ac.commutative-algebra Archives

This post is almost verbatim from Undergraduate Algebraic Geometry § 0.3.

The specific flavour of algebraic geometry comes from the use of only polynomial functions (together with rational functions); to explain this, if $U \subseteq \mathbb R$ is an open interval, one can reasonably consider the following rings of functions on $U$ :

$C^0(U)$ : all continuous functions $f: U \to \mathbb R$ ;
$C^\infty(U)$ : all smooth functions (that is, differentiable to any order);
$C^\omega(U)$ : all analytic functions (that is, convergent power series);
$\mathbb R[X]$ : the polynomial ring, viewed as a polynomial function on $U$ .

There are of course inclusions $\mathbb R[X] \subset C^\omega(U) \subset C^\infty(U) \subset C^0(U)$ .

These rings of functions correspond to some of the important categories of geometry: $C^0(U)$ to the topological category, $C^\infty(U)$ to the differentiable category (differentiable manifolds), $C^\infty(U)$ to real analytic geometry, and $\mathbb R[X]$ to algebraic geometry.

The point I want to make here is that each of these inclusion signs represents an absolutely huge gap, and that this leads to the main characteristics of geometry in the different categories. Although it’s not stressed very much in school and first year university calculus, any reasonable way of measuring $C^0(U)$ will reveal that the differentiable functions have measure $0$ in the continuous functions. The gap between $C^\omega(U)$ and $C^\infty(U)$ is exemplified by the behaviour of $\exp(1-1/x^2)$ , the standard function which is differentiable infinitely often, but for which the Taylor series (at $0$ ) does not converge to the function. Using this one can easily build a bump function that is $1$ on a compact set and vanishes outside any given open set containing the compact set. In contrast, an analytic function on $U$ extends (as a convergent power series) to an analytic function of a complex variable on a suitable domain in $\mathbb C$ , so that (using results from complex analysis) if $f \in C^\infty(U)$ vanishes on a real interval, it must vanish identically. This is a kind of “rigid” property which characterises analytic geometry as opposed to differential topology.

References

M. Reid, Undergraduate Algebraic Geometry, § 0.3

The post is also available as pdf.

Cayley-Hamilton theorem is usually presented in a linear algebra context, which says that an $F$ -vector space endomorphism (and its matrix representation) satisfies its own characteristic polynomial. Actually, this result can be generalised a little bit to all modules over commutative rings (which are certainly required to be finitely generated for determinant to make sense). There are many proofs available, among which is the bogus proof of substituting the matrix into the characteristic polynomial $\det (x \cdot I - A)$ and obtaining

$\det (A \cdot I - A) = \det (A - A) = 0.$

The reason it doesn’t work is because the product $x \cdot I$ in the characteristic polynomial is multiplication of a matrix by a scalar, while the product assumed in the bogus proof is matrix multiplication. In this post I will show that the intuition behind this “substitution-and-cancellation” is correct (in a narrow sense), and the proof is rescuable by using a trick, which in itself is quite useful.

I will develop the theory using language of rings and modules but if you don’t understand that, feel free to substitute “fields” and “vector spaces” in place.

Let $A$ be a commutative ring and $M$ be an $A$ -module generated by $\{m_1, \dots, m_n\}$ . Note that $M$ is naturally an $\End(M)$ -module and for all $f \in \End(M)$ , write $[f] \in \matrixring_n(A)$ for its representation with respect to the generators above, i.e. $f(m_i) = \sum_j [f]_{ij}m_j$ . In particular, there is a ring homomorphism $\mu: A \to \End(M), a \mapsto a \cdot -$ sending an element to its multiplication action. Let $A' = \mu(A)$ .

There is a technical remark to make: later we will use determinant of matrices over $\End(M)$ , which is non-commutative. However, throughout the discussion we are concerned with only one endomorphism $\varphi$ (besides multiplication, of course) so we can restrict the scalars to $A'[\varphi]$ , a subring contained in the centre of $\End(M)$ .

Given a module endomorphism $\varphi: M \to M$ , its characteristic polynomial is defined to be

$\chi_{[\varphi]}(x) = \det (x \cdot I - [\varphi]) \in A[x]$

where $I$ is the $n \times n$ identity matrix and the product $x \cdot I$ is multiplication of a matrix by a scalar. Note that $\chi$ is generator dependent in general. We have

Cayley-Hamilton Theorem

$\chi_{[\varphi]}(\varphi) = 0.$

Note that this is a relation of endomorphisms with coefficients in $A$ .

Proof: Let $[\varphi]_{ij} = a_{ij}$ and view $M$ as an $A'[\varphi]$ -module. Since

$\varphi m_i = \sum_j a_{ij}m_j,$

we have

(*) $\begin{equation*} \sum_j \underbrace{(\varphi \delta_{ij} - a_{ij})}_{\Delta_{ij}} m_j = 0 \end{equation*}$

with

$\Delta = \varphi \cdot I - N \in \matrixring_n(A'[\varphi]).$

Again, the multiplication is by scalar $\varphi$ , viewed as an element of the ring $\End(M)$ .

Claim that if $\det \Delta = 0 \in A'[\varphi]$ then we are done: consider the ring homomorphism

$\begin{align*} A[x] &\to \End(M) \\ x &\mapsto \varphi \end{align*}$

which maps $\chi_{[\varphi]}(t) \mapsto \chi_{[\varphi]}(\varphi) = \det \Delta$ since $\det$ is a polynomial function. So done.

To show this, recall that

$(\adj \Delta) \cdot \Delta = \det \Delta \cdot I \in \matrixring_n(A'[\varphi])$

where multiplication on the left is between matrices. Let $(\adj \Delta)_{ij} = b_{ij}$ . Then multiply $(*)$ by $b_{ki}$ and apply the identity,

$\sum_{i, j} (b_{ki} \Delta_{ij}) m_j = \sum_j (\det \Delta \delta_{kj}) m_j = (\det \Delta) m_k = 0.$

so $\det \Delta = 0$ as required.

If you feel that little work is done in the proof and suspect it might be tautological somewhere (which I had when I first saw this proof), go through it again and convince yourself it is indeed a bona fide proof. There are two tricks used here: firstly we extend the scalars by recognising $M$ as an $A'[\varphi]$ ring so that action by $\varphi$ becomes multiplication. This is essentially since it allows us to treat $\varphi$ and scalar multiplication by $A$ on an equal footing. Secondly, the ring homomorphism $A[x] \to A'[\varphi]$ substitutes $\varphi$ in place of $x$ . Aha, the intuition in the bogus proof is correct, but we need a little extra work to sort out the notation to express precisely what we mean.

The key idea in the proof, sometimes called the determinant trick, has many applications in commutative algebra:

Let $M$ be an $A$ -module generated by $n$ elements and $\varphi: M \to M$ a homomorphism. Suppose $I$ is an ideal of $A$ such that $\varphi(M) \subseteq IM$ , then there is a relation

$\varphi^n + a_1 \varphi^{n - 1} + \dots + a_{n - 1} \varphi + a_n = 0$

where $a_i \in I^i$ for all $i$ .

Proof: Let $\{m_1, \dots, m_n\}$ be a set of generators of $M$ . Since $\varphi(m_i) \in IM$ , we can write

$\varphi m_i = \sum_j a_{ij}m_j$

with $a_{ij} \in I$ . Multiply

$\sum_j \underbrace{(\varphi \delta_{ij} - a_{ij})}_{\Delta_{ij}} m_j = 0$

by $\adj \Delta$ , we deduce that $(\det \Delta) m_j = 0$ so $\det \Delta = 0 \in \End(M)$ . Expand.

An immediate corollary is Nakayama’s Lemma, which alone is an important result in commutative algebra:

Nakayama’s Lemma

If $M$ is a finitely generated $A$ -module and $I \ideal R$ is such that $M = IM$ then there exists $x \in A$ such that $x - 1 \in I$ and $xM = 0$ .

Proof: Apply the trick to $\id_M$ . Since $\id_M^i = \id_M$ and $a_n = a_n\id_M$ , we get

$\left(1 + \sum_{i = 1}^n a_i\right) \id_M = 0.$

We use the result to prove a rather interesting fact about module homomorphism:

Let $M$ be a finitely generated $A$ -module. Then every surjective module homomorphism on $M$ is also injective.

Proof: Let $\varphi: M \to M$ be surjective. Let $M$ be an $A'[\varphi]$ module and $I = (\varphi) \ideal A'[\varphi]$ . Then $M = IM$ by surjectivity of $\varphi$ . Thus by Nakayama’s Lemma, there exists $x = 1 + \varphi\psi$ , $\psi \in A'[\varphi]$ such that $(1 + \varphi\psi)M = 0$ , i.e. for all $m \in M$ , $(1 + \varphi\psi)m = 0$ . It follows that $\varphi^{-1} = -\psi$ .