1.1 Some Algebra

Notations.

• $k$ is an algebraically closed field
• $R$ is a commutative ring with identity

NOTE

On the other hand, we assume known the following topics in algebra:

1. The essentials of field theory (Galois theory, separability, transcendence degree).
2. Localization of a Ring, the behaviour of ideals in localization, the concept of a local ring.
3. Noetherian rings, and the decomposition theorem of ideals in these rings.
4. The concept of integral dependence, (cf., for example, Zariski-Samuel, vol. 1).

Definition

Define (affine) variety $V=\{z:z\in k^n,f_1(z)=\cdots =f_n(z)=0\}$ where $f_i\in k[x_1,\cdots ,x_n]$.

Define $R:=k[x_1,\cdots ,x_n]/(f_1,\cdots ,f_n)$. Then for any $\overline{g}\in R$, $z_1,z_2\in V$, $\overline{g}(z_1)=\overline{g}(z_2)$.

Noetherian's Normalization Lemma

Let $R$ be a finitely generated integral domain over $k$ with transcendence degree $\mathrm{tr}{\mathrm{deg}}_{k}R=n$. Then there exists $x_1,\cdots ,x_n\in R$ algebraically independent over $k$ such that $R$ is integral over $k[x_1,\cdots ,x_n]$, i.e. for any $y\in R$, there exists $g(t)\in k[x_1,\cdots ,x_n][t]$ such that $g(t)$ is monic and $g(y)=0$.

\begin{proof} Since $R$ is finitely generated, we can assume that $R=k[y_1,\cdots ,y_m]/I$.

If $m=n$, then the images of $y_1,\cdots ,y_m$ are algebraically independent and so $I=(0)$. Thus, $x_i:=y_i$ is what we desire.

If $m>n$, by induction on $m$, it suffices to show $R$ is integral over $S=k[z_1,\cdots ,z_{m-1}]$. If it holds, then by induction hypothesis there exists $x_1,\cdots ,x_n\in S$ such that $S$ is integral over $k[x_1,\cdots ,x_n]$. Since $R$ is integral over $S$ and $S$ is integral over $k[x_1,\cdots ,x_n]$, we have $R$ is integral over $k[x_1,\cdots ,x_n]$ by the transitivity of integral dependent.

Now we prove that $R$ is integral over $S$. Since $m>n$, the $m$ generators $y_1,\cdots ,y_m$ cannot be algebraically independent and so there exists $0\ne f\in k[y_1,\cdots ,y_m]$ such that $f(y_1,\cdots ,y_m)=0$. Let $r_1,\cdots ,r_m$ be positive integers, and let $z_2=y_2-y_1^{r_2}$, $z_3=y_3-y_1^{r_3}$, $\cdots$, $z_m=y_m-y_1^{r_m}\in R$. Then

$$0=f(y_1,z_2+y_1^{r_2},\cdots ,z_m+y_1^{r_m})=a{y}_1^N+(\text{mbox}termsofdegree<N)$$

and so there exists $g(t)\in k[z_2,\cdots ,z_m](t)$ such that $g(y_1)=0$. Hence $y_1$ is integral over $k[z_2,\cdots ,z_m]$. Note that $y_2=z_2+y_1^{r_2}\in k[z_2,\cdots ,z_m][y_1]$, which yields $y_2$ is integral over $k[z_2,\cdots ,z_m]$. Similarly, $y_3,\cdots ,y_m$ are integral over $k[z_2,\cdots ,z_m]$. Therefore, $R$ is integral over $k[z_2,\cdots ,z_m]$. \end{proof}

Geometric Meaning. Consider the curve $C$ in the affine plane defined by the equation $y^2=x^3$. Its coordinate ring is $R=k[x,y]/(y^2-x^3)$, and its function field is isomorphic to $k(t)$ via the parametrization $x=t^2$, $y=t^3$. This means the transcendence degree of $R$ over $k$ is $1$, since $t$ is algebraically independent over $k$ and generates the function field. By ^d13b03, $R$ is integral over $k[t]$, that is, everything in $R$ can be expressed in terms of $t$: $x=t^2$ and $y=t^3$. Geometrically, this corresponds to a normalization map from the affine line ${\mathbb{A}}^1$ (with coordinate ring $k[t]$) to the singular curve $C$, resolving the cusp at the origin $(0,0)$.

Check

The core idea is:

• Any finitely generated $k$-algebra (such as the coordinate ring of a complicated algebraic variety) can be mapped to an affine space ${\mathbb{A}}^n$ via a finite map.
• This “finite map” corresponds to a finite morphism in geometry, and to an integral extension / $R$ being a finitely generated module over $k[x_1,\dots ,x_n]$ in algebra.

Lemma

Let $K$ be a field, and let $S\subseteq K$ be a ring. If $K$ is integral over $S$, then $S$ is a field.

\begin{proof} It suffices to show for any $0\ne a\in S$, there is $1/a\in S$. Since $1/a\in K$ and $K$ is integral over $S$, there exists $b_i\in S$ such that

$$(1/a{)}^n+b_{n-1}(1/a{)}^{n-1}+\cdots +b_0=0$$

and it deduces that

$$1/a+b_{n-1}+b_{n-2}a+\cdots +b_0{a}^{n-1}=0$$

and $b_{n-1}+\cdots +b_0{a}^{n-1}\in S$. Thus, $1/a\in S$. \end{proof}

Going-Up

Let $R$ be a ring, and let $S\subseteq R$ be a subring. If $R$ is integral over $S$, then for any prime ideal $p\subseteq S$, there exists prime ideal $q\subseteq R$ such that $p=q\cap S$.

\begin{proof} Let $M=S-p$ be a multiplicative system. Let $S_M$ be the localization, i.e. $S_M=\{s/m:s\in S,m\in M\}$. Denote $R_M$ as the localization of $R$ w.r.t. $M$, then $R_M$ is integral over $S_M$. Let $p_M$ be the unique maximal ideal of $S_M$.

If there exists a prime ideal $q_M\subseteq R_M$ such that $q_M\cap S_M=p_M$, then we have done by the following commutative diagram.

$$\text{require}AMScd\begin{array}{ccc}R& \to & R_M\\ \uparrow & & \uparrow j\\ S& \to & S_M\end{array}$$

We claim that any maximal ideal $q_M\subseteq R_M$ satisfies that $q_M\cap S_M=p_M$. Since $q_M$ is maximal, $R_M/q_M$ is a field. Since $S/S\cap q_M$ can be seen as a subring of $R_M/q_M$, by ^815254, $S_M/S_M\cap q_M$ is also a field. It deduces that $S_M\cap q_M$ is a maximal ideal and so $S_M\cap q_M=p_M$. \end{proof}

Remark. “Going down” is natural, as $S/q\cap S$ is a subring of $R/q$ and is an integral domain. Hence $q\cap S$ is a prime ideal. This theorem tells us the map $q\to q\cap S$ is surjective.

Weak Nullstelle

$M$ is a maximal ideal of $k[x_1,\cdots ,x_n]$ iff $M=\{(x-a_1,\cdots ,x_n-a_n):a_i\in k\}$.

\begin{proof} Define

$$\phi :k[x_1,\cdots ,x_n]/(x_1-a_1,\cdots ,x_n-a_n)\to k,[f]\mapsto f(a_1,\cdots ,a_n),$$

and we can prove that $\phi$ is an isomorphism. Hence $k[x_1,\cdots ,x_n]/(x_1-a_1,\cdots ,x_n-a_n)$ is a field and so $(x_1-a_1,\cdots ,x_n-a_n)$ is a maximal ideal.

Conversely, suppose $M\subseteq k[x_1,\cdots ,x_n]$ is a maximal ideal. Then $R=k[x_1,\cdots ,x_n]/M$ is a field, which is finitely generated. By ^d13b03, we have $R\supseteq S$ and $S=k[y_1,\cdots ,y_{\ell }]$ with algebraically independent $y_i$ such that $R$ is integral over $S$. By ^815254, $S$ is a field and so $\ell =0$. Thus $R$ is integral over $k$ and so $R=k$ by $\overline{k}=k$. It deduces that $k[x_1,\cdots ,x_n]/M=k$ and $k+M=k[x_1,\cdots ,x_n]$. Assume that $x_i=m_i+a_i$, then $M\supseteq (x_1-a_1,\cdots ,x_n-a_n)$. Recall that we have proved $(x_1-a_1,\cdots ,x_n-a_n)$ is maximal, thus $M=(x_1-a_1,\cdots ,x_n-a_n)$. \end{proof}