Note on dimension of affine rings

The main purpose of this post is to prove the dimension theorem for affine rings, which is "Let $A$ be an affine ring over a field $k$, then $\dim(A) = tr.deg_kA$". In the first section, we will point out the invariance of transcendental degree via integral extensions. And from this, we can see any affine ring has the same transcendental degree with its Noether's normalization. The normalization theorem is proved in our second sections, together with its applications in the proof of Hilbert's nullstellensatz. The third section is devoted to prove the going up theorem and its interesting application in constituting the one-to-one correspondence between chains of prime ideals in an integral domain and their corresponding chains in an integral extension domain. Finally, all results will be glued together in the proof of our main theorem in Section 4.

There are some questions left, and I have not seen solution in familiar textbooks. They can be considered open questions or actual challenges for us.

(1) Given an affine ring, compute its Noether's normalization.
(2) Given an affine domain $B$ and its integral extension domain $A$. Let $P$ be a prime ideal of $B$, is there any method to find $Q\in Spec(A)$, that lies over $P$ (i.e. $Q\cap B=P$)?
(3) (Weaker than (1)) Given a variety, compute its dimension.

1. The invariance of transcendental degree via integral extensions.

Let $k[X] = k[x_1,...,x_n]$ be the polynomial ring of $n$ variables over a field $k$, $I$ be the ideal of $k[X]$, $\min(I)$ be the set of minimal prime ideals that contain $I$. A familiar result is that if $P$  is prime ideal that contains $I$, then $P$ contains $\sqrt{I}$.

By the Hilbert's nullstenlensatz, there exists finite minimal prime ideals that contain $\sqrt{I}$. We denote the set $\min(I)$, then $\sqrt{I}=\cap_{P\in \min(I)}P$. Let $A = k[X]/I$ be an affine ring. There is a 1-1 correspondence between the set $\min(A)$ that contains all minimal prime ideals of $A$ and the set $\min(I)$ by the correspondence theorem of rings. We denote $tr.deg_k A$ the transcendental degree of $A$ over $k$. Let us prove first

Theorem 1.1. For any affine ring $A$, we have

(i) $tr.deg_kA/Q\le tr.deg_kA$

(ii) $tr.deg_kA=\max\{tr.deg_kA/P|P\in \min(A)\}$

Proof. (i) is obvious.

(ii) Let $\{y_1,..,y_d\}$ be the any subset of algebraic independence of $A$ over $k$. Then $k[y_1,...,y_d]$ is the domain. It can be seen that $\sqrt{0}\cap k[y_1,...,y_d]=0$. Furthermore,

$\sqrt{0}=\cap_{P\in \min(A)P}$. Hence, $$\prod_{P\in \min(A)} (P\cap k[y_1,...,y_d]) \subset \cap_{P\in \min(A)}(P\cap k[y_1,...,y_d])=0$$. 

Hence, there exists $P\in \min(A)$, s.t. $P\cap k[y_1,...,y_d]=0$. Otherwise, for all, $P_i\in \min(A)$, if $P_i\cap k[y_1,...,y_d]=a_i\ne 0$, we have $\prod_{i}a_i\ne 0$, and the product lies in $k[y_1,...,y_d]$. This leads the contradiction. 

From this, we can embed $k[y_1,...,y_d]$ into $A/P$, via the natural homomorphism. And we have proved there exists $P\in \min(A)$ s.t. $tr. deg_kA\le tr.deg_kA/P$. (Q.E.D)

Using this result, we will give the proof for the following

Theorem 1.2. Let $A,B$ be affine rings, and $B\subset A$, and $A$ is integral over $B$. Then $tr. deg_kA=tr.deg_kB$.

Proof. Let $d = tr.deg_kB$, and the maximal set of algebraic independence elements in $B$ over $k$ is $\{y_1,...,y_d\}$. First, if $A$ is an integral domain. We denote $K$ the fraction field of $A$, and $L$ the fraction field of $B$, then $L$ is algebraic over $k(y_1,...,y_d)$, and $K$ is algebraic over $L$. Hence, $tr.deg_kA=tr.deg_kK=tr.deg_kL=tr.deg_kB=d$.

If $A$ is not the integral domain, we can see first that $d = tr.deg_kB\le tr.deg_kA$ because $B\subset A$. Besides, by the previous theorem, there exists $P\in \min(A)$ such that $tr.deg_kA=tr.deg_kA/P$, Let $p=P\cap B$, then $A/P$ is integral over $B/p$. $A/P$ is now integral domain, and applying the first case, we obtain $tr.deg_kA/P=tr.deg_kB/p\le d$. Hence, $tr.deg_kA\le d$. This inequality finishes our proof. (Q.E.D).

We now measure the "complexity" of a ring via their dimension. Let $A$ be a notherian ring, let us define the dimension of $A$ by the supremum of the finite chain of prime ideals in $A$, $P_0\subset P_1\subset ... \subset P_n$. And we denote $\dim A = n$.

If $A$ is any affine ring, then by the theorem below, there exists an integral domain $B\subset A$, such that $B$ is isomorphic to the multi-variable polynomial rings over $k$, and $A$ is integral over $B$. We call $B$ the Noether's normalization of $A$. By the previous theorem, $tr.deg_kA=tr.deg_kB$. We next prove the dimension of $B$ is actually its transcendental degree, and hence, the dimension of $A$ is actually its transcendental degree.

2. Noether's normalization theorem.

Before proving the main theorem, we need a following

Lemma 2.1. Let $f\in k[x_1,...,x_n]$, the polynomial ring over $k$. Let $N$ be an integer such that $N$ is strictly bigger than $a_i(i=1,...,n)$, where $c_ax_1^{a_1}...x_n^{a_n}$ is any monomial term of $f(X)$. Then by the change of variables $x_i \mapsto x_i + x_n^{N^i}$ for all $i = 1,...,n-1$, and $x_n\mapsto x_n$, $f(x)$ can be expressed as $\lambda x_n^M + g_1x_n^{M-1}+...+g_M$, where $\lambda \in k^\times, g_i\in k[x_1,...,x_{n-1}]$.

Proof. Let $f = f_r + ... + f_0$, where $f_i$ is the homogeneous component of $f$ of order $i$. Then for any monomial term $c_ax_1^{a_1}...x_n^{a_n}$ in $f_r$, via the transform, we obtain the degree of $x_n$ under the transformation is $c_ax_n^{a_1N}x_n^{a_2N^2}...x_n^{a_n} = c_ax_n^{a_n+a_1N+a_2N^2+...+a_{n-1}N^{n-1}}$. We note that the representation of the exponent in base $N$ is unique, because all $a_i<N$.

For any $c_bx_1^{b_1}...x_n^{b_n}$ be any other monomial term of $f_r$, via the transformation, the highest exponent of $x_n$ will be $b_n+b_1N+b_2N^2+...+b_{n-1}N^{n-1}$, which is different from the previous exponent, because $(b_1,...,b_n)\ne(a_1,...,a_n)$.

Hence, via the transformation, we can see $f(x) \mapsto \lambda x_n^M + \{\text{lower degree of } x_n\}$, where $\lambda\in k^\times$. (Q.E.D)

Using this lemma we will give the proof for the Nother's normalization theorem.

Theorem 2.2. Let $A$ be an affine ring, then there exists an integral domain $B\subset A$, such that $B$ is isomorphic to the ring of multi-variable polynomials over $A$, and $A$ is integral over $B$.

Proof. Let $A=k[z_0,...,z_n]=k[x_0,...,x_n]/I$, we will prove by induction by the number of generators in $A$ over $k$. If $r=0$, then there is nothing to prove. For $n>0$, we distinguish two cases

(i) If there is no algebraic relation between $z_1,...,z_n$ (in this case $I=0$), then $k[z_0,...,z_n]\cong k[x_0,...,x_n]$

(ii) Assume there exists $f\ne0$ in $I$, then by the previous lemma, there exists via a transformation $x_1,...,x_{n-1}$ to $y_1,...,y_{n-1}$ such that $\lambda f(y_1,...,y_{n-1},x_n)$ is zero in $A$, and the leading term of $f$ is $x_n^M$. That means $x_n$ is integral over $k[y_1,...,y_{n-1}]$. By induction hypothesis, there exists the integral domain $B \subset k[y_1,...,y_{n-1}]$ such that $B$ is isomorphic to the polynomial ring over $k$, and $k[y_1,...,y_{n-1}]$ is integral over $B$. And $A$ is also integral over $B$. This concludes our proof. (Q.E.D)

Before proving the Hilbert's nullstellensatz, we need the following result of Oscar Zarisky and an important

Lemma 2.3. Let $B\subset A$ be two integral domains, and $A$ is integral over $B$. Then $A$ is a field iff $B$ is a field.

Proof. If $A$ is a field, then for any $b\ne0$ in B, we have $\frac{1}{b}\in A$, and it is integral over $B$. That means, there exists the minimum polynomial $f(x) = b_0+b_1x +...+x^n$ of $\frac{1}{b}$, where $b_i\in B$. From this, by clearing the denominator, we have $b_0b^r+...+b_{n-1}b + 1=0$, which implies $b(b_0b^{r-1}+...+b_{n-1})=-1$. Hence, $\frac{1}{b}=-b_0b^{r-1}-...-a_{n-1}$. And $B$ is a field.

Conversely, if $B$ is a field, then for any $a\ne 0$ in A, the minimum polynomial of $A$ is $x^n+b_{n-1}x^{n-1}+...+b_1x+b_0$, where $b_i\in B$. We then have $b_0\ne 0$, otherwise, $f(x) = x(b_1+b_2x +...+b_nx^{n-1})$, and $a\ne0$, this implies $b_1+b_2x +...+b_nx^{n-1}$ has root $a$. And this leads the contradiction. Hence, $a(b_1+b_2a+...+b_na^{n-1})=-b_0$. Because $B$ is field, and $b_0\ne0$, $\frac{b_i}{b_0}\in B$, and we have $a(-\frac{b_1}{b_0}-\frac{b_2}{b_0}a-...-\frac{b_n}{b_0}a^{n-1})=1$. And the inverse of $a$ is $-\frac{b_1}{b_0}-\frac{b_2}{b_0}a-...-\frac{b_n}{b_0}a^{n-1}$. (Q.E.D)

Together with the theorem of Noether, the previous lemma allow us to prove the result of Zarisky.

Theorem 2.4. Let $k$ be a field, and $A$ is an finitely generated algebra over $k$, and $A$ is a field. Then $A$ is finite field extension of $k$. When $k$ is algebraically closed, $A\equiv k$.

Proof. We can consider $A$ an affine ring over $k$. By Theorem 2.2, there exists the normalization of $A$, which is an integral domain $B$, s.t. $A$ is integral over $B$. By Lemma 2.3, $B$ is a field. Furthermore, $B$ is isomorphic to the polynomial ring $k[x_1,...,x_n]$. This implies $B\cong k$, otherwise, $\frac{1}{x_i}\in k$, and $x_i\frac{1}{x_i}=1$, which means $x_i$ is algebraic over $k$.  Hence $A=k[z_1,...,z_n]$, and each $z_i$ is algebraic over $k$. That means, $[A:k]<\infty$. 

When $k$ is algebraically closed, then it is obvious to see that any algebraic extension of $k$ is itself. In this case $A\equiv k$. (Q.E.D)

We now prove the Hilbert's nullstellensatz.

Theorem 2.5. Let $k$ be an algebraically closed field, then any maximal ideal of $k[x_1,...,x_n]$ is of the form $(x-a_1,...,x-a_n)$.

Proof. Let $\mathfrak{m}$ be any maximal ideal of $k[x_1,...,x_n]$. Then $A=k[x_1,...,x_n]/\mathfrak{m}$ is a field, and it is also a finitely generated algebra over $k$. By Theorem 2.4, $A\cong k$. Let $\phi: k\rightarrow A$ is the field isomorphism, and the inverse image of $bar{x_i}$ is $a_i$. Then $x_i-a_i\in \mathfrak{m}$. Hence, the ideal $(x_1-a_1,...,x_n-a_n) \subset \mathfrak{m}$. Besides, it is obvious to see that $(x_1-a_1,...,x_n-a_n)$ is also a maximal ideal of $k[x_1,...,x_n]$, which implies $(x_1-a_1,...,x_n-a_n)\equiv\mathfrak{m}$ . (Q.E.D)

3. Going up theorem and the invariance of dimension via integral extension. 

To constitute the correspondence between chains of prime ideals, as our the introduction have mentioned, we first recall some basic facts about localizations. In this section, $A$ and $B$ are integral domain, and $A$ is integral extension of $B$. Let $P$ be prime ideal of $B$, we denote $S=B-P$, $\phi$ the canonical homomorphism from $B$  to $B_S$.

Lemma 3.1. The localization of $A$ at $P$ is defined and $A_S$ is integral over $B_S$.

Proof. Left as an exercise.

From this, we can see $\phi$ induces a canonical map $\psi$ from $A$ to $A_S$ that makes the following diagram commutes

$$
\begin{matrix}
B &\subset &A \\
\downarrow{\phi} & & \downarrow{\psi}  \\
B_S &\subset &A_S
\end{matrix}
$$

Lemma 3.2. Let $Q_S$ be a  prime ideal of $A_S$, and $P_S=Q_S\cap B_S$, then $P_S$ is also prime ideal of $B_S$. And $Q_S$ is maximal iff $P_S$ is maximal.

Proof. Left as an exercise.

Theorem 3.3 (Going up).  Let $P$ is a prime ideal of $B$, then there exists a prime ideal $Q$ of $A$ such that $Q\cap B = P$.

Proof. Let $S=B-P$. Looking at the commutative diagram

$$
\begin{matrix}
B &\subset &A &&&&& b &\rightarrow &b\\
\downarrow{\phi} & & \downarrow{\psi} &&&&& \downarrow & &\downarrow \\
B_S &\subset &A_S &&&&& \frac{b}{1} &\rightarrow &\frac{b}{1}
\end{matrix}
$$

We can see that for all $b\in B$, $\phi(b)\in Q_S\cap B_S$ iff $\psi(b)\in Q_S$. Hence, $\phi^{-1}(Q_S\cap B_S)=\psi^{-1}(Q_S)\cap B$. Because $Q_S$ the maximal ideal of $A_S$, from the previous lemma, $P_S=Q_S\cap B_S$ is the only maximal in $B_S$, and its inverse image $\phi^{-1}(P_S)=P$. Furthermore, there exists $Q\in Spec(A)$ such that $\psi^{-1}(Q_S) = Q$, and $Q\cap S = \emptyset$. Hence, $P=Q\cap B$. (Q.E.D)

Using this theorem, we can easily prove the first part of our target.

Theorem 3.4. Let $P_0\subset P_1\subset...\subset P_n$ be a chain of prime ideals in $B$, then there exists the correspondence chain of prime ideal $Q_0\subset Q_1\subset...\subset Q_n$ in $A$, such that $Q_i\cap B= P_i$. 

As a corollary, we have $\dim(B)\le \dim(A)$, if both are notherian. And the remaining part is pointed out via the following

Lemma 3.5. Let $Q$ be a non-zero prime ideal of $A$, then $P = Q\cap B$ is a non-zero prime ideal of $B$.

Proof. Let $b_1,b_2\in B$, and $b_1b_2\in Q\cap B$, then $b_1b_2\in Q$, which implies $b_1\in Q$ or $b_2\in Q$. And hence, $b_1\in Q\cap B$, or $b_2\in Q\cap B$. Hence, $Q\cap B$ is also prime ideal. Let $0\ne q\in Q$, and $f(x)=a_0+a_1x+...+x^n$ is minimum polynomial of $q$, where $a_i\in B$. Then by previous argument, $a_0\ne 0$. And $a_0=-q^n-...-a_1q\in Q\cap B$. (Q.E.D)

The previous lemma allows us to prove

Theorem 3.6. Let $Q \subset Q' \in Spec(A)$, and $P = Q\cap B, P'=Q'\cap B$, then $P\subset P' \in Spec(B)$, and $Q\subsetneq Q'$ implies that $P\subsetneq P'$.

Proof. Consider the quotient ring $B/P$, we can embed this into $A/Q$, whose $Q'/Q$ is non-zero prime ideal. And $A/Q$ is also an integral extension of $B/P$. Applying the previous Lemma, we have $Q'/Q\cap B/P = P'/P$ is also non-zero ideal of $B/P$. And hence, $P \subsetneq P'$. (Q.E.D).

From Theorem 3.6, we can see $\dim(A)\le\dim(B)$ if both are notherian. And the corollary of Theorem 3.4 lets us know, in this case, $\dim(A)=\dim(B)$. Via three sections, we can now know that the dimension of an affine domain is equal to the dimension of its Nother's normalization, which is isomorphic to a polynomial rings over fields. Therefore, to compute the dimension of an affine domain, it is remaining to compute the dimension for the polynomial rings $k[x_1,...,x_n]$.

4. The last step.

We will first compute the dimension of an affine domain.

Theorem 4.1. Let $A$ be an affine domain, then $\dim(A)=tr.deg_k(A)$.

Proof.  By our previous argument, $\dim(A) =\dim(B)$, where $B$ is the normalization of $A$. It is stated that $B\cong k[x_1,...,x_n]$, and it is sufficient to compute the dimension of $R=k[x_1,...,x_n]$. First, we can see that $0\subset (x_1)\subset (x_1,x_2)\subset...\subset(x_1,...,x_n)$ is a chain of prime ideal of $R$. Hence, $\dim(R)\ge n$. We will prove by induction, in fact, that $\dim(k[x_1,...,x_n])$=n. And from this, we can deduce $\dim(A)=tr.deg_k(A)$.

For $n=1$, $k[x]$ is P.I.D, which has dimension 1.

For $n>1$, we now consider the maximal chain $0 = P_0\subset P_1\subset ...\subset P_m$ of prime ideal of $k[x_1,...,x_n]$. Let $R_1=R/P_1$, and the images of $P_i$ via the canonical map $Q_i$ then it is obvious to see that $\dim(R)-1 = \dim(R_1)$, which is a domain. Let $B_1$ be the normalization of $R_1$, then $\dim(B_1)=\dim(B)-1=m-1$. Hence $B_1\subsetneq B$, and assume that $B_1\cong k[x_1,...,x_r]$ where $r<n$. By our induction hypothesis, $\dim(B_1) = r = tr.deg_k(B_1)\le n-1$.

Continue this process, by replacing the role of $B$ by $B_1$, and $P_i$ by $B_1/(B_1\cap Q_i)$, we can see that the number of variables after each step is reduced at least 1. This implies $\dim(k[x_1,...,x_n])\le n$.

From this, we can see $\dim(k[x_1,...,x_n])=n$. And the result of our theorem follows. (Q.E.D)

And it's time for our main theorem.

Theorem 4.2. Let $A$ be an affine ring, then $\dim(A)=tr.deg_k(A)$.

Proof. It can be seen that $\dim(A) = \sup \{\dim A/P|P\in \min(A)\}$, because any chain of prime ideals $P_0 \subset ...\subset P_n$ in $A$ is 1-1 correspondence to the chain of prime ideals $0=P_0/P_0\subset...\subset P_n/P_0$ in the integral domain $A/P_0$.

Furthermore, because $A/P$ is an integral domain, by previous theorem $\dim A/P = tr.deg_k(A/P)$. As we have proved in the first section, $\max\{tr.deg_k(A/P)|P\in \min(A)\}=tr.deg_k(A)$. Hence,

$$
\begin{gather}
\dim A = \sup \{\dim A/P|P\in \min(A)\}=\\

 = \max\{tr.deg_k(A/P)|P\in \min(A)\}=tr.deg_k(A)
\end{gather}
$$

(Q.E.D)