Editing Weyl algebra (section)

== Properties of ''A<sub>n</sub>'' ==
Many properties of <math>
A_1
</math> apply to <math>
A_n
</math> with essentially similar proofs, since the different dimensions commute.

=== General Leibniz rule ===
{{Main|General Leibniz rule}}

{{Math theorem
| name = Theorem
| note = general Leibniz rule
| math_statement = <math display="block">
p^k q^m = \sum_{l=0}^k \binom{k}{l} \frac{m!}{(m-l)!} q^{m-l} p^{k-l} = q^mp^k + mk q^{m-1}p^{k-1} + \cdots 
</math>
}}

{{Math proof|title=Proof|proof=
 Under the <math>
p \mapsto x, q \mapsto \partial_x
</math> representation, this equation is obtained by the general Leibniz rule. Since the general Leibniz rule is provable by algebraic manipulation, it holds for <math>
A_1
</math> as well.
}}In particular, <math display="inline">[q, q^m p^n] = -nq^mp^{n-1}</math> and <math display="inline">[p, q^mp^n] = mq^{m-1}p^n</math>.
{{Math theorem
| math_statement = The [[Center (ring theory)|center]] of Weyl algebra <math>A_n</math> is the underlying field of constants <math>F</math>.
| name = Corollary
}}

{{Math proof|title=Proof|proof= 
If the commutator of <math>f</math> with either of <math>p, q</math> is zero, then by the previous statement, <math>f</math> has no monomial <math>p^nq^m</math> with <math>n > 0</math> or <math>m > 0</math>.
}}
=== Degree ===
{{Math theorem
| name = Theorem
| note = 
| math_statement = <math>
A_n
</math> has a basis <math>
\{q^m p^n : m, n \geq 0\}
</math>.{{Sfn|Coutinho|1995|p=9|loc=Proposition 2.1}}
}}

{{Math proof|title=Proof|proof=
 By repeating the commutator relations, any monomial can be equated to a linear sum of these. It remains to check that these are linearly independent. This can be checked in the differential operator representation. For any linear sum <math>
\sum_{m, n} c_{m,n} x^m \partial_x^n
</math> with nonzero coefficients, group it in descending order: <math>
p_N(x) \partial_x^N + p_{N-1}(x) \partial_x^{N-1} + \cdots + p_M(x) \partial_x^M
</math>, where <math>
p_M
</math> is a nonzero polynomial. This operator applied to <math>
x^M
</math> results in <math>
M! p_M(x) \neq 0
</math>.
}}

This allows <math>
A_1
</math> to be a [[graded algebra]], where the degree of <math>
\sum_{m, n} c_{m,n} q^m p^n
</math> is <math>
\max (m + n)
</math> among its nonzero monomials. The degree is similarly defined for <math>
A_n
</math>.

{{Math theorem
| name = Theorem
| math_statement = For <math>A_n</math>:{{sfn | Coutinho | 1995 | pp=14-15}}
 
* <math>
\deg(g + h) \leq \max(\deg(g), \deg(h)) 
</math>
* <math>
\deg([g, h]) \leq \deg(g) + \deg(h) - 2 
</math>
* <math>
\deg(g h) = \deg(g) + \deg(h) 
</math>
}}

{{Math proof|title=Proof|proof=
We prove it for <math>A_1</math>, as the <math>A_n</math> case is similar.

The first relation is by definition. The second relation is by the general Leibniz rule. For the third relation, note that <math>\deg(g h) \leq \deg(g) + \deg(h)</math>, so it is sufficient to check that <math>gh</math> contains at least one nonzero monomial that has degree <math>\deg(g) + \deg(h)</math>. To find such a monomial, pick the one in <math>g</math> with the highest degree. If there are multiple such monomials, pick the one with the highest power in <math>q</math>. Similarly for <math>h</math>. These two monomials, when multiplied together, create a unique monomial among all monomials of <math>gh</math>, and so it remains nonzero.
}}

{{Math theorem
| name = Theorem
| math_statement = <math>A_n</math> is a [[Simple algebra|simple]] [[Domain (ring theory)|domain]].{{sfn | Coutinho | 1995 | p=16}}
}}
That is, it has no [[Ideal (ring theory)|two-sided nontrivial ideals]] and has no [[zero divisor]]s.
{{Math proof|title=Proof|proof=
Because <math>\deg(gh) = \deg(g) + \deg(h)</math>, it has no zero divisors.

Suppose for contradiction that <math>I</math> is a nonzero two-sided ideal of <math>A_1</math>, with <math>I \neq A_1</math>. Pick a nonzero element <math>f \in I</math> with the lowest degree. 

If <math>f</math> contains some nonzero monomial of form <math>xx^m\partial^n = x^{m+1} \partial^n</math>, then 
<math display=block>
[\partial, f] = \partial f - f \partial
</math>
contains a nonzero monomial of form 
<math display=block>
\partial x^{m+1} \partial^n -  x^{m+1} \partial^n \partial = (m+1) x^m \partial^n.
</math>
Thus <math>[\partial, f]</math> is nonzero, and has degree <math>\leq \deg(f)-1</math>. As <math>I</math> is a two-sided ideal, we have <math>[\partial, f] \in I</math>, which contradicts the minimality of <math>\deg(f)</math>.

Similarly, if <math>f</math> contains some nonzero monomial of form <math>x^m\partial^n\partial</math>, then <math>[x, f] = xf - fx</math> is nonzero with lower degree.
}}

=== Derivation ===
{{See|Derivation (differential algebra)}}{{Math theorem|
| math_statement = The derivations of <math display="inline">A_n</math> are in bijection with the elements of <math display="inline">A_n</math> up to an additive scalar.{{sfn | Dirac | 1926 | pp=415–417}}}}
That is, any derivation <math display="inline">D</math> is equal to <math display="inline">[\cdot, f]</math> for some <math display="inline">f \in A_n</math>; any <math display="inline">f\in A_n</math> yields a derivation <math display="inline">[\cdot, f]</math>; if <math display="inline">f, f' \in A_n</math> satisfies <math display="inline">[\cdot, f] = [\cdot, f']</math>, then <math display="inline">f - f' \in F</math>.

The proof is similar to computing the potential function for a conservative polynomial vector field on the plane.{{sfn | Coutinho | 1997 | p=597}}

{{Collapse top|title=Proof}}
Since the commutator is a derivation in both of its entries, <math display="inline">[\cdot, f]</math> is a derivation for any <math display="inline">f\in A_n</math>. Uniqueness up to additive scalar is because the center of <math display="inline">A_n</math> is the ring of scalars.

It remains to prove that any derivation is an inner derivation by induction on <math display="inline">n</math>.

Base case: Let <math display="inline">D: A_1 \to A_1</math> be a linear map that is a derivation. We construct an element <math display="inline">r</math> such that <math display="inline">[p, r] = D(p), [q,r] = D(q)</math>. Since both <math display="inline">D</math> and <math display="inline">[\cdot, r]</math> are derivations, these two relations generate <math display="inline">[g, r] = D(g)</math> for all <math display="inline">g\in A_1</math>.

Since <math display="inline">[p, q^mp^n] = mq^{m-1}p^n</math>, there exists an element <math display="inline">f = \sum_{m,n} c_{m,n} q^m p^n</math> such that <math display="block">
      [p, f] =  \sum_{m,n} m c_{m,n} q^m p^n = D(p)
      </math> 

<math display="block">
      \begin{aligned}
      0 &\stackrel{[p, q] = 1}{=} D([p, q]) \\
      &\stackrel{D \text{ is a derivation}}{=} [p, D(q)] + [D(p), q] \\
      &\stackrel{[p,f] = D(p)}{=}  [p, D(q)] + [[p,f], q] \\
      &\stackrel{\text{Jacobi identity}}{=} [p, D(q) - [q, f]]
      \end{aligned}
      </math> 

Thus, <math display="inline">D(q) = g(p) + [q, f]</math> for some polynomial <math display="inline">g</math>. Now, since <math display="inline">[q, q^m p^n] = -nq^mp^{n-1}</math>, there exists some polynomial <math display="inline">h(p)</math> such that <math display="inline">[q, h(p)] = g(p)</math>. Since <math display="inline">[p, h(p)] = 0</math>, <math display="inline">r = f + h(p)</math> is the desired element.

For the induction step, similarly to the above calculation, there exists some element <math display="inline">r \in A_n</math> such that <math display="inline">[q_1, r] = D(q_1), [p_1, r] = D(p_1)</math>.

Similar to the above calculation, <math display="block">
 [x, D(y) - [y, r]] = 0
 </math> for all <math display="inline">x \in \{p_1, q_1\}, y \in \{p_2, \dots, p_n, q_2, \dots, q_n\}</math>. Since <math display="inline">[x, D(y) - [y, r]]</math> is a derivation in both <math display="inline">x</math> and <math display="inline">y</math>, <math display="inline">[x, D(y) - [y, r]] = 0</math> for all <math display="inline">x\in \langle p_1, q_1\rangle</math> and all <math display="inline">y \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>. Here, <math display="inline">\langle \rangle</math> means the subalgebra generated by the elements.

Thus, <math display="inline">\forall y \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>, <math display="block">
 D(y) - [y, r] \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle
 </math>

Since <math display="inline">D - [\cdot, r]</math> is also a derivation, by induction, there exists <math display="inline">r' \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math> such that <math display="inline">D(y) - [y, r] = [y, r']</math> for all <math display="inline">y \in \langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>.

Since <math display="inline">p_1, q_1</math> commutes with <math display="inline">\langle p_2, \dots, p_n, q_2, \dots, q_n\rangle</math>, we have <math display="inline">D(y) = [y, r + r']</math> for all <math>y \in \{p_1, \dots, p_n, q_1, \dots, q_n\}</math>, and so for all of <math>A_n</math>.
{{Collapse bottom}}