Proving the Canonical Basis Is a Basis

Q: Why is spanning not enough?

Because a spanning set can be redundant. The whole point of the proposition is that the coefficients in ∑ c_αβ x^α ∂^β are determined by the operator, which is what allows equality to be decided and the degree to be defined. In characteristic p the same spanning argument works and the same set fails to be a basis, which shows spanning cannot imply independence.

Q: Does the order of the variables inside x^α matter?

No. The x_i commute among themselves and the ∂_i commute among themselves, so x^α and ∂^β are unambiguous. Only the relative position of the x block and the ∂ block matters.

Q: Is ∂^β x^α also a basis?

Yes, by the same argument run in the other direction. But the coefficients differ. In A_1 , the operator ∂ x has canonical form x∂ + 1 in one convention and the single term ∂ x in the other, so any statement about coefficients must name its convention.

Q: Where exactly does the proof use that K has characteristic zero?

In one place only: the conclusion D(x^σ) = σ! ∑ _α c_ασ x^α ≠ 0 requires σ! to be invertible in K . If char K = p and some entry of σ is at least p , then σ! = 0 and the argument collapses - correctly so, because the statement is then false.

Q: Does the proof need K to be algebraically closed or infinite?

Neither. Only the characteristic matters. Every field of characteristic zero is infinite anyway, but nothing beyond σ! ≠ 0 and the independence of monomials in K[X] is used.

Q: How do I actually put a given expression into canonical form?

Repeatedly rewrite ∂_i x_i as x_i∂_i + 1 and swap commuting pairs, as in the computational notes; or, for a product ∂^β x^α , apply the closed formula (1.8) in one step. Both give the same answer, because the answer is unique.

Q: What does the result say about the action on functions other than polynomials?

The proposition is proved using the action on K[X] , but faithfulness transfers to any module containing K[X] compatibly - for instance the ring of holomorphic functions over C . It does not transfer to arbitrary modules: plenty of A_n -modules are not faithful, since a module can be annihilated by a left ideal without being annihilated by a two-sided one.

Overview

The Weyl algebra $A_n$ is defined as the subalgebra of $End_K(K[X])$ generated by the multiplication operators $x_1, … ,x_n$ and the partial derivatives $∂_1, … ,∂_n$ , where $K[X] = K[x_1, … ,x_n]$ and $K$ has characteristic zero. That definition tells you what the elements of $A_n$ are - finite $K$ -linear combinations of products of generators - but it does not tell you when two such expressions denote the same operator. Since $A_n$ is not commutative, the question is not idle: $x∂$ and $∂ x$ are different operators.

The answer is the canonical basis. Every element of $A_n$ can be written as $∑ c_αβ, x^α ∂^β$ with the variables to the left of the derivatives, and that expression is unique. Two operators are equal exactly when the coefficient families $(c_αβ)$ agree, so equality in $A_n$ becomes a finite comparison of scalars. That is what makes canonical form the working representation of the algebra, on paper and in software.

This page proves the statement. The proof has two halves of very different character. Spanning is a rewriting argument: the relation $[∂_i, f] = ∂ f/∂ x_i$ lets you move each $∂$ to the right past each $x$ at the cost of a term of strictly smaller degree, and an induction on degree finishes it. Independence is where the content sits, and it is proved by exploiting the fact that elements of $A_n$ are operators: given a non-zero combination, one constructs a single monomial $x^σ$ on which it does not vanish.

The independence half also quietly settles something else, and it is worth isolating. It says that the action of $A_n$ on $K[X]$ is faithful in a very strong, degree-by-degree way. Nothing in the definition of $A_n$ as a subalgebra of an endomorphism ring guarantees that the obvious spanning set is not badly redundant; in positive characteristic it is redundant, because $∂_i^,p$ annihilates every polynomial. Characteristic zero is used in exactly one place in the argument, and this page points at it.

Definition

Fix notation first. A multi-index is an element $α = (α_1, … ,α_n) ∈ N^n$ ; it has length $|α| = α_1 + … + α_n$ and factorial $α! = α_1! ⋅ s α_n!$ . Write $x^α = x_1^α_1 ⋅ s x_n^α_n$ and $∂^β = ∂_1^β_1 ⋅ s ∂_n^β_n$ . A pair $(α,β)$ is itself a multi-index in $N^2n$ , of length $|α| + |β|$ . The conventions are set out in full on the multi-index page.

Canonical basis and canonical form

The canonical basis of $A_n$ is the set

B = , x^α ∂^β ;:; α, β ∈ N^n ,.

An operator is in canonical form when it is written as a finite $K$ -linear combination of elements of $ℬ$ .

The canonical basis is a basisCoutinho (1.2.1)

Let $K$ be a field of characteristic zero. Then $ℬ$ is a basis of $A_n$ as a vector space over $K$ : every element of $A_n$ is a finite $K$ -linear combination of the $x^α∂^β$ , and the coefficients in such a combination are uniquely determined.

Note

Uniqueness is the substantive claim. Spanning would hold for any algebra generated by the same symbols subject to the same relations; independence is a statement about this particular algebra of operators, and it is what pins the $K$ -dimension of the degree- $≤ m$ part of $A_n$ to $m+2n2n$ rather than merely bounding it.

Core Concepts

Two ideas carry the proof, and it helps to name them before the calculations start.

Straightening: a rewriting system that terminates

Any product of generators can be viewed as a word in the letters $x_i$ and $∂_i$ . The relations of the algebra say that swapping two adjacent letters either changes nothing (for $x_ix_j$ , $∂_i∂_j$ , and $∂_i x_j$ with $i ≠ j$ ) or changes the word into the swapped word plus one strictly shorter word (for $∂_i x_i$ , where the correction is the constant $1$ ). So repeatedly moving $x$ 's leftwards past $∂$ 's must stop: each swap either reduces the number of inversions at the same length, or produces a shorter word, and there are only finitely many words of bounded length. This is a rewriting argument, and the honest statement of what it gives is spanning, not uniqueness.

Separating operators by testing them on monomials

To show a non-zero combination $D = ∑ c_αβ x^α ∂^β$ is not the zero operator, it suffices to exhibit one polynomial $f$ with $D(f) ≠ 0$ . The right choice is a monomial $x^σ$ where $σ$ has the shortest length among the $β$ 's that actually occur. Why that choice works is a triangularity phenomenon: $∂^β$ kills $x^σ$ whenever $β$ is not componentwise below $σ$ , and among the $β$ with $|β| ≥ |σ|$ the only survivor is $β = σ$ itself, which returns the non-zero scalar $σ!$ . Terms with $|β| < |σ|$ would survive, which is why they are excluded by the minimality of $σ$ .

Where characteristic zero enters

Exactly one step needs it: the scalar $σ!$ must be non-zero in $K$ . If $K$ has characteristic $p$ and some $σ_i ≥ p$ , then $σ! = 0$ in $K$ , and indeed $∂_i^,p$ is the zero operator on $K[X]$ . The set $ℬ$ is then not independent, and the proposition is false as stated. This is a property of the operator algebra, not of the relations; see the section on limits of validity.

Construction and Proof

Throughout, $K$ has characteristic zero and $f, g$ denote polynomials in $K[X]$ .

Step 0: the commutator with a polynomial

Leibniz as a commutator

For every $f ∈ K[X]$ and every $i$ , the operator identity $[∂_i, f] = ∂ f/∂ x_i$ holds in $A_n$ , where $f$ denotes multiplication by $f$ .

Proof

Apply both sides to an arbitrary $g ∈ K[X]$ . The product rule gives $∂_i(fg) = (∂ f/∂ x_i) g + f, ∂_i(g)$ , that is $∂_i f (g) - f ∂_i (g) = (∂ f/∂ x_i)(g)$ . Since $g$ was arbitrary and both sides are operators on $K[X]$ , the identity holds in $End_K(K[X])$ , hence in $A_n$ .

Specialising $f = x^α$ gives the straightening rule used repeatedly below:

∂_i, x^α = x^α, ∂_i + α_i, x^α - e_i,

(1.1)

where $e_i$ is the multi-index with $1$ in position $i$ and $0$ elsewhere, and the second term is read as $0$ when $α_i = 0$ .

Step 1: the canonical basis spans

Spanning

By definition $A_n$ is spanned over $K$ by the words in the generators, so it is enough to rewrite a single word $w$ as a combination of elements of $ℬ$ . Induct on the length of $w$ . If $w$ has no $∂$ immediately to the left of an $x$ , then all the $x$ 's already precede all the $∂$ 's and $w ∈ B$ . Otherwise $w$ contains a factor $∂_i x_j$ . If $i ≠ j$ the two commute and the swap leaves a word of the same length with one fewer inversion. If $i = j$ , then $∂_i x_i = x_i ∂_i + 1$ , so $w$ becomes the swapped word plus a word of length two less. In both cases the process strictly decreases the pair (length, number of inversions) in lexicographic order, so it terminates, and it terminates at a $K$ -combination of elements of $ℬ$ .

Step 2: how derivatives act on monomials

Derivative of a monomialCoutinho (1.2.2)

Let $σ, β ∈ N^n$ with $|σ| ≤ |β|$ . Then

∂^β (x^σ) = cases σ! & if β = σ, 0 & otherwise. cases

(1.2)

Proof

In one variable, $∂^b(x^s)$ equals $s(s-1) ⋅ s(s-b+1),x^s-b$ if $b ≤ s$ and $0$ if $b > s$ . Multiplying these over the $n$ variables gives $∂^β(x^σ) = 0$ unless $β_i ≤ σ_i$ for every $i$ , and in that case $∂^β(x^σ) = ( ∏ _i (σ_i!)/((σ_i-β_i)!) ) x^σ-β$ . Now impose $|σ| ≤ |β|$ . Together with $β_i ≤ σ_i$ for all $i$ this forces $|β| = |σ|$ and hence $β = σ$ , in which case the displayed value is $σ!$ , a non-zero element of $K$ because $char K = 0$ .

Step 3: the canonical basis is independent

Independence

Let $D = ∑ _α,β c_αβ, x^α ∂^β$ be a finite combination with at least one $c_αβ ≠ 0$ . We produce $f$ with $D(f) ≠ 0$ , which shows $D ≠ 0$ in $End_K(K[X])$ .

Among all $β$ for which some $c_αβ ≠ 0$ , choose one, call it $σ$ , of minimal length. So $c_ασ ≠ 0$ for at least one $α$ , while $c_αβ = 0$ for every $α$ whenever $|β| < |σ|$ .

Apply $D$ to the test polynomial $x^σ$ . Every surviving term has $|β| ≥ |σ|$ , so Lemma (1.2) applies to it and kills it unless $β = σ$ . Hence

D(x^σ) = ∑ _α,β c_αβ, x^α, ∂^β(x^σ) = σ! ∑ _α c_ασ, x^α.

(1.3)

The polynomial $∑ _α c_ασ x^α$ is non-zero, because the monomials $x^α$ are independent in $K[X]$ and at least one $c_ασ$ is non-zero. Since $σ! ≠ 0$ in $K$ , we get $D(x^σ) ≠ 0$ , so $D ≠ 0$ .

Remark

Note what the two halves used. Spanning used only the commutation relations. Independence used that the elements of $A_n$ are honest operators on $K[X]$ and that $K[X]$ is infinite dimensional - there is always a monomial of large enough degree to test against. Both facts are supplied by the definition of $A_n$ as a subalgebra of $End_K(K[X])$ .

Key Equations

The relations that drive the rewriting are the standard ones (see commutation relations):

[∂_i, x_j] = δ_ij, [x_i, x_j] = 0, [∂_i, ∂_j] = 0 (1 ≤ i,j ≤ n).

(1.4)

Their consequence for a whole polynomial, which is the form actually used in the rewriting, is

∂_i, f = f, ∂_i + (∂ f)/(∂ x_i) (f ∈ K[X]).

(1.5)

Counting the pairs $(α,β) ∈ N^2n$ with $|α| + |β| ≤ m$ gives the dimension of the degree- $≤ m$ part of $A_n$ , a number that is a consequence of the proposition and not available before it:

_K , D ∈ A_n : D ≤ m , = m+2n2n.

(1.6)

Finally, the identity that makes the whole algebra readable: every $D ∈ A_n$ has one and only one expression

D = ∑ _α, β ∈ N^n c_αβ, x^α ∂^β, c_αβ ∈ K, almost all zero.

(1.7)

Variable Definitions

$K$: the ground field, of characteristic zero
$K[X]$: the polynomial ring $K[x_1, … ,x_n]$ , on which $A_n$ acts
$A_n$: the $n$ -th Weyl algebra, the subalgebra of $End_K(K[X])$ generated by the $x_i$ and the $∂_i$
$α, β, σ$: multi-indices in $N^n$
$|α|$: the length $α_1 + … + α_n$ of a multi-index
$α!$: the multi-index factorial $α_1! ⋅ sα_n!$
$e_i$: the multi-index with a $1$ in position $i$ and zeros elsewhere
$x^α ∂^β$: a canonical basis element: all variables to the left, all derivatives to the right
$c_αβ$: the coefficients of an operator written in canonical form
$δ_ij$: the Kronecker delta, equal to $1$ if $i = j$ and $0$ otherwise

Properties and Behaviour

Canonical form is well defined

Two elements of $A_n$ are equal if and only if their canonical forms have identical coefficients. In particular the degree $D = |α|+|β| : c_αβ ≠ 0$ is well defined, with $0 = - ∈ fty$ .

Freeness over the polynomial ring

$A_n$ is free as a left $K[X]$ -module with basis $∂^β : β ∈ N^n$ , and free as a right $K[X]$ -module with basis $∂^β$ as well. Grouping (1.7) by $β$ writes any $D$ uniquely as $∑ _β f_β(x), ∂^β$ with $f_β ∈ K[X]$ ; this is the presentation used by most computer algebra systems.

The reversed order also gives a basis

$∂^β x^α$ is also a $K$ -basis of $A_n$ . Straightening in the opposite direction proves spanning, and independence follows because the change-of-order matrix is triangular with respect to total degree with $1$ 's on the diagonal. The two bases are genuinely different: for instance $∂ x = x∂ + 1$ in $A_1$ , so the coefficient families do not match.

Faithfulness and nesting

The action of $A_n$ on P1 is faithful: only the zero operator acts as zero, which is precisely the independence half restated. Also, for $m ≤ n$ the inclusion $A_m ⊂ eq A_n$ realises $A_m$ as the span of those basis elements involving only the first $m$ variables, so it is injective.

Dimension count

$A_n$ has countably infinite dimension over $K$ , and the degree filtration has finite-dimensional pieces of size $m+2n2n$ as in (1.6). This finiteness is the input to every later dimension-theoretic argument in the subject, including Bernstein's inequality.

Examples and Special Cases

The smallest non-trivial straightening

In $A_1$ , $∂ x = x∂ + 1$ . Both sides applied to $f$ give $(xf)' = xf' + f$ . The canonical forms of $∂ x$ and $x ∂$ differ in the constant coefficient, so they are different operators - the first is the identity on constants, the second annihilates them.

A degree-four operator in canonical form

In $A_1$ , repeated use of (1.1) gives $∂^2 x^2 = x^2 ∂^2 + 4x∂ + 2$ . Applying both sides to $1$ gives $2 = 2$ ; to $x$ gives $6x = 6x$ ; to $x^2$ gives $12x^2 = 12x^2$ . The worked example below recovers the same canonical form by the method of the independence proof, without any commutator manipulation.

Two variables

In $A_2$ , $∂_1 x_2 = x_2 ∂_1$ because the indices differ, while $∂_1 x_1 x_2 = x_1 x_2 ∂_1 + x_2$ . So $∂_1 x_1 x_2 ∂_2$ has canonical form $x_1 x_2 ∂_1 ∂_2 + x_2 ∂_2$ .

Minimality of the test index is not optional

Take $D = x∂ - 1$ in $A_1$ . The coefficient of $x∂$ is non-zero, so one might try the test monomial $x^σ$ with $σ = 1$ . But $D(x) = x ⋅ 1 - x = 0$ : the term of lower order in $∂$ cancels the contribution. The proof avoids this by insisting that $σ$ has minimal length among the occurring $β$ 's, which here forces $σ = 0$ and the test polynomial $1$ , giving $D(1) = -1 ≠ 0$ .

Worked Example

Recovering a canonical form by evaluation, in $A_1$

Step 1 - the operator and what we may assume
Let $D ∈ A_1$ be the operator $f ↦ (x^2 f)''$ , that is $D = ∂^2 x^2$ . Its degree is $4$ , and straightening will express it as $∑ _a,b c_ab x^a ∂^b$ with $b ≤ 2$ , since $D$ contains only two derivatives. We recover the $c_ab$ using nothing but Lemma (1.2), which is exactly the mechanism of the independence proof.
Step 2 - test against $1$ to read off the $b = 0$ coefficients
Because $∂^b(1) = 0$ for $b ≥ 1$ , applying $D$ to $1$ leaves only the $b = 0$ terms:

$D(1) = ∑ _a c_a0, x^a.$

Directly, $D(1) = (x^2)'' = 2$ . Hence $c_00 = 2$ and $c_a0 = 0$ for $a ≥ 1$ .
Step 3 - test against $x$ to read off the $b = 1$ coefficients
Here $∂^0(x) = x$ , $∂^1(x) = 1$ and $∂^b(x) = 0$ for $b ≥ 2$ , so

$D(x) = ∑ _a c_a0, x^a+1 + ∑ _a c_a1, x^a.$

Directly, $D(x) = (x^3)'' = 6x$ . The first sum is $2x$ by Step 2, so $∑ _a c_a1 x^a = 4x$ , giving $c_11 = 4$ and all other $c_a1 = 0$ .
Step 4 - test against $x^2$ to read off the $b = 2$ coefficients
Now $∂^0(x^2) = x^2$ , $∂^1(x^2) = 2x$ and $∂^2(x^2) = 2 = 2!$ , matching Lemma (1.2) with $σ = 2$ . So

$D(x^2) = ∑ _a c_a0, x^a+2 + 2 ∑ _a c_a1, x^a+1 + 2 ∑ _a c_a2, x^a.$

Directly, $D(x^2) = (x^4)'' = 12x^2$ . The first sum contributes $2x^2$ and the second contributes $2 ⋅ 4x ⋅ x = 8x^2$ , so $2 ∑ _a c_a2x^a = 12x^2 - 2x^2 - 8x^2 = 2x^2$ , giving $c_22 = 1$ and all other $c_a2 = 0$ .
Step 5 - cross-check against straightening
The evaluation method has produced $D = x^2∂^2 + 4x∂ + 2$ . Independently, (1.1) gives $∂ x^2 = x^2 ∂ + 2x$ , hence

$∂^2 x^2 = ∂(x^2∂ + 2x) = (x^2∂ + 2x)∂ + 2(x∂ + 1) = x^2∂^2 + 4x∂ + 2.$

The two computations agree, as uniqueness of canonical form requires.

Result

$∂^2 x^2 = x^2∂^2 + 4x∂ + 2$ , and the coefficients were determined one order of $∂$ at a time by evaluating on $1, x, x^2$ . The triangular pattern - each new test monomial exposing exactly one new layer of coefficients, scaled by $σ!$ - is the independence proof turned into an algorithm.

Applications and Industry Use

In a mathematics topic, this section covers downstream use inside mathematics, computing and engineering rather than a manufactured product.

The proposition is used far more often than it is cited, because it is what licenses every routine manipulation in the subject.

Deciding equality. Comparing two operators becomes comparing finitely many scalars. Without uniqueness, every identity in $A_n$ would need an argument.
Defining invariants by formula. The degree, the Bernstein filtration and the order filtration are all defined by reading exponents off the canonical form, and each is well defined only because the form is unique.
Symbols and associated graded rings. The symbol of an operator is the top-degree part of its canonical form with $∂_i$ replaced by a commuting variable $ξ_i$ . This map is well defined for the same reason.
Module presentations. A cyclic module $A_n/A_n D$ is specified by writing $D$ in canonical form; see presentations of cyclic modules. Linear algebra over $K$ on the coefficient vectors is how such modules are computed with.
Data structures. Every implementation stores an operator as a sparse map from exponent pairs to coefficients; that this representation is faithful and unambiguous is precisely the proposition.

Computational Notes

Read this as the manufacturing section of the template: how the object is actually built by machine, at what cost, and where the computation stops being decidable.

The spanning proof is an algorithm, usually called normalisation or straightening.

Represent the input as a $K$ -combination of words in the $2n$ generators.
Scan each word for an occurrence of $∂_i x_j$ . If $i ≠ j$ , swap. If $i = j$ , replace the word $u,∂_i x_i, v$ by $u, x_i∂_i, v + u v$ .
Repeat until no $∂$ precedes an $x$ . Collect equal exponent pairs and drop zero coefficients.

Termination is guaranteed by the argument in Step 1 of the construction. The cost is dominated by the correction terms: normalising $∂^k x^k$ in $A_1$ produces $k+1$ canonical monomials, and in the worst case the number of intermediate terms grows like the number of ways to pair off derivatives with variables, so naive straightening of a product of two operators of degree $d$ costs on the order of $d^2n$ coefficient operations. Implementations therefore multiply directly in the coefficient representation, using the closed formula

∂^β, x^α = ∑ _γ ≤ (α,β) βγ (α!)/((α-γ)!), x^α - γ, ∂^β - γ,

(1.8)

where $γ$ ranges over multi-indices bounded componentwise by both $α$ and $β$ .

Non-commutative Groebner basis engines for the Weyl algebra all take canonical form as their normal form for terms: Macaulay2's Dmodules package, Singular's dmod.lib together with its Plural kernel, and the ore_algebra package in SageMath. Because the leading exponent pair of a product is the sum of the leading exponent pairs, the canonical form behaves well enough under multiplication for term orders to exist at all.

Limits of Validity

The proposition as proved here needs all three of the following, and the first is the one most often forgotten.

Characteristic zero. Independence uses $σ! ≠ 0$ in $K$ . Over a field of characteristic $p$ the operator $∂_i^,p$ is zero on $K[X]$ , so $x^α∂^β = 0$ whenever some $β_i ≥ p$ and the set $ℬ$ is very far from independent. See the positive characteristic page.
The algebra must be the operator algebra. The abstract algebra presented by the relations (1.4) does have the analogous basis in every characteristic, by the diamond lemma. In characteristic $p$ the two algebras differ: the operator algebra is a proper quotient of the abstract one.
A fixed order. The statement is about the specific ordering 'variables left, derivatives right'. Any other fixed order gives a different basis and different coefficients for the same operator.

What survives in general

For rings of differential operators on a general smooth affine variety there is no global canonical basis of this shape; one gets a local statement in etale coordinates, and the invariant surviving object is the associated graded ring. That is why the later theory is phrased in terms of filtrations rather than bases.

Failure Modes and Common Mistakes

Treating spanning as though it settled uniqueness

The rewriting argument shows that the $x^α∂^β$ generate $A_n$ as a vector space. That is compatible with massive collapse - in characteristic $p$ it is exactly what happens. A proof of the proposition that stops after straightening has proved half the statement, and it is the easy half.

Using Lemma (1.2) without its hypothesis

The lemma says $∂^β(x^σ) = 0$ for $β ≠ σ$ only when $|σ| ≤ |β|$ . Drop that and it is plainly false: in $A_1$ , $∂(x^2) = 2x ≠ 0$ although $β = 1 ≠ 2 = σ$ . The hypothesis is what makes the test monomial argument triangular.

Choosing the test monomial by the wrong index

The test index $σ$ must minimise $|β|$ over the occurring terms, not maximise it and not be chosen arbitrarily. The counterexample above, $D = x∂ - 1$ with $D(x) = 0$ , shows that a careless choice produces a value of zero for a non-zero operator and destroys the argument.

Reading coefficients off an unstraightened expression

Expressions such as $∂ x$ or $∂^2 x^2$ look like they have degree-two and degree-four canonical forms with a single term, but their canonical forms have three terms and are not obtained by reordering. In particular the constant term of $∂^2 x^2$ is $2$ , not $0$ , and any conclusion drawn from the unstraightened expression - about the degree in $∂$ , about whether the operator kills constants, about its symbol - can be wrong.

Historical Notes

The relations $[∂_i, x_j] = δ_ij$ are the canonical commutation relations written down by Born, Heisenberg and Jordan in 1925 and studied systematically by Hermann Weyl in his 1928 book on group theory and quantum mechanics; the connection is the subject of the page on quantum origins. The name 'Weyl algebra' became standard after Dixmier's 1968 paper, which is also where much of the structure theory of $A_n$ was first assembled.

The basis statement itself is an instance of a much older pattern. The Weyl algebra is the quotient of the universal enveloping algebra of the $(2n+1)$ -dimensional Heisenberg Lie algebra by the ideal identifying the central element with $1$ , and the Poincare-Birkhoff-Witt theorem gives ordered monomial bases for enveloping algebras. Bergman's diamond lemma of 1978 recast such results as confluence statements about rewriting systems, which is the modern way to see that the abstract algebra defined by generators and relations has the ordered monomials as a basis in any characteristic.

Coutinho's proof, followed here, deliberately avoids that machinery. It uses only the product rule and the infinite-dimensionality of $K[X]$ , which is appropriate for an introduction and has the incidental virtue of making the role of characteristic zero completely visible.

Comparison

Three routes to the same basis statement.
	Test polynomial (this page)	Diamond lemma	PBW for the Heisenberg algebra
What is being proved	$ℬ$ is a basis of the operator algebra	$ℬ$ is a basis of $Kz/J$	$ℬ$ is a basis of $U(h_n)/(z-1)$
Main input	Product rule and infinite-dimensionality of $K[X]$	Confluence of overlapping rewrites	PBW theorem
Characteristic zero needed	Yes, for $σ! ≠ 0$	No	No
Length	About half a page	Short, given the lemma	Short, given PBW
Gives faithfulness of the action	Yes, directly	No	No
Prerequisites	None beyond calculus	Rewriting theory	Lie theory

Key Takeaways

Key points

The monomials $x^α∂^β$ , variables left and derivatives right, form a $K$ -basis of $A_n$ when $char K = 0$ .
Spanning is a terminating rewriting argument driven by $∂_i f = f∂_i + ∂ f/∂ x_i$ .
Independence is proved by evaluating a putative relation on a monomial $x^σ$ where $σ$ has minimal length among the derivative exponents occurring.
The key computation is $∂^β(x^σ) = σ!$ if $β = σ$ and $0$ otherwise, valid when $|σ| ≤ |β|$ .
Characteristic zero is used exactly once, to know $σ! ≠ 0$ ; in characteristic $p$ the statement fails for the operator algebra because $∂_i^,p = 0$ .
Consequences: canonical form is unique, degree and symbols are well defined, $A_n$ is free over $K[X]$ on the $∂^β$ , and $A_n$ acts faithfully on $K[X]$ .

FAQs

Why is spanning not enough?

Because a spanning set can be redundant. The whole point of the proposition is that the coefficients in $∑ c_αβ x^α ∂^β$ are determined by the operator, which is what allows equality to be decided and the degree to be defined. In characteristic $p$ the same spanning argument works and the same set fails to be a basis, which shows spanning cannot imply independence.

Does the order of the variables inside

x^α

matter?

No. The $x_i$ commute among themselves and the $∂_i$ commute among themselves, so $x^α$ and $∂^β$ are unambiguous. Only the relative position of the $x$ block and the $∂$ block matters.

∂^β x^α

also a basis?

Yes, by the same argument run in the other direction. But the coefficients differ. In $A_1$ , the operator $∂ x$ has canonical form $x∂ + 1$ in one convention and the single term $∂ x$ in the other, so any statement about coefficients must name its convention.

Where exactly does the proof use that

K

has characteristic zero?

In one place only: the conclusion $D(x^σ) = σ! ∑ _α c_ασ x^α ≠ 0$ requires $σ!$ to be invertible in $K$ . If $char K = p$ and some entry of $σ$ is at least $p$ , then $σ! = 0$ and the argument collapses - correctly so, because the statement is then false.

Does the proof need

K

to be algebraically closed or infinite?

Neither. Only the characteristic matters. Every field of characteristic zero is infinite anyway, but nothing beyond $σ! ≠ 0$ and the independence of monomials in $K[X]$ is used.

How do I actually put a given expression into canonical form?

Repeatedly rewrite $∂_i x_i$ as $x_i∂_i + 1$ and swap commuting pairs, as in the computational notes; or, for a product $∂^β x^α$ , apply the closed formula (1.8) in one step. Both give the same answer, because the answer is unique.

What does the result say about the action on functions other than polynomials?

The proposition is proved using the action on $K[X]$ , but faithfulness transfers to any module containing $K[X]$ compatibly - for instance the ring of holomorphic functions over $ℂ$ . It does not transfer to arbitrary modules: plenty of $A_n$ -modules are not faithful, since a module can be annihilated by a left ideal without being annihilated by a two-sided one.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 1 §2, Proposition (1.2.1) and Lemma (1.2.2).
J. Dixmier, Sur les algebres de Weyl, Bulletin de la Societe Mathematique de France 96 (1968), 209-242.
G. M. Bergman, The diamond lemma for ring theory, Advances in Mathematics 29 (1978), 178-218 - for the characteristic-free treatment of ordered monomial bases.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, Graduate Studies in Mathematics 30, revised edition, American Mathematical Society, 2001 - Ch. 1, for the Weyl algebra and its standard basis.
J.-E. Bjork, Rings of Differential Operators, North-Holland Mathematical Library 21, North-Holland, 1979 - Ch. 1.
H. Weyl, Gruppentheorie und Quantenmechanik, Hirzel, Leipzig, 1928 - the source of the commutation relations in this form.
M. Saito, B. Sturmfels and N. Takayama, Groebner Deformations of Hypergeometric Differential Equations, Algorithms and Computation in Mathematics 6, Springer, 2000 - Ch. 1, for normal forms and term orders in $A_n$ .
ISO 80000-2:2019, Quantities and units - Part 2: Mathematics, International Organization for Standardization.

AI Suggested Questions

Verify formula (1.8) for $n = 1$ , $α = β = 2$ , and check it against the straightening of $∂^2 x^2$ .
Write $∂_1^2 x_1 x_2 ∂_2$ in canonical form in $A_2$ and state its degree.
Exhibit explicitly a non-trivial linear relation among the $x^a ∂^b$ in the operator algebra generated by $x$ and $∂$ over a field of characteristic $3$ .
Show that the change of basis from $x^α ∂^β$ to $∂^β x^α$ is triangular with respect to total degree and has $1$ 's on the diagonal.
Give the coefficient-extraction algorithm of the worked example in general form, for an operator of $∂$ -degree at most $d$ in $A_1$ .
Prove that $A_n$ is free as a right $K[X]$ -module on $∂^β$ , and compare with the left module statement.
Explain why the proposition implies that $A_m → A_n$ is injective for $m ≤ n$ .