Spring 2026: Math 831 Daily Update

Tuesday, January 20

We began the semester with some basic terminology, giving proofs and examples where appropriate.

Terminology.

(i) $I\subseteq R$ is an ideal if $a, b\in I$ implies $a+b\in I$ and $r\in R$ implies $ra\in I$.
(ii) $P\subseteq R$ is a prime ideal if $P$ is an ideal and $ab\in I$ implies $a\in I$ or $b\in I$. We showed that if $R := \mathbb{Q}[x,y]$, then $P := \{fx\ |\ f\in R\}$ is a prime ideal.
(iii) $M\subseteq R$ is a maximal ideal if $M$ is a proper ideal not contained in any other proper ideal. We showed that for $R$ as in (ii), the set of polynomials with zero constant term is a maximal ideal. Using Zorn's lemma, we proved
Proposition. Maximal ideals exist in $R$. In particular, every proper ideal of $R$ is contained in a maximal ideal.
(iv) $0 \not = a$ is a zerodivisor if there exists $0\not = b$ such that $ab = 0$. Otherwise $a$ is a non-zerodivisor.
(v) $R$ is an integral domain if every non-zero element in $R$ is a non-zerodivisor.
(vi) $u\in R$ is a unit if there exists $v\in R$ such that $uv = 1$.
(vii) $R$ is a field if every non-zero element of $R$ is a unit.
(viii) The sum of ideals $I, J$ is the ideal $I+J := \{i+j\ | \ i\in I\ \text{and}\ j\in J\}$.
(ix) For the subset $X\subseteq R$, $\langle X\rangle$, the ideal generated by $X$ consists of all finite $R$-linear combinations of the elements of $X$. This was observed to the same as the intersection of all ideals containing $X$. Special cases were $\langle a\rangle$, a principal ideal, and $X = \{x_1,\ldots, x_n\}$ so that $\langle X\rangle := \{r_1x_1+\cdots + r_nx_n\ |\ r_i\in R\}$.
(x) Given ideals $I, J\subseteq R$, the product of $I$ and $J$ is the ideal $IJ := \langle \{ij\ |\ i\in I\ \text{and}\ j \in J\}\rangle$.
(xi) For $n\geq 1$, we have $I^n$, the $n$th power $I$, which is the ideal consisting of all finite sums of $n$-fold products of elements from $I$.
(xii) A map $\phi :R\to S$ between rings is a ring homomorphism if $\phi (a+b) = \phi (a)+\phi (b)$ and $\phi (ab) = \phi (a)\phi(b)$, for all $a, b\in R$.
(xiii) Given an ideal $I\subseteq R$, the factor ring $R/I$ consists of all cosets of the form $a+I$, with $a\in R$, subject to:
- $a+I = b+I$ if and only if $a-b\in I$.
- $(a+I)+(b+I) := (a+b)+I$.
- $(a+I)\cdot (b+I) := ab+I$.

We then proved that every maximal ideal in $R$ is a prime ideal and gave an example where the converse fails, namely $P = \langle x\rangle \subseteq \mathbb{Q}[x,y]$. From this we initiated our discussion of the process of localization, by giving the following definition.

Definition. A subset $S\subseteq R$ is said to be multiplicatively closed (henceforth abbreviated as mc) if: $0\not \in S$, $1\in S$ and $s_1, s_2\in S$ implies $s_1s_2\in S$.

Examples. $S$ is a multiplicatively closed set in each of the following cases:

(i) $S := \{1, f, f^2, f^3, \ldots \}$, for any $f\in R$ such that $f^n\not = 0$, for all $n\geq 0$.
(ii) $S$ is the set of non-zerodivisors in $R$.
(iii) $S := R\backslash P$, for $P$ a prime ideal of $R$.

We ended class with the following:

Krull's Lemma. Let $I\subseteq R$ be an ideal and $S\subseteq R$ a multiplicatively closed set such that $I\cap S = \emptyset$. Then there exists an ideal $P$ containing $I$ that is maximal with respect to the property $P\cap S = \emptyset$. Moreover, $P$ is a prime ideal.

Thursday, January 22

We began class by defining the nilradical $\sqrt{I}$ of an ideal $I\subseteq R$ as the set of elements $r\in R$ such that $r^n\in I$, for some $n\geq 1$. We also noted that $\sqrt{(0)}$ is called the nilradical of $R$ and consists of the nilpotent elements in $R$. We then gave a proof of the following corollary to Krull's Lemma,

Corollary. Let $I\subseteq R$ be an ideal. Then $\sqrt{I} = \bigcap \{P\ |\ P\ \text{is a prime ideal containing}\ I\}$.

For a multiplicatively closed set $S\subseteq R$ we then defined an equivalence relation on $R\times S$ as follows: $(r_1,s_1)\sim (r_2,s_2)$ if and only if there exists $s_3\in S$ such that $s_3(r_1s_2-r_2s_1) = 0$. After showing that $\sim$ is an equivalence relation, we wrote the class of $(r,s)$ as a fraction $\frac{r}{s}$ and showed that the expected operations

\frac{r_1}{s_1}+\frac{r_2}{s_2} := \frac{r_1s_2+r_2s_1}{s_1s_2} \quad\quad \text{and}\quad\quad \frac{r_1}{s_1}\cdot \frac{r_2}{s_2} := \frac{r_1r_2}{s_1s_2}

are well defined and make $R_S$ into a commutative ring whose zero element is $\frac{0}{1} = \frac{0}{s}$, for any $s\in S$ and whose multiplicative identity is $\frac{1}{1} = \frac{s}{s}$, for all $s\in S$.

We then noted the following:

(i) The natural map $\phi: R\to R_S$ taking $r$ to $\frac{r}{1}$ is a ring homomorphism.
(ii) $\frac{r}{s} = \frac{0}{1}$ in $R_S$ if and only if $s'r = 0$ in $R$ for some $s'\in S$. In particular, the kernel of $\phi$ is the set $\{r\in R\ |\ sr = 0, \text{for some}\ s\in S\}$.
(iii) $\frac{s}{1}$ is a unit in $R_S$ for all $s\in S$.

We then discussed the correspondence between the ideals of $R$ and those of $R_S$. For an ideal $I\subseteq R,$ we noted $I_S := \{\frac{i}{s}\ |\ i\in I\ \text{and}\ s\in S\}$ is an ideal of $R_S$. Similarly, we noted that $\phi^{-1}(J)$ is an ideal of $R$ for every ideal $J\subseteq R_S$. Later in the lecture we further noted that $\phi^{-1}(J)$ has the property that $sr\in \phi^{-1}(J)$, for some $s\in S$ implies $r\in \phi^{-1}(J)$.

We then showed that this correspondence was not quite a perfect one-to-one correspondence between the ideals of $R$ disjoint from $S$ and the ideals in $R_S$. On the one hand we do have $J = \phi^{-1}(J)_S$, for all ideals $J\subseteq R_S$, which shows that any ideal $J\subseteq R_S$ is $I_S$, for some ideal $I\subseteq R$. On the other hand, it is not the case that $\phi^{-1}(I_S) = I$. In fact, we saw that $\phi^{-1}(I_S) = \{r\in R\ |\ sr\in I, \ \text{for some}\ s\in S\}$, which often strictly contains $I$. For example, if $R = \mathbb{Z}$, $I = \langle 6\rangle$, and $S = \{1,2,2^2,2^3,2^4,\ldots\}$, then $\phi^{-1}(I_S) = \langle 3\rangle$ which is strictly larger than $I$. We ended class by proving the following theorem which shows the correspondence in question is one-to-one for prime ideals.

Theorem. Let $R$ be a commutative ring and $S\subseteq R$ a multiplicatively closed subset. Then there is a one-to-one correspondence between the prime ideals of $R$ disjoint from $S$ and the prime ideals of $R_S$ given by $P \leftrightarrow P_S$.

Tuesday, January 27

We began class by proving the following local-global principle:

Proposition. Let $I, J\subseteq R$ be ideals. The following are equivalent:

(i) $I = J$.
(ii) $I_P = J_P$, for all prime ideals $P\subseteq R$.
(iii) $I_{\mathfrak{m}} = J_{\mathfrak{m}}$, for all maximal ideals $\mathfrak{m} \subseteq R$.

We then began our study of unique factorization domains, UFDs. We discussed with details, the following

Preliminaries. Let $R$ be an integral domain with quotient field $K$.

(i) Cancellation holds in $R$, i.e., if $a\not = 0$ and $ab = ac$, then $b = c$.
(ii) $u\in R$ is a unit if there exists $v\in R$ such that $uv = 1$.
(iii) If $a\not = 0$, we say a divides b, written $a\mid b$, if there exists $r\in R$ such that $b = ra$. This happens if and only if $\langle b\rangle \subseteq \langle a\rangle$.
(iv) $a\mid b$ and $b\mid a$ if and only if $\langle a\rangle = \langle b\rangle$ if and only if $b= ua$, for some unit $u\in R$. In this case we say that $a$ and $b$ are associates.
(v) The non-zero, non-unit $q\in R$ is said to be irreducible if whenever $q = ab$, for $a,b\in R$, then either $a$ or $b$ is a unit.
(vi) The non-zero, non-unit $p\in R$ is said to be prime if whenever $p\mid ab$, then $p\mid a$ or $p\mid b$. By induction, if $p$ is prime and $p$ divides $a_1\cdots a_n$, then $p$ divides $a_i$, for some $i$.
(vii) For $p, q, u\in R$ with $u$ a unit, $q$ is irreducible if and only if $qu$ is irreducible and $p$ is prime if and only if $pu$ is prime.
(viii) For $p, q\in R$, $q$ is irreducible if and only if $\langle q\rangle$ is maximal among principal ideals and $p$ is prime if and only if $\langle p\rangle$ is a prime ideal.
(ix) Every prime element is irreducible, but not conversely.
(x) In the ring $R = \mathbb{Q}[t^2, t^3]$, with $t$ an indeterminate, the element $t^2$ is irreducible, but not prime. Another classic example: In the ring $\mathbb{Z}[\sqrt{-5}]$, $3, 2\pm \sqrt{-5}$ are irreducible elements that are not prime.

Proposition. Let $R$ be an integral domain and suppose $p_1\cdots p_r = q_1\cdots q_s$, with each $p_i, q_j$ prime. Then $r = s$ and after re-indexing $q_i = u_ip_i$, with $u_i$ a unit in $R$.

We noted that for the classical example from algebraic number theory, with $R = \mathbb{Z}[\sqrt{-5}]$ can be shown that $3\cdot 3 = 9 = (2+\sqrt{-5})\cdot (2-\sqrt{-5})$ gives two distinct factorizations into irreducible elements in the ring $\mathbb{Z}[\sqrt{-5}]$, showing that the proposition fails for irreducible elements. It is precisely this failure that prevents most of the rings in commutative algebra, algebraic geometry and algebraic number theory from being unique factorization domains. This led to the following

Theorem. Let $R$ be an integral domain. The following are equivalent:

(i) Every non-zero, non-unit can be written uniquely (up to order and unit multiple) as a product of irreducible elements, i.e., if $a\in R$ is a non-zero, non-unit element and $a = q_1\cdots q_r = q'_1\cdots q'_s$ with each $q_i, q'_i$ irreducible, then $r = s$ and after re-indexing, for all $i$ we have $q'_i = u_iq_i$, for units $u_i\in R$.
(ii) Every non-zero, non-unit $a\in R$ can be written as a product of primes.

A crucial component in the proof of the theorem was that if $R$ is a UFD then every irreducible element is a prime element. Thus, for a UFD, an element is prime if and only if it is irreducible.

We ended class by presenting, but not verifying, the following examples to illustrate how subtle the UFD property can be.

Examples. 1. If $R$ is a UFD, then $R[x]$ is a UFD. Thus, the polynomial rings $\mathbb{Z}[x_1, \ldots, x_n]$ and $K[x_1, \ldots, x_n]$ are UFDs, for $K$ any field.

2. $\mathbb{C}[x,y]/\langle x^2+y^2-1\rangle$ is a UFD, but $\mathbb{R}[x,y]/\langle x^2+y^2-1\rangle$ is not a UFD.

3. $\mathbb{C}[x,y,z]/\langle x^2+y^2+z^2-1\rangle$ is not a UFD, but $\mathbb{R}[x,y,z]/\langle x^2+y^2+z^2-1\rangle$ is a UFD.

4. $\mathbb{C}[x,y,z]/\langle x^2+y^3+z^5\rangle$ is a UFD, but $\mathbb{C}[x,y,z]/\langle x^2+y^3+z^4\rangle$ is not a UFD.

Thursday, January 29

We continued our discussion of UFDs with the following sequence of results.

Definition+Proposition. Let $R$ be an integral domain. $R$ is said to satisfy ACC on principal ideals (ACC for ascending chain condition) if given a chain of principal ideals $\langle a_1\rangle\subseteq \langle a_2\rangle \subseteq \cdots$, there exists $n\geq 1$ such that $\langle a_n\rangle = \langle a_{n+1}\rangle = \cdots$. The following statements are equivalent:

(i) $R$ satisfies ACC on principal ideals.
(ii) Every non-empty collection of principal ideals has a maximal element.

With this Definition+Proposition in hand, we were able to prove the following important characterization of UFDs.

Theorem A. Let $R$ be an integral domain. The following are equivalent:

(i) $R$ is a UFD.
(ii) $R$ satisfies ACC on principal ideals and irreducible elements are prime.
(iii) Every non-zero prime ideal contains a prime element.

An immediate and easy corollary was that any PID is a UFD: After all, any prime ideal is principal, necessarily generated by a prime element.

We then turned to the following proposition, the second part of which is known as Nagata's Lemma.

Proposition. Let $R$ be an integral domain and $S\subseteq R$ a multiplicatively closed set in which each element is a product of prime elements. Write $\mathcal{P}$ for the set of prime factors of the elements of $S$. Then:

(i) If $R$ is a UFD, then $R_S$ is a UFD.
(ii) If every element of $R$ is divisible by at most finitely many primes in $\mathcal{P}$ and $R_S$ is a UFD, then $R$ is a UFD.

The surprising condition (iii) in Theorem A played a key role in the proof of this proposition. We then stated and proved the following major theorem.

Theorem B. Suppose $R$ is a UFD. Then the polynomial ring $R[x]$ is a UFD.

The proof of the theorem relied on Nagata's Lemma and the following facts which we did not prove: (i) For $a\in R$, $a\mid f(x)$ in $R[x]$ if and only if, in $R$, $a$ divides every coefficient of $f(x)$ and (ii) If $p\in R$ is a prime element, then $p$ is still a prime element in $R[x]$. The proof of the theorem then followed using Nagata's Lemma: Let $S$ be the non-zero elements of $R$. Thus every element of $S$ is a product of prime elements. By the facts just stated, every element of $S$ is a product of prime elements in $R[x]$, thus $R[x]$ is a UFD, if $R[x]_S = R_S[x]$ is a UFD. But $R_S = K$, the quotient field of $R$ and $R_S[x] = K[x]$ is a PID, and hence a UFD.

As corollaries we noted that the polynomial rings $\mathbb{Z}[x_1, \ldots, x_n]$ and $K[x_1, \ldots, x_n]$ are UFDs, for $K$ any field.

Tuesday, February 3

We spent most of the class working through the following examples.

Example A. The complex coordinate ring of the unit circle $S := \mathbb{C}[x,y]/\langle x^2+y^2-1\rangle$ is a UFD.

We did this by first showing that $S \cong \mathbb{C}[u, v]/\langle uv-1\rangle$, where $u,v$ are a new set of indeterminates. We then showed that $\mathbb{C}[u,v]/\langle uv-1\rangle \cong \mathbb{C}[u, u^{-1}]$. Since the latter ring is the localization of a UFD, namely $\mathbb{C}[u]$, at a multiplicatively closed set whose elements are products of primes, $\mathbb{C}[u, u^{-1}]$, and hence $S$ is a UFD.

Example B. The real coordinate ring of the unit circle $R := \mathbb{R}[x, y]/\langle x^2+y^2-1\rangle$, is not a UFD.

This fact follows because the image of $x$ in $R$ is an irreducible element that is not a prime element. Crucial steps in the proof required showing:

(i) $R$ is an integral domain.
(ii) $\overline{x}$, the image of $x$ in $R$, is not a prime element. This followed since $R/\langle \overline{x}\rangle \cong \mathbb{R}[y]/\langle y^2-1\rangle$, which is not an integral domain.
(iii) Every polynomial in $f\in \mathbb{R}[x, y]$ can be written uniquely as $f = f_0+f_1+\cdots + f_n$, where each $f_j\in \mathbb{R}[x,y]$ is homogeneous of degree $j$.
(iv) $x^2+y^2$ is an irreducible polynomial in $\mathbb{R}[x,y]$.
(v) $\overline{x}\in R$ is an irreducible element.

We then addressed the question: Why doesn't this same approach show that $S := \mathbb{C}[x,y]/\langle x^2+y^2-1\rangle$ is not a UFD? The answer is that while the image of $x$ in $S$ is still not prime, the image of $x$ in $S$ is no longer irreducible. In fact, in $S$ we have

x\equiv \frac{1}{2}(x-iy)\cdot \{(x+iy+i)(x+iy-i)\},

where $x-iy$ is a unit in $S$, but neither $x+iy+i$ nor $x+iy-i$ are units in $S$.

We then recorded the following result, which appears in the celebrated paper of R. Swan, Vector bundles and projective modules. A write up of this theorem appears in the Supplementary Materials folder on our Canvas page and our course webpage.

Theorem (Swan). Let $F$ denote either the real numbers or the complex numbers and set

R := F[x_1, \ldots, x_n]/\langle x_1^2+\cdots + x_n^2-1\rangle.

If $F = \mathbb{R}$, then $R$ is a UFD for $n\geq 3$ and if $F = \mathbb{C}$, then $R$ is a UFD for $n\geq 4$.

We ended class with the following definition and comments.

Definition. Let $R$ be an integral domain. Given non-zero, non-unit elements $a, b\in R$, we say that the non-zero, non-unit element $d$ in $R$ is a greatest common divisor of a and b if: (i) $d\mid a$ and $d\mid b$ and (ii) If $e\mid a$ and $e\mid b$, then $e\mid d$. In this case we will write $d = \textrm{GCD}(a,b)$. $R$ is a GCD domain if every pair of non-zero, non-unit elements has a GCD.

Comments. Let $R$ be an integral domain.

(i) Using Bezout's principal, it is easy to see that concept of GCD in $\mathbb{Z}$ is the same as GCD above.
(ii) GCDs are unique up to unit multiple.
(iii) If GCDs exist, then LCMs (least common multiples) exist and one has $\textrm{GCD}(a,b)\cdot \textrm{LCM}(a,b) = ab$.
(iv) UFDs are GCD domains. In particular, if $a, b$ have prime factorizations $a = p_1^{e_1}\cdots p_n^{e_n}$, $b = p_1^{f_1}\cdots p_n^{f_n}$, with each $e_i, f_j\geq 0$, then

\textrm{GCD}(a,b) = p_1^{\textrm{min}(e_1,f_1)} \cdots p_n^{\textrm{min}(e_n,f_n)}\quad\quad \textrm{and}\quad\quad \textrm{LCM}(a,b) = p_1^{\textrm{max}(e_1,f_1)} \cdots p_n^{\textrm{max}(e_n,f_n)}.

Thursday, February 5

We began class with the following observation showing the connection of the GCD property with homological considerations.

Homological Connection. Let $R$ be an integral domain and take $a, b$ non-zero, non-units in $R$. Consider the surjective $R$-module map

R^2\overset{\phi}\to \langle a,b\rangle

given by $\phi \begin{pmatrix} r\\s\end{pmatrix} = ra+sb$. Write $K$ for the kernel of $\phi$. If $K$ is a cyclic module, then $\textrm{GCD}(a,b)$ exists. In this case if $K = \left\langle \begin{pmatrix} c\\e\end{pmatrix}\right\rangle$ and $\begin{pmatrix} -b\\a\end{pmatrix} = d\cdot \begin{pmatrix} c\\e\end{pmatrix}$, then $d = \textrm{GCD}(a,b)$.

We then presented the following theorem which shows the relationship between the UFD property and the homological connection above.

Theorem A. Let $R$ be an integral domain satisfying ACC on principal ideals. The following are equivalent.

(i) $R$ is a UFD.
(ii) $R$ is a GCD domain.
(iii) In the notation above, $K$ is a cyclic module, for all non-zero, non-unit $a, b\in R$.

After defining what it means for an ideal to have a finite free resolution, we then mentioned, but did not yet prove, the following major theorem that plays a key role in our quest to show that regular local rings are UFDs.

Theorem B. Let $R$ be an integral domain and $I\subseteq R$ a locally principal ideal. If $I$ has a finite free resolution, then $I$ is principal.

We then noted how these components are relevant for our goal of showing that regular local rings are UFDs, and the course we need to take to achieve this result. We will show that a regular local ring $(R, \mathfrak{m})$ has the property that every ideal has a finite free resolution. By induction on the dimension of the ring, condition (iii) holds locally in the ring $R_S$, for $S = \{1, x, x^2, \ldots \}$ with $x \in \mathfrak{m}\backslash \mathfrak{m}^2$. Such an $x$ is a prime element. But any $K$ in $R_S^2$ in part (iii) is isomorphic to the ideal $(aR_S:b)$, which is then locally principal. Applying Theorem B shows $(aR_S:b)$ is principal, and hence $K$ is cyclic. By Theorem A, $R_S$ is a UFD and by Nagata's lemma, $R$ is a UFD.

We finished class by recalling various facts and terminology concerning modules over a commutative ring, including:

(i) The definition of a module, noting the axioms for a module over a ring are the same as those for a vector space over a field.
(ii) The notions of submodule, homomorphism between modules, factor modules, submodules generated by a subset, and direct sums are all completely analogous to the same for vector spaces over a field.
(iii) We also defined exact sequences, short exacts sequences and split exact sequences, showing that if $0 \to A\overset{f}\to B\overset{g}\to C\to 0$ is a split exact sequence with splitting map $j: C\to B$, then $B = f(A)\oplus j(C)$.
(iv) We defined an $R$-module to be projective, if it satisfies any of the following three equivalent conditions:
1. (a) Given a diagram of $R$-modules and $R$-module homomorphisms $f, g$ $P f h M g N 0$ there exists $h$ as above so that the diagram commutes.
2. (b) $P$ is a direct summand of a free $R$-module $F$, i.e., there exists a free $R$-module $F$ such that (up to isomorphism) $F = P\oplus K$, for some $K$.
3. (c) Every short exact sequence $0\to A\overset{f}\to B\overset{g}\to P \to 0$ splits, i.e., there exists $j: P\to B$ such that $gj = 1_P$.
(v) We finished class by defining what it means for a module to be Noetherian or Artinian, noting that $R$ is a Noetherian ring or Artinian ring if it is Noetherian or Artinian as a module over itself.

Tuesday, February 10

We began class by finishing our list of facts concerning modules over the commutative ring $R$.

(vi) Given a short exact sequence of $R$-modules $0\to A\overset{f}\to B\overset{g}\to C\to 0$, $B$ is Noetherian (resp., Artinian) if and only if $A$ and $C$ are Noetherian (resp., Artinian).
(vii) If $R$ is a Noetherian (resp., Artinian) ring and $M$ is a finitely generated $R$-module, then $M$ is a Noetherian (resp., Artinian) $R$-module.
(ix) If $S\subseteq R$ is a multiplicatively closed subset and $M$ is an $R$-module, then we may extend localization to $M$ to obtain an $R_S$-module $M_S$. The construction of $M_S$ is essentially the same as the construction of $R_S$ given in the lecture of January 22.

We then began a sequence of lemmas leading to a proof of Theorem B stated in the previous lecture. We first noted the following: Suppose $R$ is an integral domain with quotient field $K$, and $I\subseteq R$ is a non-zero ideal. We set $I^{-1} := \{\alpha \in K\ |\ I\alpha\subseteq R\}$. $I^{-1}$ is called a fractional ideal. When $R$ is Noetherian, $I^{-1}$ is a finitely generated submodule of $K$. Moreover, $II^{-1}$ is an ideal of $R$ containing $I$ and $I$ is said to be invertible if $II^{-1} = R$.

Lemma A. Suppose $R$ is an integral domain and $I\subseteq R$ is an invertible ideal. Then $I$ is a projective $R$-module.

The proof of Lemma A required showing that $I$ is a finitely generated ideal and there is a surjective map $R^n\to I$ that splits.

Lemma B. Suppose $P$ is a projective $R$-module admitting a finite free resolution. Then there exist free $R$-modules $F$, $G$ such that $F = P\oplus G$.

We noted that a projective $R$-module as in Lemma B is called a stably free module and that the non-free projective module defined as the kernel of the surjective map $R^3\xrightarrow{\begin{pmatrix} x & y & z\end{pmatrix}} R$, with $R$ the coordinate ring of the real two-sphere, is stably free.

Lemma C. Let $I\subseteq R$ be an ideal that is stably free as an $R$-module. Then $I$ is principal.

The key idea behind the proof of Lemma C is the following. Suppose $R^n = G\oplus K$, where $G$ is a free summand of $R^n$ of rank $n-1$. If we let $A$ denote the $n\times (n-1)$ matrix whose columns are a basis for $G$, then we may extend $A$ to an invertible matrix by adding a column that consists of the signed minors of $A$.

The following major theorem now easily follows.

Theorem. Let $R$ be an integral domain and $I\subseteq R$ a locally principal ideal. If $I$ has a finite free resolution, then $I$ is a principal ideal.

Proof. By Lemma A, $I$ is a projective module. By Lemma B, $I$ is stably free of rank one, so by Lemma C, $I$ is a principal ideal.

Thursday, February 12

In this lecture we began an initial discussion of dimension theory in Noetherian rings, beginning with the following lemma:

Chinese Remainder Theorem. Let $I, J\subseteq R$ be comaximal ideals, i.e., $I+J = R$. Then $I\cap J = IJ$ and $R/IJ\cong R/I\oplus R/J$.

We then proved the following first step in the dimension theory of Noetherian rings.

Theorem. The commutative ring $R$ is Artinian if and only if $R$ is Noetherian and every prime ideal is a maximal ideal.

The proof was broken down into the following steps. We wrote $J$ for the Jacobson radical of $R$, i.e., the intersection of all maximal ideals of $R$. We noted that $x\in J$ if and only if $1+rx$ is a unit for all $r\in R$.

Step 1: If $R$ is Artinian, then $R$ has finitely many maximal ideals.
Step 2: If $R$ is Artinian there exists $n\geq 1$ such that $J^n = 0$, i.e., $J$ is a nilpotent ideal.
Step 3: If $R$ is Artinian, with maximal ideals $\mathfrak{m}_1, \ldots, \mathfrak{m}_t$ and $J^n = 0$, then $R\cong R/\mathfrak{m}_1^n \oplus \cdots\oplus R/\mathfrak{m}_t^n$.
Step 4: If $R$ is Noetherian and every prime ideal is maximal, then $R$ has finitely many prime ideals.
Step 5: If $R$ is as in Step 4 with maximal ideals $\mathfrak{m}_1, \ldots, \mathfrak{m}_t$, then $J^n = 0$ for some $n$, and $R\cong R/\mathfrak{m}_1^n \oplus \cdots\oplus R/\mathfrak{m}_t^n$.
Step 6: If $(R,\mathfrak{m})$ is quasi-local and $\mathfrak{m}^n = 0$, then $R$ is Artinian if and only if $R$ is Noetherian.
Step 7: In an Artinian ring, every prime ideal is a maximal ideal.

Tuesday, February 17

In today's lecture we presented two fundamental results concerning Noetherian rings. We began class with the celebrated:

Hilbert's Basis Theorem. If $R$ is Noetherian, then the polynomial ring $R[x]$ is Noetherian.

We noted that Hilbert's basis theorem has as immediate corollaries that finitely generated algebras over a Noetherian ring are Noetherian, and in particular, the coordinate ring of an algebraic variety is a Noetherian ring.

We then presented the

Artin-Rees Lemma. Let $R$ be a Noetherian ring, $I\subseteq R$ an ideal, $M$ a finitely generated $R$-module and $N\subseteq M$ a submodule. Then there exists $k\geq 1$ such that $I^nM\cap N = I^{n-1}(I^kM\cap N)$, for all $n\geq k$.

The proof we gave is modeled after the one given by Artin: If $I = \langle a_1, \ldots,a_t\rangle$, one considers the $S$-module $\mathcal{M} := M[x_1, \ldots, x_t]$, where $S := R[x_1, \ldots, x_t]$ is the polynomial ring in $t$ variables over $R$. The submodule $\mathcal{N}$ of $\mathcal{M}$ generated by all homogeneous $f(x_1, \ldots, x_t)\in \mathcal{M}$ such that $f(a_1, \ldots, a_t) \in N$ is a finitely generated $S$-module. Taking $k$ to be the maximum of the degrees of the generators of $\mathcal{N}$ works.

We then discussed and presented a proof of the following linear algebra-like statement.

Key Fact. Suppose $R$ is a commutative ring, $M$ is an $R$-module with $x_1, \ldots, x_n\in M$. If we have a system of equations in $M$:

\begin{align*} a_{11}x_1+\cdots +a_{1n}x_n &= 0\\ \vdots\quad\quad\quad\quad &= \vdots\\ a_{n1}x_1+\cdots + a_{nn}x_n &= 0, \end{align*}

with $a_{ij}\in R$. Let $A$ denote the coefficient matrix. Then $(\det A)\cdot x_i = 0$, for $1\leq i\leq t$.

We ended class with the following immediate consequence of the Key Fact.

Nakayama's Lemma, first version. Let $I\subseteq R$ be an ideal contained in the Jacobson radical. If $IM = M$, then $M = 0$.

Thursday, February 19

We began class with

Nakayama's Lemma, second version. Let $M$ be a finitely generated $R$-module, $N\subseteq M$ a submodule and $I\subseteq R$ an ideal contained in the Jacobson radical. If $M = N+IM$, then $M = N$.

We noted that one can interpret this latter statement as saying that if a set of elements in $M$ generate $M$ modulo an ideal contained in the Jacobson radical, then these elements generate $M$. This comment immediately gave rise to:

Consequences of Nakayama's Lemma. Let $(R, \mathfrak{m}, k)$ be a (Noetherian) local ring and $M$ a finitely generated $R$-module.

(i) Suppose $M = \langle x_1, \ldots, x_n\rangle$. Then $x_1, \ldots, x_n$ is a minimal generating set for $M$ if and only if the images of $x_1, \ldots, x_n$ in $M/\mathfrak{m} M$ form a basis for $M/\mathfrak{m} M$ as a vector space over $k$.
(ii) Suppose $M = \langle x_1, \ldots, x_n\rangle$ and $\pi : R^n\to M\to 0$ is the natural surjection taking the column vector $\begin{pmatrix} r_1\\\vdots\\r_n\end{pmatrix}$ to $r_1x_1+\cdots + r_nx_n$. Then $x_1, \ldots, x_n$ is a minimal generating set for $M$ if and only if for every column vector $v\in K$, $v$ has all of its entries in $\mathfrak{m}$.
(iii) Suppose $K$ is minimally generated by $v_1,\ldots, v_t$. Then we have an exact sequence of finitely generated $R$-modules $R^t\overset{f_1}\to R^n\overset{\pi}\to M\to 0$, in which we may think of $f_1$ as an $n\times t$ matrix with entries in $\mathfrak{m}$.
(iv) It follows that $M$ has a minimal free resolution of the form $\cdots \longrightarrow F_2 \overset{f_2}\longrightarrow F_1\overset{f_1}\longrightarrow F_0\overset{\pi}\to M\to 0,$ in which each $F_i$ is a finitely generated free $R$-module and each $f_i$ is a matrix with entries in $\mathfrak{m}$.

This was followed by

Krull's Intersection Theorem. Let $R$ be a Noetherian ring, $I\subseteq R$ an ideal and $M$ a finitely generated $R$-module. Set $N := \bigcap_{n\geq 1}I^nM$. Then there exists $a\in I$ such that $(1+a)\cdot N = 0$. In particular,

(i) If $I$ is contained in the Jacobson radical of $R$, then $\bigcap_{n\geq 1}I^nM = (0)$.
(ii) If $R$ is an integral domain, then $\bigcap_{n\geq 1}I^n = (0)$.

We then presented one of the most important theorems in Noetherian ring theory. But first we defined the height of a prime ideal: The height of the prime ideal $P$ is $n$ if there exists a chain of prime ideals $P_0\subsetneq \cdots \subsetneq P_n = P$, and no longer such chain exists. Here we allow for the height of a prime ideal to be infinite, but this cannot happen in a Noetherian ring. The dimension of $R$ is defined to be the maximum height of any prime ideal in the ring. There are Noetherian rings that have infinite dimension.

Krull's Principal Ideal Theorem. Let $R$ be a Noetherian ring, $\langle a\rangle \subseteq R$ a principal ideal and $P$ a prime ideal minimal over $\langle a\rangle$. Then $\textrm{height}(P) \leq 1$.

A key component of the proof of this theorem was the main theorem from the lecture of February 12: A commutative ring is Artinian if and only if it is Noetherian and every prime ideal is a maximal ideal.

We finished class by mentioning a consequence of Krull's principal ideal theorem, often called Krull's height theorem.

Krull's Generalized Principal Ideal Theorem. Let $R$ be a Noetherian ring, $I = \langle a_1, \ldots, a_n\rangle$ an ideal of $R$ and $P$ a prime ideal minimal over $I$. Then $\textrm{height}(P)\leq n$.

Tuesday, February 24

We began class by proving the generalized version of Krull's principal ideal theorem mentioned at the end of the previous lecture. We then mentioned the consequence that the defining ideal of a codimension $c$ affine variety in $\mathcal{A}^n_K$ needs at least $c$ generating polynomials. However, such a variety could be defined set-theoretically by $c$ equations. We also mentioned that in characteristic zero, it is not known if every algebraic curve is the set-theoretic intersection of two algebraic surfaces. Algebraically: If $P\subseteq K[x,y,z]$ is a height two prime ideal, are there $f, g\in K[x,y,z]$ such that $P$ is the nilradical of $\langle f, g\rangle$?

We then stated the following corollary of Krull's height theorem.

Corollary. Let $R$ be a Noetherian ring and $P$ a prime ideal of height $d$. Then there exist $a_1, \ldots, a_d\in P$ such that $P$ is minimal prime of the ideal $I := \langle a_1, \ldots, a_d\rangle$. Moreover, if $Q$ is any minimal prime of $I$, $Q$ also has height $d$.

The proof of the corollary required the following lemma, whose proof was deferred to a future lecture.

Prime Avoidance Lemma. Let $J, I_1, I_2, P_1, \ldots, P_n\subseteq R$ be ideals with $P_1, \ldots, P_n$ prime. If

J\subseteq I_1\cup I_2\cup P_1\cup \cdots \cup P_n,

then $J\subseteq I_1$ or $J\subseteq I_2$ or $J\subseteq P_j$, for some $1\leq j\leq n$.

This led to the important definition of a system of parameters in case $(R, \mathfrak{m})$ is local and $d$-dimensional, namely elements $a_1, \ldots, a_d\in \mathfrak{m}$ form a system of parameters if $\mathfrak{m}$ is the only prime ideal containing $\langle a_1, \ldots, a_d\rangle$, i.e., $R/\langle a_1, \ldots, a_d\rangle$ is zero-dimensional (and hence Artinian). We also noted the

Important Observation. Suppose $(R, \mathfrak{m})$ is a local ring and $a_1, \ldots, a_d$ is a system of parameters. Then for every $1\leq i\leq d$, $\dim (R/\langle a_1, \ldots, a_i\rangle) = d-i$.

We then gave the following key definition for this portion of the course.

Definition. Let $(R, \mathfrak{m})$ be a local ring having dimension $d$. Then $R$ is a regular local ring if $\mathfrak{m}$ can be generated by $d$ elements. Any $d$-generating set for $\mathfrak{m}$ is called a regular system of parameters.

We ended class with the following comments and proofs of parts (i), (ii).

Comments. Suppose $(R,\mathfrak{m})$ is a regular local ring with regular system of parameters $a_1, \ldots, a_d$.

(i) Each $a_i \in \mathfrak{m}\backslash \mathfrak{m}^2$.
(ii) For each $1\leq i\leq d$, $R/\langle a_1, \ldots, a_i\rangle$ is a regular local ring.
(iii) $R$ is an integral domain.
(iv) For $0\neq x\in \mathfrak{m}$, $R/\langle x\rangle$ is a regular local ring if and only if $x\in \mathfrak{m}\backslash \mathfrak{m}^2$.

Thursday, February 26

We began class by reviewing the definition of regular local ring, and re-stated Comments (i)-(iv) from the previous lecture. We also provided proof of items (ii) and (iv). We then gave a proof of the Prime Avoidance Lemma, as stated in the previous lecture. We also mentioned that there are a number of variations on the PAL, including the lovely result due to S. McAdam stating that the lemma is true without any of the ideals in the union being prime, if the ring $R$ contains an infinite field.

We then stated that our next immediate goal is to prove the following

Theorem. Let $(R, \mathfrak{m}, k)$ be a local ring. Then $R$ is a regular local ring if and only if $k$ admits a minimal finite free resolution, in which case the length of the resolution equals $\dim (R)$.

We then gave three examples of minimal finite free resolutions of $k$, for low dimensional regular local rings.

Examples. The following are minimal finite free resolutions. In fact, each case is an example of a Koszul complex. $(R,\mathfrak{m},k)$ is a regular local ring.

(i) $\mathfrak{m} = \langle x\rangle$, yields: $0\to R\overset{\cdot x}\longrightarrow R\to k\to 0$.
(ii) $\mathfrak{m} = \langle x,y\rangle$, yields: $0\to R\overset{\begin{pmatrix} -y\\x\end{pmatrix}}\longrightarrow R^2\overset{\begin{pmatrix} x & y\end{pmatrix}}\longrightarrow R \to k\to 0$.
(iii) $\mathfrak{m} = \langle x, y, z\rangle$, yields: $0 \rightarrow R \xrightarrow{\begin{pmatrix} z\\-y\\x\end{pmatrix}} R^3 \xrightarrow{\begin{pmatrix}-y & - z & 0\\x & 0 & -z\\0 & x & y\end{pmatrix}} R^3 \xrightarrow{\begin{pmatrix} x & y & z\end{pmatrix}} R \overset{\pi} \longrightarrow R/\mathfrak{m}\rightarrow 0.$

We proved (i), (ii) and left (iii) as an exercise.

We then looked at the following

Definition. Let $R$ be a commutative ring and $x := x_1, \ldots, x_n \in R$. Write $K(x_1, \ldots, x_n)$ or $K(\underline{x})$ to denote the Koszul complex on $x_1, \ldots, x_n$. $K(x)$ is the complex whose $r$th module is $K_r := R^{\binom{n}{r}}$, for $0 \leq r \leq n$ and is zero otherwise. For each $1 \leq i_1 < \cdots < i_r \leq n$, let $e_{i_1} \wedge \cdots \wedge e_{i_r}$ denote the standard basis element so that these basis elements are written in their standard order (and hence indexed 'lexicographically' by the multi-index). For each $r$, the boundary map $\partial : K_r\to K_{r-1}$ is defined by the equation

\partial(e_{i_1} \wedge \cdots \wedge e_{i_r}) = x_1 e_{i_2} \wedge \cdots \wedge e_{i_r} + \cdots + (-1)^{r+1} x_r e_{i_1} \wedge \cdots \wedge e_{i_{r-1}}.

We noted that it is easy to check that $\partial^2 = 0$, so we do have a complex (of finitely generated, free $R$-modules). We also noted (but did not prove) that $x_j\cdot \textrm{H}_i(K(\underline{x}))= 0$, for all $i, j$.

We ended class with a discussion and proof of the following

Proposition. Let $(R, \mathfrak{m}, k)$ be a local ring and $M$ a finitely generated $R$-module. Consider the start of two minimal free resolutions of $M$

\begin{CD} F_1 @>{\phi_1}>> F_0 @>{\pi}>> M @>>> 0\\ @. @. @VV{\alpha}V @.\\ G_1 @>{\phi_1'}>> G_0 @>{\pi'}>> M @>>> 0, \end{CD}

where $\alpha$ is an isomorphism. Taking $K := \textrm{ker} (\pi)$ and $L := \textrm{ker}(\pi')$, we have $K\cong L$.

Tuesday, March 3

After recalling the proposition presented at the end of the previous lecture, we presented the following corollary to the proposition.

Corollary. Let $(R, \mathfrak{m}, k)$ be a local ring and $M$ a finitely generated $R$-module. If $M$ has a minimal FFR of length $n$, then every minimal free resolution of $M$ has length $n$.

The proof of the Corollary was by induction on $n$. The base case $n = 0$ followed from the observation that if some kernel in a minimal free resolution is zero, then all subsequent modules in the resolution are zero. The corollary enabled us to make the following definition in the special case that $(R, \mathfrak{m}, k)$ is a local ring.

Definition. Suppose $(R, \mathfrak{m}, k)$ is a local ring and $M$ is a finitely generated $R$-module. Then $M$ has projective dimension $n$ if $M$ has a minimal FFR of length $n$. Here we allow $n$ to be infinite. We will write $\mathrm{pd}(M)$ to denote the projective dimension of $M$ over $R$.

We briefly, and informally, discussed the general concept of projective dimension, and noted that our definition above is consistent with the general definition, since finitely generated projective modules over a local ring are free. In fact, we showed how this followed from Nakayama's Lemma. We then presented the following proposition.

Proposition A. Let $(R, \mathfrak{m}, k)$ be a local ring and $M$ a finitely generated $R$-module. Suppose $x\in R$ is a non-zerodivisor on both $M$ and $R$. Set $R_1 := R/\langle x\rangle$. Then $\mathrm{pd}(M) = \mathrm{pd}_{R_1}(M/xM)$.

The proof followed from a slightly stronger result, namely if $\cdots \longrightarrow F_1\overset{\phi_1}\longrightarrow F_0\overset{\pi}\longrightarrow M\to 0$ is a minimal free resolution of $M$, then $\cdots \longrightarrow F_1/xF_1\overset{\phi_1'}\longrightarrow F_0/xF_0\overset{\pi'}\longrightarrow M/xM\to 0$ is a minimal free resolution of $M/xM$ over $R_1$, where the primed maps are just those maps from the original resolution reduced mod $xR$.

We ended the class by discussing the following proposition, and proving the base case $\mathrm{pd}(M) = 1$. We also noted how this proposition will play a key role in the homological characterization of regular local rings — namely the implication that if $R$ is a regular local ring of dimension $d$, then $\mathrm{pd}(k) = d$.

Proposition B. Let $(R,\mathfrak{m}, k)$ be a local ring and $M$ a finitely generated $R$-module with finite projective dimension. Suppose $x\in R$ satisfies the following conditions: $x$ is a non-zerodivisor in $R$, $xM = 0$, and $x\in \mathfrak{m}\backslash \mathfrak{m}^2$. Then $\mathrm{pd}_{R_1}(M) = \mathrm{pd}(M) -1$, for $R_1 := R/\langle x\rangle$.

Thursday, March 5

We began class by finishing the proof of Proposition B from the previous lecture. Two key points in the case $\mathrm{pd}(M) > 1$ were the following: Assuming that $0\to K\to F_0 \to M\to 0$ is the start of a minimal FFR of $M$ and $F_0 = R^n$, the first point was that just as in the proof of the base case, $xe_1, \ldots, xe_n$ are part of a minimal generating set for $K$. This is where the $x\in \mathfrak{m}\backslash \mathfrak{m}^2$ hypothesis is used. The second point was that we had to show $\mathrm{pd}(K/xF_0) = \mathrm{pd}(K)$. This was obtained by modifying the maps in the minimal FFR of $K$ (derived from the minimal FFR of $M$) as follows: Given the minimal FFR, with $F_1 = R^{s+n}$,

0\to F_n\overset{\phi_n}\longrightarrow F_{n-1} \overset{\phi_{n-1}}\longrightarrow \cdots \longrightarrow F_3\overset{\phi_3}\longrightarrow F_2\overset{\phi_2} \longrightarrow F_1 \overset{\phi_1}\longrightarrow K\to 0,

then

0\to F_n\overset{\phi_n}\longrightarrow F_{n-1} \overset{\phi_{n-1}}\longrightarrow \cdots \longrightarrow F_3\overset{\phi_3}\longrightarrow F_2\overset{\phi_2'} \longrightarrow F_1' \overset{\phi_1'}\longrightarrow K/xF_0\to 0,

is a minimal free resolution of $K/xF_0$, where $F_1' = R^s$, $\phi_1'$ takes the standard basis of $R^s$ to the images in $K/xF_0$ of the first $s$ generators of $K$ and $\phi_2'$ is the matrix $\phi_2$ with its last $n$ rows removed.

After finishing the proof of the proposition, as a final bit of preparation for the homological characterization of regular local rings, we presented the following special case of the Auslander-Buchsbaum formula.

Lemma. Let $(R, \mathfrak{m})$ be a local ring and suppose there exists $0\neq a\in R$ such that $a\cdot \mathfrak{m} = 0$. Then the only $R$-modules with a minimal FFR are free $R$-modules.

At last, the homological characterization of regular local rings:

Theorem. Let $(R, \mathfrak{m}, k)$ be a local ring with Krull dimension $d$. The following are equivalent.

(i) $R$ is a regular local ring.
(ii) Every finitely generated $R$-module has finite projective dimension.
(iii) $k$ has finite projective dimension.
(iv) $\mathrm{pd}(k) = d$.

Sketch of proof. The proof was by induction on $d$, with the base case easily handled by the previous lemma. For the inductive step, assuming (i), one applies Proposition A from the previous lecture taking the module to be $K$, the kernel of a minimal presentation of $M$ and $x\in \mathfrak{m}\backslash \mathfrak{m}^2$. (ii) clearly implies (iii). That (iii) implies (iv) follows by induction from Proposition B in the previous lecture taking the module to be $k$ and $x\in \mathfrak{m}\backslash \mathfrak{m}^2$. The proof of (iv) implies (i) is similar: taking the module to be $k$ in Proposition B, and $x\in \mathfrak{m}\backslash \mathfrak{m}^2$, by induction $R/\langle x \rangle$ is a regular local ring, so $R$ is a regular local ring.

In preparation for arriving at a proof that regular local rings are UFDs, we finished class with the following lemma, which used Krull's principal ideal theorem.

Proposition. Suppose $R$ is a Noetherian domain. Then $R$ is a UFD if and only if every height one prime is principal.

Tuesday, March 10

We began class by stating the following proposition which is key for the theorem that follows it. The proof of the proposition was delayed until later in the lecture.

Proposition. Let $(R, \mathfrak{m})$ be a local ring and $M$ a finitely generated $R$-module. If $M$ admits a FFR, then $M$ admits a minimal FFR.

We then stated and proved the following important property of regular local rings.

Theorem. Let $(R, \mathfrak{m}, k)$ be a regular local ring and $P\subseteq R$ be a prime ideal. Then $R_P$ is a regular local ring.

Proof. Since $R$ is a regular local ring, the $R$-module $R/P$ admits a (minimal) FFR. Thus, the $R_P$-module $(R/P)_P$ admits a FFR. Note that even if the FFR of $R/P$ over $R$ is minimal, the resolution localized at $P$ need no longer be minimal. By the Proposition, $(R/P)_P$ admits a minimal free resolution. Since $(R/P)_P$ is the residue field of $R_P$, it follows from the homological characterization of regular local rings that $R_P$ is a regular local ring. $\square$

We then noted the following

Remark. After the concept of regular local ring was defined in purely algebraic terms, it was assumed that if $R$ is a regular local ring, then $R_P$ would also be a regular local ring, based upon the validity of this statement in geometric contexts. It wasn't until regular local rings were characterized in homological terms that a proof of this fact was given. Regarding the need for the Proposition, we noted that the proof of the special case of the Auslander-Buchsbaum formula given prior to the theorem characterizing regular local rings required the existence of a minimal FFR.

We were then able to formally state and prove the main theorem of the first half of this course.

Major Theorem. Let $(R, \mathfrak{m})$ be a regular local ring. Then $R$ is a UFD.

Proof. By induction on $d = \dim R$. If $\dim R = 0$, $R$ is a field, and there is nothing to prove. Suppose $d > 0$. Take $x\in \mathfrak{m}\backslash \mathfrak{m}^2$ and set $S := \{1, x, x^2, \ldots\}$. Since $x$ is a prime element, it suffices to show that $R_S$ is a UFD, by Nagata's Lemma. By the proposition from the end of the previous lecture, it suffices to show that any height one prime in $R_S$ is principal. Such a prime is of the form $P_S$, for $P\subseteq R$ a height one prime not containing $x$. Now, let $Q\subseteq R_S$ be a maximal ideal, so that $Q = Q'_S$, for $Q'\subseteq R$ a prime ideal disjoint from $S$. Note that $(R_S)_Q = R_{Q'}$, so $(R_S)_Q$ is a regular local ring of dimension less than $d$. Thus $(R_S)_Q$ is a UFD, so $P_Q$ is principal. Thus $P$ is a locally principal ideal of $R_S$. On the other hand, $P = P'_S$, for $P'\subseteq R$ a prime ideal. $P'$ has a FFR as an $R$-module, and upon localizing at $S$, we see that $P$ has a FFR as an $R_S$-module. Thus, $P$ is a locally principal ideal admitting a FFR, therefore $P$ is principal by the main theorem from the lecture of February 10. Thus, every height one prime of $R_S$ is principal, so $R_S$ is a UFD, and therefore $R$ is a UFD. $\square$

We followed this by giving a proof of the Proposition above. The proof was by induction on the length of the FFR of $M$ and relied upon the following lemma.

Lemma. Suppose the module $K\oplus L$ has a FFR, where $L$ is a finite free $R$-module. Then $K$ has a FFR.

We ended with a brief discussion of some of the topics we'll cover next. These included: (i) The two forms of Hilbert's Nullstellensatz; (ii) A refinement of (i) stating that if $P$ is a prime ideal in the polynomial ring $R$ in finitely many variables over an algebraically closed field of characteristic zero, then $P$ is the intersection of the maximal ideals $M\subseteq R$ such that $(R/P)_M$ is a regular local ring; (iii) The Zariski-Nagata theorem, which states that if $P\subseteq R$ are as in (ii), then the $n$th symbolic power of $P$ is the set of polynomials whose partial derivatives of order less than $n$ vanish on the algebraic variety defined by $P$.

Thursday, March 12

Today's lecture was a brief interlude devoted to a discussion of associated primes and regular sequences. Throughout $R$ will denote a Noetherian ring and $M$ a non-zero $R$-module. For a non-zero element $x\in M$, we write $(0:x)$ or $(0:_R x)$ to denote the annihilator of $x$, i.e., $(0:x) = \{r\in R\ |\ rx = 0\}$. We began with

Basic Definitions and Properties.

1. A prime ideal $P\subseteq R$ is an associated prime of $M$ if $P = (0:x)$, for some (non-zero) $x\in M$. We write $\mathrm{Ass}(M)$ or $\mathrm{Ass}_R(M)$ to denote the associated primes of $M$.
2. Suppose $P$ is an ideal maximal among ideals of the form $(0:x)$, for $0\not= x\in M$. Then $P\in \mathrm{Ass}(M)$.
3. $\mathrm{Ass}(M)$ is non-empty (since $R$ is Noetherian and $M\not= 0$).
4. $r\in R$ is a zero-divisor on $M$ if and only if $r\in P$, for some $P\in \mathrm{Ass}(M)$. Thus the set of zero-divisors on $M$ equals the union of the associated primes of $M$.
5. Let $S\subseteq R$ be multiplicatively closed and $P\subseteq R$ a prime ideal disjoint from $S$. Then $P_S\in \mathrm{Ass}_{R_S}(M_S)$ if and only if $P\in \mathrm{Ass}(M)$.
6. Suppose the prime ideal $P\subseteq R$ is minimal over $(0:x)$, for $x\in M$. Then $P\in \mathrm{Ass}(M)$. Moreover, if $P$ is minimal over the $\mathrm{ann}(M)$, then $P\in \mathrm{Ass}(M)$. In particular, if $P$ is minimal over the ideal $I$, $P\in \mathrm{Ass}(R/I)$.

We then proceeded to prove the following propositions.

Proposition A. Suppose $0\to L\to M\to N\to 0$ is a short exact sequence of $R$-modules. Then $\mathrm{Ass}(M)\subseteq \mathrm{Ass}(L)\bigcup \mathrm{Ass}(N)$.

Proposition B. If $R$ is a Noetherian ring and $M$ a finitely generated $R$-module, then $\mathrm{Ass}(M)$ is finite.

The proof of Proposition B was by induction on the number of generators of $M$, which easily reduced the statement to the case $M$ is cyclic (by using Proposition A). Thus, it sufficed to show that $R/I$ has finitely many associated primes. For this, we noted that an ideal $I$ maximal with respect to the property that $\mathrm{Ass}(R/I)$ is infinite must be irreducible, i.e., $I$ is not the intersection of two proper ideals. One then invokes the following key theorem (with proof), which provides the necessary contradiction.

Theorem. Let $R$ be a Noetherian ring and $I\subseteq R$ an irreducible ideal. Then $\mathrm{Ass}(R/I) = \{P\}$ for some prime ideal $P\subseteq R$. Moreover, whenever $ab\in I$, then $a\in I$ or $b\in \sqrt{I}$.

This led to the following important definitions.

Definition. Let $I\subseteq R$ be an ideal.

(i) The ideal $Q\subseteq R$ is said to be a primary ideal if $ab\in Q$ implies $a\in Q$ or $b\in \sqrt{Q}$. Thus, by the theorem above and its proof, if $Q\subseteq R$ is a primary ideal, $\mathrm{Ass}(R/Q) = \{P\}$, for some prime ideal $P\subseteq R$, and $P = \sqrt{Q}$. In this case we say $Q$ is a $P$-primary ideal.
(ii) A primary decomposition of $I$ is an intersection $I = Q_1\cap \cdots \cap Q_r$, where each $Q_i$ is $P_i$-primary for $P_i = \mathrm{Ass}(R/Q_i)$.
(iii) The primary decomposition in (ii) is said to be reduced or irredundant if $I$ is not the intersection of a proper subset of $Q_1, \ldots, Q_n$ and the primes $P_1, \ldots, P_n$ are distinct.

We then quickly observed that if an ideal has a primary decomposition, it has an irredundant primary decomposition.

This then led to the following important

Primary Decomposition Theorem. Let $R$ be a Noetherian ring and $I\subseteq R$ an ideal. Then $I$ has an irredundant primary decomposition, $I = Q_1\cap \cdots \cap Q_n$. Moreover, if $\sqrt{Q_i} = P_i$, for $1\leq i\leq n$, then $\mathrm{Ass}(R/I) = \{P_1, \ldots, P_n\}$.

Proof. By the theorem above, it suffices to show that $I$ can be written as a finite intersection of irreducible ideals. The proof of this is the following classic argument due to E. Noether: If this were false, one could choose $I$ maximal with respect to the property that it cannot be written as a finite intersection of irreducible ideals. Then $I$ is not an irreducible ideal, so $I = K\cap L$ for ideals properly containing $I$. But then $K, L$ are intersections of finitely many irreducible ideals, which immediately shows that $I$ is a finite intersection of irreducible ideals, which is a contradiction. Thus, every ideal in a Noetherian ring is a finite intersection of irreducible ideals and hence has an irredundant primary decomposition.

For the second statement, first suppose $P_1 = \sqrt{Q_1}$. We may localize at $P_1$. Find $z\in Q_2\cap \cdots \cap Q_n\backslash Q_1$, which we can do because the decomposition is irredundant. Now, $P_1^h\subseteq Q_1$ for some $h\geq 1$, thus $P_1^hz\subseteq I$. Choose $h$ least with this property. Then $tz\not\in I$, for some $t\in P_1^{h-1}$. But then $P_1(zt)\subseteq I$, so $P_1$ consists of zero-divisors modulo $I$, i.e., $P_1\subseteq Q$, for some $Q\in \mathrm{Ass}(R/I)$. Since $P_1$ is prime, $P_1 = Q$, so $P_1\in \mathrm{Ass}(R/I)$. We may repeat the argument for $P_2, \ldots, P_n$, so $\{P_1, \ldots, P_n\}\subseteq \mathrm{Ass}(R/I)$.

Now take $P\in \mathrm{Ass}(R/I)$. Then we can write $P = (I:x)$, for some $x\not\in I$. We then have

P = (I:x) = (Q_1\cap \cdots \cap Q_n : x) = (Q_1:x)\cap \cdots \cap (Q_n:x),

so $P = (Q_i:x)$, for some $1\leq i\leq n$. This shows $P\in \mathrm{Ass}(R/Q_i) = \{P_i\}$, so $P = P_i$, since $P_i$ is minimal over $Q_i$. Thus, $\mathrm{Ass}(R/I)\subseteq \{P_1, \ldots, P_n\}$, and the proof is complete. $\square$