# Evaluation Homomorphism

Is it true that for any unital ring (not necessarily commutative), that we have a ring homomorphism for all a = (a_1, ....,a_n)$\in R, from R[X_1,.....,X_n] → R given by sending a polynomial f to f(a)? I have only ever seen this for fields. I cannot think of any possible reason it would be false for general rings but I just wanted to make sure. ## Answers and Replies Related Linear and Abstract Algebra News on Phys.org No, it is not true. Take for example the Weyl-algebra. This is [itex]W=\mathbb{Z}<a,b> / (ab-ba-1)$. 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).

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