Why the Weyl Algebra Is a Domain

Q: Does "domain" here include commutativity?

No. In noncommutative ring theory a domain is a ring with 1 ≠ 0 and no zero divisors; commutative domains are called integral domains. A_n is a domain that is as noncommutative as possible, in the sense that its centre is just K .

Q: Can I divide one operator by another the way I would divide polynomials?

Not in A_n . There is no division algorithm, because leading coefficients are polynomials in x rather than scalars and cannot generally be inverted. In B_1(K) , where the coefficients are rational functions, a left division algorithm does exist and makes every left ideal principal - Coutinho sets this out in the exercises to Ch. 2. The contrast is precisely the reason A_1 is harder than B_1 .

Overview

A domain is a ring in which a product of non-zero elements is non-zero. The word is used here without any assumption of commutativity: $A_n$ is a domain and is thoroughly noncommutative. The result is Corollary (2.1.2) in Coutinho, and it costs almost nothing once the degree is available.

The argument is the same one that shows a polynomial ring over a field has no zero divisors. If $D$ and $D^{'}$ are non-zero, their degrees are non-negative integers; additivity of degree gives $(DD') = (D)+ (D') ≥ 0$ , and only the zero operator has degree $- ∈ fty$ . So $DD' ≠ 0$ .

What deserves attention is not the proof but its reach. Being a domain gives cancellation, forces the units to be the non-zero scalars, rules out non-trivial idempotents, and - combined with the Noetherian property - guarantees that $A_n$ embeds in a division ring, the Weyl skew field $D_n$ . It is also the reason the theory of modules over $A_n$ looks nothing like the theory of modules over a matrix algebra, despite $A_n$ being simple.

Finally, the result is a genuine test case for the standing hypothesis on the ground field. Over a field of characteristic $p$ the Weyl algebra splits into two non-isomorphic candidates, and one of them has nilpotent elements. Tracking which step of the proof breaks is the cleanest way to see why characteristic zero is assumed throughout; that is done in the last sections here and in full on the positive characteristic page.

Definition

Fix a field $K$ of characteristic zero and let $A_n = A_n(K)$ be the $n$ -th Weyl algebra, defined as the subalgebra of $End_K(K[X])$ generated by multiplication by $x_1, … ,x_n$ and the partial derivatives $∂_1, … ,∂_n$ .

Domain

A ring $R$ with $1 ≠ 0$ is a domain if $ab ≠ 0$ whenever $a ≠ 0$ and $b ≠ 0$ . Equivalently, $R$ has no zero divisors on either side. A commutative domain is an integral domain; a domain in which every non-zero element is a unit is a division ring.

The Weyl algebra is a domainCoutinho (2.1.2)

For every $n ≥ 0$ and every field $K$ of characteristic zero, $A_n(K)$ is a domain. More precisely, for non-zero $D, D' ∈ A_n$ the product $D D^{'}$ is non-zero and $(DD') = (D) + (D')$ .

The statement is stronger than "no zero divisors"

The theorem records the degree of the product as well as its non-vanishing. That refinement is what is actually used downstream - in the classification of units, in bounding the degree of elements produced inside an ideal, and in Hilbert function estimates - so it is worth carrying the equality rather than the inequality.

Core Concepts

Three different-looking explanations all reduce to the same fact, and each is useful in a different context.

Leading symbols multiply

Assign to a non-zero $D$ of degree $m$ its leading symbol $σ_m(D)$ , a non-zero homogeneous polynomial of degree $m$ in the $2n$ commuting variables $x_1, … ,x_n,ξ_1, … ,ξ_n$ . Reordering $∂$ past $x$ changes an operator only in degree $m-2$ and below, so $σ_m+m'(DD') = σ_m(D)σ_m'(D')$ . Since $K[x,ξ]$ is a polynomial ring over a field, the right-hand side is non-zero, hence so is $D D^{'}$ .

A filtered ring inherits the domain property from its associated graded ring

The previous paragraph is the special case of a general principle: if $R$ carries an exhaustive filtration by finite steps and $gr R$ is a domain, then $R$ is a domain. For the Bernstein filtration, $gr,A_n ≅ K[x_1, … ,x_n,ξ_1, … ,ξ_n]$ , which is a domain. This is the version that generalises: the same argument shows the ring of differential operators on a smooth irreducible affine variety is a domain, because its associated graded ring is the coordinate ring of the cotangent bundle, which is irreducible.

Operators on a polynomial ring cannot annihilate everything

A more hands-on reading: $A_n$ acts on $K[X]$ , and if $DD' = 0$ then $D$ kills the image of $D^{'}$ . In characteristic zero the image of a non-zero operator is an infinite-dimensional subspace of $K[X]$ , large enough that no non-zero operator can vanish on it. This is the picture that actually breaks in characteristic $p$ , where $∂_i^p$ annihilates all of $K[X]$ while being a perfectly good non-zero word in the generators.

Construction and Proof

Proof that $A_n$ is a domain

Let $D, D' ∈ A_n$ be non-zero. Writing each in canonical form, both have a well-defined degree, and that degree is a non-negative integer because a non-zero canonical monomial $x^α∂^β$ contributes $|α|+|β| ≥ 0$ . By (2.7), $(DD') = (D) + (D') ≥ 0$ . The only element of $A_n$ of degree $- ∈ fty$ is $0$ , so $DD' ≠ 0$ .

Cancellation

If $D ≠ 0$ and $DE = DF$ , then $D(E-F) = 0$ , so $E - F = 0$ . The same works on the right. Cancellation is what allows one to divide out common factors in computations with operators, and it is used constantly without comment.

The units of $A_n$ are exactly the non-zero constantsCoutinho, Ch. 2 §2

Suppose $DD' = 1$ . Then $(D) + (D') = (1) = 0$ , and since both degrees are non-negative integers, $(D) = (D') = 0$ . A degree-zero operator is a scalar. Conversely every non-zero scalar is invertible. Hence $A_n^ × = K^ ×$ , which is (2.10).

The consequence is worth stating plainly: every non-constant operator generates a proper non-zero left ideal. So although $A_n$ has no proper non-zero two-sided ideals, it is drowning in one-sided ideals, and it is very far from being a division ring.

No non-trivial idempotents, no nilpotents

If $e^2 = e$ then $e(e-1) = 0$ , so $e = 0$ or $e = 1$ . If $D^k = 0$ with $k$ minimal and $D ≠ 0$ , then $D ⋅ D^k-1 = 0$ contradicts the domain property. In particular $A_n$ admits no non-trivial direct sum decomposition as a left module over itself, so $A_n$ is indecomposable as a module over itself - unlike a matrix algebra, which is the other standard example of a simple ring.

The skew field of fractions exists

$A_n$ is a Noetherian domain. A Noetherian domain satisfies the Ore condition on both sides, by Goldie's theorem, so it embeds in a division ring $D_n$ in which every non-zero element of $A_n$ becomes invertible. $D_n$ is called the $n$ -th Weyl skew field. Gelfand and Kirillov proved that $D_n$ determines $n$ : for $m ≠ n$ , $D_m$ and $D_n$ are not isomorphic as $K$ -algebras.

This is a strictly stronger statement than the domain property and is quoted here, not proved. It is included because it explains a common source of confusion: one may invert operators, but only after leaving $A_n$ .

Key Equations

The single input to the proof:

(D D') = (D) + (D') for all D, D' ∈ A_n,

(2.7)

with the convention $(0) = - ∈ fty$ and $- ∈ fty + m = - ∈ fty$ .

Multiplicativity of symbols is the graded form of the same statement:

σ_m+m'(DD') = σ_m(D),σ_m'(D') in K[x_1, … ,x_n,ξ_1, … ,ξ_n].

(2.8)

Cancellation follows at once, on both sides:

DE = DF,; D ≠ 0 ; ⟹ ; E = F; ED = FD,; D ≠ 0 ; ⟹ ; E = F.

(2.9)

And the group of units collapses:

A_n^ × = K^ × = K ∖ 0.

(2.10)

In positive characteristic the relation that destroys the argument is

∂_i^,p = 0 as an operator on Z_p[x_1, … ,x_n],

(2.11)

because $k(k-1) ⋅ s(k-p+1)$ is a product of $p$ consecutive integers and so is divisible by $p$ .

Variable Definitions

$K$: the ground field, of characteristic zero unless stated otherwise
$A_n$: the $n$ -th Weyl algebra over $K$
$D, D^{'}, E, F$: elements of $A_n$ , that is differential operators with polynomial coefficients
$(D)$: the degree of $D$ in the Bernstein weighting, where every generator has weight $1$
$σ_m(D)$: the leading symbol of $D$ , its degree- $m$ part with $∂_i$ replaced by the commuting variable $ξ_i$
$gr,A_n$: the associated graded algebra of $A_n$ for the Bernstein filtration, isomorphic to a polynomial ring in $2n$ variables
$A_n^ ×$: the group of invertible elements of $A_n$
$D_n$: the skew field of fractions of $A_n$ , the Weyl skew field
$Z_p$: the field with $p$ elements, used only in the positive characteristic comparison

Properties and Behaviour

Once $A_n$ is known to be a domain, a long list of structural facts follows, and an equally interesting list does not follow.

What the domain property gives, and what it does not.
Statement	Holds in $A_n$ ?	Reason
No zero divisors	yes	additivity of degree
Cancellation on both sides	yes	immediate from the above
Only units are $K^ ×$	yes	degrees of $D$ and $D^-1$ must sum to $0$
Only idempotents are $0$ and $1$	yes	$e(e-1)=0$
No non-zero nilpotents	yes	domain
Simple as a ring	yes	separate theorem, uses characteristic $0$
Left Noetherian	yes	separate theorem, via $gr,A_n$
Left principal ideal ring	no	$A_1∂^2 + A_1(x∂-1)$ is not cyclic
Division ring	no	$x$ has no inverse
Finite dimensional over $K$	no	the canonical monomials are infinite in number
Every left ideal $2$ -generated	yes	Stafford's theorem, quoted not proved

Not a principal ideal ringCoutinho, Ch. 2 §2 and Exercises 4.1, 4.9

For $n ≥ 2$ the left ideal of $A_n$ generated by $∂_1, … ,∂_n$ is not principal: a generator would have to have degree $1$ , and no single degree- $1$ operator generates all of the $∂_i$ . Even for $n = 1$ the algebra is not a left principal ideal ring - the ideal $A_1∂^2 + A_1(x∂ - 1)$ is not cyclic. What is true, by a theorem of Stafford, is that every left ideal of $A_n$ is generated by two elements; the proof is technical and is not given in the Primer.

Simple plus domain is a strong combination

The two standard examples of simple rings are matrix algebras over a field and $A_n$ . Matrix algebras are full of zero divisors and idempotents; $A_n$ has none. Any argument that treats "simple ring" as a synonym for "matrix algebra" will give wrong answers here, and that mismatch is exactly why $A_n$ has no non-zero finite-dimensional representations.

Worked Example

Two operators in $A_1$ , and the same computation over $Z_3$

Step 1 - choose the operators and record their degrees
Work in $A_1 = A_1(Q)$ and take

$D = ∂^2 + x, E = x∂ - 1.$

Both are in canonical form. $D$ has summands of degree $2$ and $1$ , so $(D) = 2$ ; $E$ has summands of degree $2$ and $0$ , so $(E) = 2$ . The theorem predicts $(DE) = (ED) = 4$ and, in particular, that neither product vanishes.
Step 2 - compute $D E$ in canonical form
Only $∂^2 x = x∂^2 + 2∂$ is needed, which one checks on a test function: $∂^2(xf) = xf'' + 2f'$ . Then

$DE = (∂^2 + x)(x∂ - 1) = ∂^2 x∂ - ∂^2 + x^2∂ - x,$

$DE = (x∂^2 + 2∂)∂ - ∂^2 + x^2∂ - x = x∂^3 + ∂^2 + x^2∂ - x.$

The top-degree term is $x∂^3$ , of degree $4$ . As predicted, $DE ≠ 0$ and $(DE) = 2 + 2$ .
Step 3 - compute $E D$ and compare
Using $∂ x = x∂ + 1$ , so that $x∂ x = x^2∂ + x$ :

$ED = (x∂ - 1)(∂^2 + x) = x∂^3 + x∂ x - ∂^2 - x = x∂^3 + x^2∂ - ∂^2.$

Both products have the same top term $x∂^3$ , whose symbol is $xξ^3 = σ_2(D)σ_2(E) = ξ^2 ⋅ xξ$ , confirming (2.8). Their difference is

$[D, E] = (x∂^3 + ∂^2 + x^2∂ - x) - (x∂^3 + x^2∂ - ∂^2) = 2∂^2 - x,$

of degree $2 = 2 + 2 - 2$ , in line with the commutator estimate. Notice the products are different but both non-zero: the domain property says nothing about commutativity.
Step 4 - use cancellation
Suppose someone hands you $F ∈ A_1$ with $DF = DE$ . Since $D ≠ 0$ , cancellation gives $F = E$ immediately, with no need to look at $F$ at all. Equivalently, left multiplication by any non-zero operator is an injective $K$ -linear map $A_1 → A_1$ . It is never surjective unless $D$ is a scalar, because degrees are shifted up by $(D)$ .
Step 5 - repeat over $Z_3$ and watch it fail
Now let the ground field be $Z_3$ and let $R_1$ be the algebra of operators on $Z_3[x]$ generated by multiplication by $x$ and by $∂$ . Here $∂ ≠ 0$ (it sends $x$ to $1$ ) and $∂^2 ≠ 0$ (it sends $x^2$ to $2$ ). But for every $k$ ,

$∂^3(x^k) = k(k-1)(k-2),x^k-3 = 0 in Z_3[x],$

because three consecutive integers always include a multiple of $3$ . Hence $∂ ⋅ ∂^2 = ∂^3 = 0$ in $R_1$ with both factors non-zero: $R_1$ is not a domain. Concretely, the failure is that the monomials $x^a∂^b$ are no longer linearly independent as operators, so the degree is not even well defined - the very first line of the proof is unavailable.

Result

In $A_1(Q)$ : $DE = x∂^3 + ∂^2 + x^2∂ - x$ and $ED = x∂^3 + x^2∂ - ∂^2$ , both of degree $4 = (D)+ (E)$ , and $[D,E] = 2∂^2 - x$ . Over $Z_3$ the analogous operator algebra contains the nilpotent $∂$ with $∂^3 = 0$ , so it is not a domain. The domain property is a characteristic-zero statement about the operator realisation, not a formal consequence of the relations alone.

Applications and Industry Use

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

The domain property is used quietly and constantly:

Presentations of modules. A cyclic module $A_1/A_1P$ is non-zero for every non-constant $P$ , precisely because $P$ is not a unit. Without this, cyclic presentations of solution modules of differential equations could collapse.
Order reduction. Algorithms that divide one operator by another and argue by descent on degree rely on cancellation and on additivity of degree to guarantee termination.
Torsion arguments. Over a domain the notion of a torsion element is meaningful, and over an Ore domain the torsion submodule of a module is a submodule. This is what lets one speak of the rank of an $A_n$ -module and of holonomic modules being torsion.
Ruling out finite-dimensional models. A finite-dimensional representation would make some non-zero operator act as a nilpotent matrix; the interaction between that and the domain property is one route into the non-existence theorem.
Localisation. Constructing $A_n[f^-1]$ or $B_n(K)$ requires the Ore condition, which for $A_n$ follows from being a Noetherian domain; see localisation.

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.

Computationally, the domain property has a very practical face: it means that non-zero remainders in a division-like process cannot be created by accidental cancellation of leading terms of different degrees.

Concretely, when a Gröbner basis engine for $A_n$ reduces $D$ by $E$ , it cancels the leading monomials by construction, and the guarantee that the remainder has strictly smaller leading monomial is what makes the reduction terminate. That guarantee is exactly multiplicativity of symbols in $gr,A_n$ , that is (2.8). Implementations in Macaulay2 (Dmodules), Singular (dmod.lib) and SageMath all rely on it.

One caveat for exact computation: the argument uses that the coefficient ring is a domain, so leading coefficients never multiply to zero. If a system is configured to compute over a modular ring such as $Z/m$ with $m$ composite - occasionally done for speed - leading coefficients can vanish, degree bookkeeping becomes unsound, and results must be certified over $ℚ$ before being trusted.

There is also a decidability boundary worth naming. Deciding whether a given operator is a unit is trivial here (check whether it is a non-zero constant). Deciding whether a given operator is a left divisor of another, or whether a given left ideal is principal, is a genuine computation requiring Gröbner methods, and no shortcut follows from the domain property.

Limits of Validity

The theorem as stated is about $A_n(K)$ with $K$ a field of characteristic zero. Each part of that hypothesis matters differently.

Characteristic zero is needed for the operator definition, not for the degree argument. If $A_n$ is defined by generators and relations, the canonical monomials are a basis by construction, degree is additive, and the algebra is a domain over any field. If $A_n$ is defined as operators on $K[X]$ and $char,K = p$ , the two definitions disagree and the operator version acquires nilpotents.
The ground ring should be a domain. Over a commutative ring $R$ that is not a domain - say $R = Z/6$ - the constants already contain zero divisors, and $A_n(R)$ inherits them. Over $ℤ$ the algebra $A_1(Z)$ is a domain but is not simple.
The result does not say $A_n$ is a division ring. It says nothing at all about invertibility beyond ruling it out for non-constants. Localisations such as $A_1[x^-1]$ or the ring $B_1(K)$ of operators with rational coefficients are strictly larger rings.

What generalises to other varieties

For a smooth irreducible affine variety $X$ over a field of characteristic zero, $D(X)$ is a domain, by the graded argument: $gr,D(X)$ is the coordinate ring of the cotangent bundle, which is irreducible, hence a domain. Irreducibility cannot be dropped. If $X$ is a disjoint union of two points, $D(X) ≅ K × K$ , which has zero divisors. On singular varieties $D(X)$ may fail to be Noetherian or finitely generated, and no naive statement should be assumed.

Failure Modes and Common Mistakes

Concluding that $A_n$ has few one-sided ideals

Being a domain and being simple both sound like scarcity of ideals, and neither says anything about left ideals. $A_n$ has an enormous supply of them - one for every non-constant operator, at least - and the study of $A_n$ -modules is precisely the study of that supply. Simplicity constrains only two-sided ideals.

Assuming a domain has a division ring of fractions for free

It does not. Noncommutative domains need not satisfy the Ore condition, and free algebras on two or more generators are the standard counterexample. $A_n$ does have a skew field of fractions, but only because it is additionally Noetherian; the implication runs through Goldie's theorem, not through the domain property alone.

Applying the degree proof to the order filtration

The proof can be run with the order filtration instead, since $gr$ is again a polynomial ring - but the bookkeeping changes: order-zero operators are all of $K[X]$ , not just the constants, so the classification of units does not follow the same way. Fix one filtration for the whole of an argument and say which.

Believing the relations alone force the operator picture

In characteristic zero the surjection from the algebra given by generators and relations onto the algebra of operators is an isomorphism, so the two pictures coincide and either can be used. In characteristic $p$ it is not injective - $∂^p$ maps to zero - and the two pictures give genuinely different rings, only one of which is a domain. Statements proved in one picture must be re-examined in the other.

Historical Notes

That the algebra of position and momentum operators has no zero divisors was folklore in the early quantum-mechanical literature of the late 1920s, where it appeared as the observation that a product of two non-trivial observables is never identically zero. It became a theorem in a modern algebraic form with the systematic study of the algebra $A_1$ by Jacques Dixmier in 1968, who determined its units, its automorphism group and its simple modules, and who asked whether every endomorphism of $A_n$ is an automorphism - now the Dixmier conjecture.

The existence of the skew field of fractions $D_n$ and the fact that it remembers $n$ are due to Gelfand and Kirillov, in 1966, in the work that introduced what is now called Gelfand-Kirillov dimension. Their conjecture that the fraction field of the enveloping algebra of an algebraic Lie algebra is always some $D_n$ over a purely transcendental extension was later shown to be false in general, but it is true in many cases and remains the reason $D_n$ is a standard object.

The graded formulation - a filtered ring whose associated graded ring is a domain is a domain - belongs to the general theory of filtered rings developed in the 1960s and 1970s and codified in Björk's 1979 book. It is the form in which the result survives the passage from $A_n$ to rings of differential operators on general smooth varieties.

Comparison

It helps to place $A_n$ among neighbouring rings of operators.

Domain and simplicity properties of related operator algebras.
Ring	Domain?	Simple?	Comment
$K[x_1, … ,x_n]$ , $char,K = 0$	yes	no	commutative, many ideals
$A_n(K)$ , $char,K = 0$	yes	yes	the subject of this page
$B_n(K)$ , rational function coefficients	yes	yes	order filtration; for $n=1$ also a principal ideal ring
$M_r(K)$ , $r ≥ 2$	no	yes	simple but full of zero divisors
$M_r(A_n)$ , $r ≥ 2$	no	yes	same phenomenon over $A_n$
Operators on $Z_p[x]$	no	no	$∂^p = 0$ ; nilpotents
$Z_p$ -algebra on $z_1, z_2$ with $[z_2,z_1]=1$	yes	no	$z_1^p$ is central, generating a two-sided ideal
$A_1(Z)$	yes	no	the prime $p$ generates a proper two-sided ideal

The last two rows are the ones to remember. Being a domain and being simple are independent conditions, and in positive characteristic the Weyl algebra can lose either one depending on which of the two definitions is used. The comparison is worked through on the positive characteristic page.

Key Takeaways

Key takeaways

$A_n(K)$ has no zero divisors, for any field $K$ of characteristic zero. The proof is one line from $(DD') = (D)+ (D')$ .
Equivalently, leading symbols multiply in $gr,A_n ≅ K[x,ξ]$ , and a polynomial ring over a field is a domain.
Consequences: two-sided cancellation, only $0$ and $1$ are idempotent, no non-zero nilpotents, and $A_n^ × = K^ ×$ .
Because the units are only the scalars, every non-constant operator generates a proper non-zero left ideal - $A_n$ is simple but nothing like a division ring.
$A_n$ is a Noetherian domain, hence Ore, hence embeds in the Weyl skew field $D_n$ ; but Ore does not follow from the domain property alone.
It is not a principal ideal ring, even for $n = 1$ ; Stafford's theorem gives two generators for every left ideal.
In characteristic $p$ the operator realisation has $∂_i^p = 0$ and is not a domain, while the generators-and-relations version is a domain but not simple.

FAQs

Does "domain" here include commutativity?

No. In noncommutative ring theory a domain is a ring with $1 ≠ 0$ and no zero divisors; commutative domains are called integral domains. $A_n$ is a domain that is as noncommutative as possible, in the sense that its centre is just $K$ .

A_n

is a domain and simple, why is it not a division ring?

Simplicity is about two-sided ideals; being a division ring is about one-sided ideals as well. The classification of units shows $x$ is not invertible in $A_1$ , yet $A_1x$ is a proper non-zero left ideal that is not two-sided - indeed the two-sided ideal generated by $x$ is all of $A_1$ , since $[∂, x] = 1$ .

Is the converse true - does a simple domain have to look like

A_n

No. There are many simple Noetherian domains that are not Weyl algebras, including the ring $B_n(K)$ of differential operators with rational function coefficients, primitive quotients of enveloping algebras, and quantum tori at generic parameters. What is special about $A_n$ is the combination of simplicity, the domain property, finite Gelfand-Kirillov dimension $2n$ and a commutative associated graded ring that is a polynomial ring.

Where exactly does the proof use characteristic zero?

In the definition, not in the degree computation. Characteristic zero is what makes the canonical monomials $x^α∂^β$ linearly independent as operators on $K[X]$ , so that the degree exists at all. If instead you define $A_n$ by generators and relations, the canonical monomials are independent by construction and the domain property holds in any characteristic.

Can I divide one operator by another the way I would divide polynomials?

Not in $A_n$ . There is no division algorithm, because leading coefficients are polynomials in $x$ rather than scalars and cannot generally be inverted. In $B_1(K)$ , where the coefficients are rational functions, a left division algorithm does exist and makes every left ideal principal - Coutinho sets this out in the exercises to Ch. 2. The contrast is precisely the reason $A_1$ is harder than $B_1$ .

Does the domain property tell me anything about

A_n

-modules?

Yes, indirectly but importantly. It makes torsion a meaningful notion, so one may speak of the rank of a module over the skew field $D_n$ ; holonomic modules turn out to be exactly torsion modules in the relevant sense for $n = 1$ . It also means $A_n$ is indecomposable as a module over itself, so free modules have a well-defined rank.

A_n ⊗ _K A_m

a domain?

Yes, and this is a useful check on the theory: the external product decomposition gives $A_n ⊗ _K A_m ≅ A_n+m$ , which is a domain by the theorem. Note that this is special. A tensor product of two domains over a field need not be a domain in general - $C ⊗ _RC$ is the standard counterexample - so the conclusion here comes from the isomorphism, not from a general principle.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 2 §1, Corollary (2.1.2), and Ch. 2 §2 for the units and the failure of the principal ideal property.
J. Dixmier, Sur les algèbres de Weyl, Bulletin de la Société Mathématique de France 96 (1968), 209-242 - units, automorphisms and simple modules of $A_1$ .
I. M. Gelfand and A. A. Kirillov, Sur les corps liés aux algèbres enveloppantes des algèbres de Lie, Publications Mathématiques de l'IHÉS 31 (1966), 5-19 - the skew field $D_n$ and the invariance of $n$ .
J. T. Stafford, Module structure of Weyl algebras, Journal of the London Mathematical Society 18 (1978), 429-442 - every left ideal of $A_n$ is generated by two elements.
J.-E. Björk, Rings of Differential Operators, North-Holland Mathematical Library 21, 1979 - Ch. 1, filtered rings whose associated graded ring is a domain.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, revised edition, American Mathematical Society, 2001 - Ch. 2 for the Ore condition and Goldie's theorem, Ch. 8 for $A_n$ .
S. P. Smith, Differential operators on commutative algebras, in Ring Theory (Antwerp 1985), Lecture Notes in Mathematics 1197, Springer, 1986 - the Weyl algebra in positive characteristic.
Macaulay2 Dmodules package documentation - normal forms and Gröbner bases in $A_n$ , which depend on multiplicativity of leading symbols.

AI Suggested Questions

Show that a filtered ring whose associated graded ring is a domain must itself be a domain, with the filtration hypotheses stated precisely.
Give an explicit non-zero operator in $A_1$ whose left ideal is not two-sided, and compute the two-sided ideal it generates.
Why does the Ore condition hold for $A_n$ , and what goes wrong for the free algebra on two generators?
Prove that $A_1∂^2 + A_1(x∂ - 1)$ is not a cyclic left ideal of $A_1$ .
Compare $A_1$ with $B_1(K)$ : which one has a division algorithm, and why does that make every left ideal principal?
What is the torsion submodule of an $A_1$ -module, and how does it relate to holonomicity?
Work out the units and the zero divisors of $A_1(Z)$ and explain why it is not simple.