Is There a Ring Homomorphism for All Unital Rings in Evaluation Homomorphism?

[Monobrow]

Is it true that for any unital ring (not necessarily commutative), that we have a ring homomorphism for all a = (a_1, ...,a_n)[itex]\in R, from R[X_1,...,X_n] → R given by sending a polynomial f to f(a)? I have only ever seen this for fields. I cannot think of any possible reason it would be false for general rings but I just wanted to make sure.

 

[jbunniii]

It is true for commutative unital rings. The evaluation homomorphism exists and is unique. See for example Rotman's Advanced Modern Algebra, 2nd edition, Theorem 2.25.

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

 

[micromass]

No, it is not true.

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

Anyway, consider the polynomial ring $W[X,Y]$. 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 $R<X_1,X_2,...,X_n>$. These are polynomial rings in which the $X_i$ and the $X_j$ do not commute. In fact, there are no relations between the $X_i$ and the $X_j$ 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).

 

[Monobrow]

Thanks a lot. It was the non-commutative case I was worried about and it's good to see why it doesn't work.

 

[mathwonk]

I think you need to be a little careful even to define polynomial rings and evaluation over non commutative R. I believe one usually assumes R[X1,...,Xn] is a ring in which the elements Xi are in the center. Then, given an element a of R which may not commute with the coefficients of a given polynomial f, one has two evaluations of f at a, left evaluation and right evaluation. Nonetheless these two maps are useful; in particular there is a left and a right division algorithm, and two remainder theorems, allowing you to compute the remainder of both left and right division of f by (X-a).

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