## Properties of Noetherian modules

Suppose $A$ is a commutative ring with $1$. We say that $A$-module module $M$ is Noetherian if every ascending chain of submodules in $M$ eventually terminates. This is similar to how we define a ring to be Noetherian (a commutative ring $A$ is Noetherian if every ascending chain of ideals in $A$ eventually terminates).

In “Undergraduate Commutative Algebra” by Miles Reid, the following proposition is proved (page 53), from which some important properties of Noetherian modules can be derived.

Proposition. Let $0\longrightarrow L\overset{\alpha}{\longrightarrow} M\overset{\beta}{\longrightarrow} N\longrightarrow 0$ be a short exact sequence of $A$-modules. Then,

$\displaystyle M\textrm{ is Noetherian } \Longleftrightarrow L \textrm{ and } N \textrm{ are Noetherian}$

I have written up proof of this here. (I have filled in some details, where Miles have left to the reader).

Here are some consequences:

(1) If $M_i$ are Noetherian modules (for $i=1, 2, ..., r$), then $\bigoplus_{i=1}^{r} M_i$ is Noetherian.

(2) If $A$ is a Noetherian ring, then $A$-module $M$ is Noetherian if and only if it is finitely generated $A$-module.

(3) If $A$ is a Noetherian ring, $M$ is a finitely generated $A$-module, then any subdmoule $N\subseteq M$ is again finitely-generated $A$-module.

Proof. (1) We recall that direct sum $M=M_1\oplus M_2$ can be realized as an exact sequence $0\longrightarrow M_1\overset{\alpha}{\longrightarrow} M\overset{\beta}{\longrightarrow} M_2\longrightarrow 0$, where $\alpha(m_1)=(m_1, 0)$, and $\beta((m_1, m_2))=m_2$. Applying the Proposition, we obtain that $M= M_1\oplus M_2$ is Noetherian module. By simple induction, this is extended to direct sum of any finite number of Noetherian modules.

(2) If $A$-module $M$ is finitely generated, then we have a surjection $A^{m}\to M$ for some positive integer $m$, where $A^{m}$ is a free module of rank $m$. In other words, we have an exact sequence $A^{m}\overset{\beta}{\longrightarrow} M\longrightarrow 0$. From (1), we know $A^{m}$ is Noetherian. By the one of the directions ($\Rightarrow$) of the Proposition, we obtain that $M$ is also Noetherian. Conversely, if $M$ is a Noetherian $A$-module, it is clear that $M$ is finitely generated $A$-module, for otherwise we would obtain ascending chain of submodules

$(m_1)\subsetneq (m_1, m_2)\subsetneq (m_1, m_2, m_3)\subsetneq \cdots$

that does not terminate. Here $(m_1, m_2, ..., m_k)$ is the submodule generated by elements $m_1, m_2, ..., m_k$.

(3) Since $M$ is finitely generated $A$-module, we get from (2) that $M$ is a Noetherian $A$-module. It is clear that any submodule of a Noetherian module is again Noetherian module. So a submodule $N\subseteq M$ is Noetherian. Applying (2) again, we get that $N$ is finitely-generated $A$-module, as desired.

## Idempotent elements outside the nilradical

Here is a cute exercise in Introduction to Commutative Algebra by Atiyah & Macdonald (A&M), that I recently solved, and felt like writing it up. The problem reads as follows:

Exercise 6 (Chapter 1). A ring $A$ is such that every ideal not contained in the nilradical contains non-zero idempotent (that is, an element $e$ such that $e^2=e\neq 0$). Prove that the nilradical and Jacobson radical of $A$ are equal.

Before proceeding to solution, let’s define what each of these terms mean. Even before that, let me note that throughout the word “ring” will mean commutative ring with 1 (multiplicative identity). This is the assumption made in the beginning of A&M. The nilradical $\mathfrak{N}$ of a ring $A$ is the set of all nilpotent elements of $A$ (an element $x\in A$ is called nilpotent if $x^{k}=0$ for some positive integer $k$.) Proposition 1.8 in A&M gives the following characterization:

Proposition 1.8. The nilradical of $A$ is the intersection of all the prime ideals of $A$.

On the other hand, Jacobson radical $\mathfrak{R}$ of a ring $A$ is defined to be the intersection of all maximal ideals of $A$. Proposition 1.9 in A&M gives the following characterization:

Proposition 1.9. $x\in\mathfrak{R} \Longleftrightarrow 1-xy$ is a unit in $A$ for all $y\in A$.

Using these two results, we can proceed to the solution of the exercise above.

Solution of Exercise 6 (Chapter 1). Since every maximal ideal is a prime ideal, and nilradical is the intersection of all prime ideals, it is clear that nilradical is contained in the intersection of all maximal ideals, that is, nilradical is contained in Jacobson radical. This proves one of the inclusions (note that this inclusion, namely $\mathfrak{N}\subset\mathfrak{R}$ holds generally in any ring). Now we need to show that Jacobson radical is subset of the nilradical. Let $x\in\mathfrak{R}$. We want to show that $x\in\mathfrak{N}$. Assume, to the contrary, $x\notin\mathfrak{N}$. By assumption, the ideal generated by $x$, namely the principal ideal $x=\{xa \ : \ a\in A\}$ is not contained in the nilradical. By hypothesis, this means that there exists a nonzero idempotent element in $(x)$. Hence, there exists $a\in A$ such that $(xa)^2=xa\neq 0$. Now since $x\in\mathfrak{R}$, using Proposition 1.9 we get that $1-xa$ is a unit. But $1-xa=1-(xa)^2=(1-xa)(1+xa)$. Since $1-xa$ is a unit, we can multiply this equation from the left by its inverse $(1-xa)^{-1}$ to get $1=1+xa$, or equivalently, $xa=0$. This contradicts $xa\neq 0$, and so we conclude that $x\in \mathfrak{N}$, as desired. This completes the proof that the nilradical and Jacobson radical of ring $A$ coincide.

## Inverse image of a prime ideal is a prime ideal

Here is one of those problems that invariably shows up in homework for first course on ring theory: Assume $A$ and $B$ are commutative rings with 1. Prove that if $f: A\to B$ is a ring homomorphism, and $I$ is a prime ideal in $B$, then $f^{-1}(I)$ is a prime ideal in $A$.

Showing that $f^{-1}(I)$ is an ideal in $A$ is routine and I shall omit the details. To show that $f^{-1}(I)$ is a prime ideal in $A$, one way to proceed is to apply the definition of a prime ideal: if $xy\in f^{-1}(I)$, then we want to show that $x\in f^{-1}(I)$ and $y\in f^{-1}(I)$. I think this is the kind of approach that seems natural for a beginner in ring theory (like me). But recently, reading “Introduction to Commutative Algebra” by Atiyah & Macdonald, I learnt the following solution, which seems more insightful. The idea is as follows:

To show that $f^{-1}(I)$ is a prime ideal in $A$, it suffices to prove that $A/f^{-1}(I)$ is an integral domain. We claim that $A/f^{-1}(I)$ is isomorphic to a subring of $B/I$, and this would finish the proof immediately for the following reason: We know $B/I$ is an integral domain (because $I$ is a prime ideal in $B$) and every subring of an integral domain is again an integral domain. So we need to exhibit an injective ring homomorphism $\varphi: A/f^{-1}(I)\to B/I$. This map is explicitly given by

$\displaystyle \varphi(a+f^{-1}(I))=f(a)+I$

For every $a, a'\in A$, we have

$\varphi((a+f^{-1}(I))+(a'+f^{-1}(I)))=\varphi(a+a'+f^{-1}(I))=f(a+a')+I=f(a)+f(a')+I=(f(a)+I)+(f(a')+I)=\varphi(a+f^{-1}(I))+\varphi(a'+f^{-1}(I))$

and

$\varphi((a+f^{-1}(I))(a'+f^{-1}(I)))=\varphi(aa'+f^{-1}(I))=f(aa')+I=f(a)f(a')+I=(f(a)+I)(f(a')+I)=\varphi(a+f^{-1}(I))\varphi(a'+f^{-1}(I))$

which shows that $f$ is a ring homomorphism. We will now show that $f$ is injective. Assume $\varphi(a)=\varphi(a')$ for some $a, a'\in A$. Then, by definition,

$f(a)+I=f(a')+I$

or equivalently,

$f(a-a')=f(a)-f(a')\in I$

which implies that $a-a'\in f^{-1}(I)$ so that $a+f^{-1}(I)=a'+f^{-1}(I)$. Thus, $\varphi$ is injective. We have proved that $\varphi: A/f^{-1}(I)\to B/I$ is an injective ring homomorphism. As a result, $A/f^{-1}(I)$ is isomorphic to a subring of $B/I$, and so is an integral domain, proving that $f^{-1}(I)$ is prime ideal in $A$.

## Bounded Variation

It is natural to ask whether we can characterize those functions from interval $[a,b]$ to $\mathbb{R}$ that can be written as a difference of two increasing functions. One such characterization involves studying variation of function across the subintervals of $[a,b]$. Here are my notes from class that explains this concept.

## Euler Summation Formula

I was recently asked to prove Euler Summation Formula in an analysis homework. I guess the goal of the exercise was to get familiar with manipulating integrals. It turns out that Euler Summation Formula has an application in analytic number theory, and indeed this theorem appears in Tom Apostol’s book. If you are interested, here is the proof.

## Hilbert Basis Theorem

David Hilbert is considered to be one of the greatest mathematicians of modern times. Attached to his name are numerous theorems, and his list of 23 problems presented in International Congress of Mathematics in 1900 has been very influential. I recently learned the following elegant proof of Hilbert Basis Theorem in an algebra course. Here are my notes taken in class that leads to the statement and proof of Hilbert Basis Theorem. According to Rotman’s Advanced Modern Algebra this proof is due to Sarges. The original proof of Hilbert is different, and apparently more complicated. In the same book, Rotman gives the following anectode which I found interesting. Here it goes:

The following anectode is well known. Around 1890, Hilbert proved the famous Hilbert Basis Theorem, showing that every ideal in $\mathbb C[x_1,x_2,...,x_n]$ is finitely generated. As we will see, the proof is nonconstructive in the sense that it does not give an explicit set of generators of an ideal. It is reported that when P. Gordan, one of the leading algebraists of the time, first saw Hilbert’s proof, he said, “This is not Mathematics, but theology!” On the other hand, Gordan said, in 1899 when he published a simplified proof of Hilbert’s Theorem, “I have convinced myself that theology also has its advantages.”