One of the most important theories in Commutative Algebra and Algebraic Geometry is dimension theory, and it is also important to understand this via some points of view. This self-contained note will be the starting point to dig deeper on this beautiful theory. This note is heavily based on Gathmann's note on Commutative Algebra (Chapter IX-X-XI).
1. Some facts related to finitely generated $R$-module. Let us begin with some background knowledge related to module theory, i.e. the Cayley-Hamilton's theorem as well as Nakayama's lemmas.
Let $M$ be a $R$-module, and $\varphi: M\to M$ a $R$-homomorphism. One can consider $M$ as $R[x]$-module by defining the action $xm=\varphi(m)$, and hence, $(a_nx^n+...+a_0)m=a_n\varphi^n(m)+...+a_0m$, for $a_i\in R$. It can be easily checked that $f(x)m$ defined above satisfies the conditions of scalar multiplication. This allows us to prove the following
Theorem 1.1. Let $M$ be a finitely generated $R$-module, $\varphi: M\to M$ a $R$-homomorphism and $I$ an ideal of $R$ such that $\varphi(M)\subset IM$, then there exists a non-negative integer $n$ and $a_0,...,a_{n-1}\in I$ such that $\varphi^n+a_{n-1}x^{n-1}+...+a_0id=0\in Hom_R{M,M}$.
Proof. Let $\{m_1,...,m_n\}$ be the generators of $M$ as $R$-module. The assumption $\varphi(M)\subset IM$ implies that $\varphi(m_i)=\sum_{j}c_{ij}l_j$, where $c_{ij}\in I, l_i\in M$. For any $j$, we can represent $\varphi(l_j)=\sum_{k}c_{jk}m_k$ for some $c_{jk}\in R$. From this, one can see that $\varphi(m_i)=\sum_{j}a_{ij}m_j$, for $a_{ij}\in I$.
Now let us consider the action from $R[x]$ to $M$ defined as above, i.e. $xm=\varphi(m)$. Then it can be seen that $xm_i=\sum_{j}a_{ij}m_j$, and hence, $\sum_j(x\delta_{ij}-1)a_{ij}m_j=0$ for all $i$, and $\delta_{ij}$ is the Kronecker's delta. This defines a matrix multiplication, and let us denote the matrix $((x\delta_{ij}-1)a_{ij})_{ij}=A_x$. If we multiply $A_x$ with the matrix of cofactors of $A_x$, we obtain the matrix $(\det A_x)I_n$, and hence, $\det(A_x)m_i=0$ for all $i$.
If we expand the determinant of the matrix, it can be seen that the only places where $x$ appears is the main diagonal, and all $a_{ij}\in I$. This yields $(x^n+a_{n-1}x^{n-1}+...+xa_1+a_0)m_i=0(\forall i)$, for some $a_i\in I$. Replacing $xm$ as $\varphi(m)$, we obtain $(\varphi^n+...+a_0id)m_i=0$, which means $(\varphi^n+...+a_0id)$ acts as zero homomorphism in $Hom_R(M,M)$. (Q.E.D)
If $R\equiv k$ is a field, $M\equiv V$ a finite dimension $k$-vector space, and $I\equiv k$, $\varphi$ is an endomorphism of $V$, we then obtain the classical Cayley-Hamilton's theorem. Surprisingly, the theorem above gives us the Nakayama's lemma as an application.
Theorem 1.2 (Nakayama's lemma). Let $M$ be a finitely generated $R$-module, and $I$ an ideal of $R$ such that $M=IM$, then there exists $a\in I$ such that for all $m\in M$, we have $m=am$.
Proof. Consider the $id$ map $id: M\to M$ it can be seen that $id(M)=M\subset IM$. Hence, one can apply the Cayley-Hamiton's theorem to obtain $(id^n+a_{n-1}id^{n-1}...+a_0id)m=0$ for all $m\in M$. From this, $m+a_{n-1}m+...+a_0m=m(1+a_{n-1}+...+a_0)=0$. Let $a=-a_{n-1}-...-a_0$. We obtain $m=am$, for all $m\in M$. (Q.E.D)
From this, one can easily deduce some versions for the Nakayama's lemma.
Corollary 1.3 (Nakayama's Lemma). Let $R$ be a Noetherian ring, and $J=\cap_{P\in Max(R)}$ be the Jacobson radical of $R$, $M$ be a finitely generated $R$-module such that $JM=M$, then $M=0$.
Proof. Applying Theorem 1.2, there exists $a\in J$ such that for all $m\in M$, we have $am=m$, or equivalently $(1-a)m=0$. However, $a$ lies in all maximals ideal of $R$ implies that $1-a$ does not lie in any maximal ideal of $R$, and hence, $1-a\in R^\times$. This follows directly that $m=0$. (Q.E.D).
And one of the famous version is the Nakayama's lemma for local ring.
Corollary 1.4. Let $R$ be a local Noetherian ring with the maximal ideal $P$, $M$ is a finitely generated $R$-module such that $PM=M$, then $M=0$.
Proof. It is easy to see that the Jacobson radical in this case is $P$ itself, and we just apply the Corollary 1.3. (Q.E.D)
As an application, let us prove the following
Corollary 1.4. Let $M$ be a finitely generated $R$-module and $\varphi: M\to M$ is surjective $R$-map. Then $\varphi$ is an isomorphism.
Proof. Let us consider $M$ as $R[x]$-module by defining the scalar multiplication $xm=\varphi(m)$. It can be seen that $M$ is finitely generated $R[x]$-module. Let $I=(x)$ be an ideal of $R[x]$. We can see that $IM =\varphi(M) = M$, and hence, applying Theorem 1.2, there exists $f(x)\in (x)$ such that $f(x)m=m$ for all $m\in M$. We can represent $f(x) = xg(x)$ for some $g\in R[x]$, and $f(x)m=xg(x)m=g(x)xm=m$. If $\varphi(m)=0$, then $xm=0$, and from this $g(x)xm=m=0$. This yields $\varphi$ is injective, and hence, bijective. (Q.E.D)
An exercise below can be seen as an application of Nakayama's lemmas.
Exercise 1.5. Let $R$ be a Noetherian local domain with maximal ideal $P$. Prove that if $P/P^2\cong R/P$, then $R$ is a DVR.
We are now ready for more advanced knowledge related to dimension arguments.
2. Lying Over, Incomparability, Going-Up and Going-Down.
Let us begin with some interesting lemmas, that are directly applied to the proof of these four theorems above. These lemmas have directly consequences in Algebraic Number Theory, and one can realize that it is a generalization of Gauss' Lemma.
Lemma 2.1. Let $R\subset R'$ be a ring extension, where $R'$ is a domain, and $\overline{R}$ the integral closure of $R$ in $R'$. Let $f(x),g(x)\in R'[x]$ be two monic polynomials such that $f(x)g(x)\in R[x]$, then $f(x),g(x)\in \overline{R}[x]$.
Proof. Let $K$ be the quotient field of $R'$, and $L$ is an extension of $K$ such that $h(x)=f(x)g(x)$ splits over $L$. Assume $h(x)=\prod_i(x-\alpha_i)\prod_j(x-\beta_j)$, where $f(x)=\prod_{i}(x-\alpha_i),g(x)=\prod_{j}(x-\beta_j)$. From the assumption that $h(x)\in R[x]$ and that $h(\alpha_i)=0$ for all $i$, we have $\alpha_i$ is integral over $R$. This implies all coefficients of $f(x)\in R'[x]$ is integral over $\overline{R}$, and lie in $R'$. Because $\overline{R}$ is the integral closure of $R$ in $R'$, we obtain all coefficients of $f(x)$ lie in $\overline{R}$. And similar thing holds for $g(x)$. (Q.E.D)
In the case $R=\mathbb{Z}, R'=\mathbb{Q}$, one can see $\mathbb{Z}$ is integrally closed in $\mathbb{Q}$. Let $f(x),g(x)\in\mathbb{Z}[x]$, and $f(x)|g(x)$ in $\mathbb{Q}[x]$, then in $\mathbb{Q}[x]$, we can write $f(x)=g(x)h(x)$, where $h(x)\in\mathbb{Q}[x]$. It can be seen from the lemma above that actually, $h(x)\in\mathbb{Z}[x]$, and hence $f(x)|g(x)$ in $\mathbb{Z}[x]$. And one obtains exactly the Gauss' lemma. As an application, we next prove the following
Lemma 2.2. Let $R\subset R'$ be the integral ring extension, where $R'$ is a domain and $R$ is normal (recall that a domain is called normal if it is integrally closed in its quotient field). Let $a\in R'$.
(a) The minimal polynomial of $a$ over $R$, denoted by $f_a(x)$ lie in $R[x]$.
(b) Let $P$ be a prime ideal of $R$, and $a\in PR'$, then all non-leading coefficient of $f_a$ lie in $P$.
Proof.
(a) $a$ is integral over $R$ implies that there is a monic polynomial $g(x)\in R[x]$ such that $g(a)=0$. Let $f(x)$ be the minimal polynomial of $a$ over $K[x]$, where $K$ is the quotient filed of $R$. Then there exists $h(x)\in K[x]$ such that $g(x)=f(x)h(x)$. Now, it follows from Lemma 1, that $f(x)\in R[x]$.
(b) If $a\in PR'$, then $a$ can be expressed by $p_1a_1+...+p_na_n$, for some $p_i\in P, a_i\in R'$. Let $R_1=R[a_1,...,a_n]$, then $R_1$ can be viewed as finitely generated $R$-module. Let us consider the $R$-homomorphism, $\varphi: R_1\to R_1$ sending $x$ to $ax$, and hence $\varphi(R_1)\subset PR_1$. By the Cayley-Hamilton's theorem, one can see $\varphi$ satisfies the monic relation $\varphi^n+a_{n-1}\varphi^{n-1}+...+a_1\varphi+a_0id=0\in Hom_R(R_1,R_1)$, for some $a_i\in P$. Putting 1 in the monic relation, we have $a^n+a_{n-1}a^{n-1}+...+a_1a+a_0=0$. Let $f(x)=f_a(x)g(x)$ for some $g(x)\in R[x]$. Reduction mod $P$ yields $x^n=f_a(x)\mod pg(x)\mod p$. Because $R/P[x]$ is an integral domain, we must obtain $f_a(x)\mod p=x^m$, and hence, other coefficients lie in $P$. This finishes our lemma. (Q.E.D)
We are now ready for the proof of Lying Over theorem.
Theorem 2.3 (Lying Over). Let $R\subset R'$ be ring extension, and $P$ a prime ideal of $R$, then
(a) There exists a prime ideal $P'\subset R'$ lying over $P$, i.e. $P'\cap R=P$ iff $PR'\cap R\subset P$.
(b) If $R'$ is integral over $R$ then there always exists $P'\in Spec(R')$ lying over $P$.
Proof.
(a) The "$\Rightarrow$" is easy, since $PR'\subset P$, and hence $PR'\cap R\subset P'\cap R=P$. For "$\Leftarrow$", we need a following
Lemma 2.4. Let $R$ be a ring, and $subset R$ be a multiplicative closed subset, $I$ an ideal of $R$ such that $I\cap S=\emptyset$, then there exists $P\in Spec(R)$ such that $I\subset P, P\cap S=\emptyset$, and $P_S$ is maximal in $R_S$.
Proof. Let $I_S$ be the extension of $I$ in the localized ring $R_S$, and $P_S$ is a maximal ideal of $R_S$ containing $I_S$, then via the contraction map $P\in Spc(R)$, and $I\subset P$. Now, if $s\in P\cap S$, then via the extension map, one can see $1\in P_S$, a contradiction to the maximality of $P_S$. Hence, $P\cap S=\emptyset$. (Q.E.D)
Using this, let $S=R\setminus P$, then $PR'\cap R\subset P$ implies that $PR'\cap S=\emptyset$. By the lemma above, there exists $P'\in Spec(R')$ containing $PR'$ (1) and $P'\cap S=\emptyset$ (2). From (1), we can see that $P\subset PR'\cap R\subset P'\cap R$, and from (2), $P'\cap R\setminus P=\emptyset$, which implies $P'\cap R\subset P$. And we now obtain $P'\cap R=P$.
(b) If $R'$ is integral over $R$, we apply the first half of the proof of Lemma 2.2 (b), to see that any element $a\in PR'$ satisfies the monic relation $a^n+a_{n-1}a^{n-1}+...+a_1a+a_0=0$ where $a_i\in P$. This yields if $a\in PR'\cap R$, then actually, $a^n\in P$, and hence $a\in P$, because $P$ is prime. Now, the conclusion follows directly from Part (a). (Q.E.D).
Remark 2.5. The Lying Over theorem is the first step to obtain the equality of dimension between $R$ and $R'$, where $R'$ is integral over $R$. It, together with Going-Up Theorem, say that if we have the following chain of prime ideals $P_0\subsetneq P_1\subsetneq...\subsetneq P_n$ in $R$ then we can obtain the corresponding chain $Q_0\subseteq Q_1\subseteq ...\subseteq Q_n$ where $Q_i$ is lying over $P_i$. It can be seen that $Q_i\ne Q_{i+1}$ since $Q_i\cap R\ne Q_{i_1}\cap R$. And hence, the corresponding chain is strict. This yields $\dim R\le \dim R'$.
Now, if we start with a prime chain $Q_0\subsetneq Q_1\subsetneq ...\subsetneq Q_n$ in $R'$, we then obtain a corresponding chain $Q_0\cap R subseteq Q_1\cap R\subseteq ...\subseteq Q_n\cap R$ in $R$. The question now arises is can $P_i=P_{i+1}$? The following theorem, Incomparability will answer it.
Theorem 2.6 (Incomparability). If $R\cap R'$ be integral ring extension, and $P'\ne Q'\in Spec(R')$, and $P=P'\cap R=Q'\cap R\in Spec(R)$, then $P'\not\subset Q'$, and $Q'\not\subset P'$.
Proof. Assume that $P'\subset Q'$, then from the assumption that $P'\ne Q'$, there exists $a\in Q'\setminus P'$. We can embed $R/P'\cap R=R/P$ into the domain $R'/P'$ by sending $a\mod P$ to $a\mod P'$. It can be seen that this map is well-defined and that it is injective. Also, $R'$ is integral over $R$ implies that $R'/P'$ is integral over $R/P$ (we just need to send the monic relation for $a\in R'$ via the projection map mod $P'$). Now, let $\overline{a}$ the image of $a$ via the mod $P'$map, it is non-zero by our assumption, and satisfies the monic relation
$$\overline{a}^n+\overline{a_{n-1}}\overline{a}^{n-1}+...+\overline{a_0}=0$$
for some $\overline{a_i}\in R/P$, and we can choose smallest value for such $n$. Now, $\overline{a}\in Q'/P'$ implies that $\overline{a_0}\in Q'/P'$. That means,
$$\overline{a_0}\in Q'/P'\cap R/P=\{a\in Q'|a\text{ mod }P'\}\cap \{a\in R|a\text{ mod } P'\}=$$
$$=\{a\in Q'\cap R|a\text{ mod } P'\}=\{a\in P|a\text{ mod } P'\}=\{0\}$$
And hence, we obtain the relation in the domain $R'/P'$
$$(\overline{a}^{n-1}+\overline{a_{n-1}}\overline{a}^{n-2}+...+\overline{a_1})a=0$$
This yields $\overline{a}^{n-1}+\overline{a_{n-1}}\overline{a}^{n-2}+...+\overline{a_1}$ and we obtain the smaller $n$ for the monic relation. It is a contradiction. Hence $P'\not\subset Q'$, and similarly, $Q'\not\subset P'$. This finishes our theorem. (Q.E.D)
Remark 2.7. Let us postpone for a while to see the consequence of the Incomparability Theorem, by continue to consider our previous argument on Remark 2.6. If $Q_i\cap R=Q_{i+1}\cap R$ then by Incomparability Theorem, $Q_i\not\subset Q_{i+1}$, a contradiction. Hence, we obtain the strict chain $Q_0\cap R subsetneq Q_1\cap R\subsetneq ...\subsetneq Q_n\cap R$ in $R$. This does yield $\dim R'\le \dim R$. Now, it follows that $\dim R'=\dim R$. We are now staying away this result one theorem, the Going-Up.
Theorem 2.8. (Going-Up). Let $R\subset R'$ be integral ring extension, and $P\subset Q$ a chain of prime ideal in $R$, $P'\in Spec(R')$ lying over $P$. Then there exists $Q'\in Spec(R')$ lying over $Q$ such that $P'\subset Q'$.
Proof. Let us consider the integral extension of domains $R/P\subset R'/P'$. By Lying Over Theorem, there exists $Q'/P'$ lying over $Q/P$, that means $Q'/P'\cap R/P=Q/P$. Via the embedding $R/P$ into $R'/P'$, we have
$$Q'/P'\cap R/P=\{a\in Q'|a\text{ mod }P'\}\cap \{a\in R|a\text{ mod }P'\}=$$
$$=\{a\in Q'\cap R|a\text{ mod }P'\}$$
Also $Q/P=\{a\in Q|a\text{ mod }P'\}$, and hence, $Q'/P'\cap R/P=Q/P$ implies that $Q'\cap R=Q$, and that $Q'$ lying over $Q$, and $P'\subset Q'$. (Q.E.D)
From Going-Up Theorem, and Remark 2.5, 2.7, we obtain the following important
Theorem 2.9. Let $R\subset R'$ be integral extension of ring, then $\dim R=\dim R'$.
Its consequences will be discussed in our next parts.
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