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- Thread starter Monobrow
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I don't know anything about polynomials over noncommutative rings, but this might be worth a look:

http://math.stackexchange.com/quest...ension-for-noncommutative-rings-in-this-proof

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Take for example the Weyl-algebra. This is [itex]W=\mathbb{Z}<a,b> / (ab-ba-1)[/itex]. So it is the algebra generated by a and b such that [a,b]=1.

Anyway, consider the polynomial ring [itex]W[X,Y][/itex]. There does not exist a homomorphism T such that T(X)=a and T(Y)=b.

Indeed. If it would exist, then T(XY)=ab and T(YX)=ba. But XY=YX, and thus ab=ba. So it fails.

The problem is that polynomial rings have an inherent commutativity which is not compatible with noncommutative rings.

Very related to your question is the question about free objects of commutative R-algebras (with R commutative). The polynomial ring is a free object for this algebra. But as soon as you allow the algebras (and R) to be noncommutative, then it is not a free object anymore.

The free object in such algebra is given by [itex]R<X_1,X_2,...,X_n>[/itex]. These are polynomial rings in which the [itex]X_i[/itex] and the [itex]X_j[/itex] do not commute. In fact, there are no relations between the [itex]X_i[/itex] and the [itex]X_j[/itex] at all.

The property you mention in the OP fails for polynomial rings in noncommutative setting, but it holds for the free algebra I mentioned above (in commutative or noncommutative setting).

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mathwonk

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This is all discussed on pages 37-41 of Fundamental Concepts of Higher Algebra by A.A.Albert.

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