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I have tried to solve it like this: let F be a vector field of characteristic 0 an A1=F[T,D], T,D are from EndF[ω] over F. D is differential operator and T(f(ω))=ωf(ω).In A1 algebra [T,D]=I. I is identity operator. I must show that φ:A1→Mn(E), F⊆E is not isomrphism for any n. The Weyl algebra A1 does not contain any division ring larger then F and also is infinite dimensional over a field F. Since dimMn(E)<∞ there can not be isomorphism from an infinite dimensional space to a finite dimensional vector space Mn(E). Lets prove that A1 does not contain any division ring larger than F. It is enough to prove that any nonconstant differential operator is not invertible. The derivative D is a linear operator D:F[ω]→F[ω] and this is endomorphism. Then KerD is a set of constant functions. A general result from the Linear algebra is that a linear map is one-to-one if and only if kernel is 0. This shows that a nonconstant derivative operator is not invertible.

Im not sure if this is a good solution for this problem and I really need some help. Thanks!!

XYZ
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  • Can we not assume that you know such algebras have no nonzero zero divisors? Or maybe that is not true for $A_1$ over all fields? – rschwieb Oct 23 '16 at 21:07
  • I mention this because if that is known, there are simple arguments. Namely, that a matrix ring with $n>1$ always has zero divisors, so to be a matrix ring $n=1$, but that means it is a division ring (contradiction.) Or alternatively: a matrix ring over a division ring is Artinian, and if the Weyl algebra were an Artinian domain, it would be a division ring (contradiction.) – rschwieb Oct 24 '16 at 13:51

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Such Weyl algebra has no finite dimensional representation, but matrix rings do. In fact you can observe there is no pair of matrices over a field of zero charactristic whose commutator is the identity, since a commutator is traceless but the identity matrix is not (in charactristic zero!).

Pedro
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