})$. Is it possible to find $g$ such that $C=Z_f(\overline{k})\cap Z_g(\overline{k})$?
By computing the degree of the resultant as polynomial, one can prove that $Z_f(\overline{k})\cap Z_g(\overline{k})\le \deg f.\deg g$ when $f, g$ are coprime.
Corollary 2.11. Let $f\in\overline{k}[x,y]$ be an irreducible polynomial. Let $g,h$ be two polynomials such that $g=h$ on $Z_f(\overline{k})$, then $f|(g-h)$.
Proof. Left as an exercise.
We can also define the coordinate ring (or ring of algebraic functions) of $f$ as $C_f=\overline{k}[x,y]/(f)$. When $f$ is irreducible, then the corollary above lets us know that $C_f$ is a domain.
Let $C(Z_f(\overline{k}),\overline{k})$ denote the set of continuous maps from $Z_f(\overline{k})$ to $\overline{k}$. Corollary shows that there exists an embedding $i_f: C_f \rightarrow C(Z_f(\overline{k}),\overline{k})$.
Definition 2.12. Let $\overline{k}(Z_f)$ the field of fractions of $C_f$, we call this field the field of rational functions of the affine curve defined by $f$. The elements of $\overline{k}(Z_f)$ are called the rational functions of $Z_f(\overline{k})$.
Lemma 2.13. Let $\alpha\in\overline{k}(Z_f)^*$. Then there exists finitely many points $P_1,...,P_s$ in $Z_f(\overline{k})$ such that $\alpha$ defined a continuous map $\alpha: Z_f(\overline{k})-\{p_1,...,p_s\}\rightarrow \overline{k}$.
Remark. The proof the Lemma is not difficult, and this is worth noting that the set $\{p_1,...,p_s\}$ is not fixed, because the representative for $\alpha$ of the form $\frac{f}{g}$ may not be unique.
Proof. Left as an exercise.
We note that $Z_f(\overline{k})-\{p_1,...,p_s\}$ is an open subset of $Z_f(\overline{k})$. Let $\frac{f}{g}$ and $\frac{f'}{g'}$, and $V(g)=Z_g(\overline{k})\cap Z_f(\overline{k})$, $V(g')=Z_{g'}(\overline{k})\cap Z_f(\overline{k})$, then it is obvious to see that they are identical on the open set $Z_f(\overline{k})-(V(g)\cup V(g'))$. Remember that any non-empty subset of $Z_f(\overline{k})$ is dense in it, because $Z_f(\overline{k})$ is irreducible. That means, our two representatives agree on almost $Z_f(\overline{k})$.
It is tempting to extend this observation. Let $V$ be an irreducible algebraic set (or variety in our context), as notations above, we define $C_V$ the coordinate ring of $V$. Because $V$ is irreducible, $C_V$ is a domain. We should be familiar with the notion of regular functions.
Definition 2.14. Let $V$ be an affine variety, $a$ is a point on $V$. We call $\varphi$ a regular function at $a$ if $\varphi$ is defined on an open set $U_a$ containing $a$, and $\varphi(x)=\frac{g(x)}{f(x)}$ for all $x\in U_a$, where $f$ and $g$ are polynomials in $C_V$-the coordinate ring of $X$.
Definition 2.15. Let $V$ be an affine variety, and $U$ be an open subset of $V$, the map $\varphi:U\rightarrow k$ is called regular function if $\varphi$ is regular at any point in $U$.
We define the relation $\sim$ on the set of all regular functions on $V$. Let $\varphi_1,\varphi_2$ be two regular functions on $V$, then $\varphi_1\sim\varphi_2$ if there exists a non-empty open set $U$ of $V$, such that $\varphi_{1_{|U}}=\varphi_{2_{|U}}$. Then it is easy to check that $\sim$ is an equivalent relation. By our definition, any regular function is equivalent to a function of the form $\frac{f}{g}$, where $f,g\in C_V$. And we also have
Proposition 2.16. $\frac{f}{g}\sim\frac{f'}{g'}$ iff $fg'=gf'$ on $V$.
Proof. If $\frac{f}{g}\sim\frac{f'}{g'}$ then there exists an non-empty open subset $U$ of $V$ such that $\frac{f}{g}=\frac{f'}{g'}$. Then $fg'-gf'=0$ on $U$. Because $U$ is dense in $V$, and the set of zeros of $fg'-gf'$ is closed in $V$ that contains $U$. Hence, it contains $\overline{U}=V$. That means, $fg'=gf'$ on $V$.
Conversely, if $fg'=gf'$ on $V$, then $\frac{f}{g}=\frac{f'}{g'}$ on $V-(V(g)\cup V(g'))=V-V(gg')$. Because $C_V$ is an integral domain, $gg'\ne 0$. And hence, $V-V(gg')$ is a non-empty open subset of $V$. By our definition $\frac{f}{g}\sim\frac{f'}{g'}$ (Q.E.D).
We denote $R(V)$ the set of all equivalent classes of regular functions. Then it is easy to see that $R(V)$ have the field structure, defined by usual addition and multiplication on $C_V$. We consider the map from $\overline{k}(V)$-the fraction field of $C_V$ to $R(V)$ by sending $\frac{f}{g}$ to $[\frac{f}{g}]_\sim$. This map is actually a field isomorphism. And we have
Proposition 2.7. The set of regular functions on an affine variety $V$ forms a field, which is isomorphic to $\overline{k}(V)$.
Proof. Left as an exercise.
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