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Evaluation Homomorphism

  1. Feb 20, 2013 #1
    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.
     
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  3. Feb 20, 2013 #2

    jbunniii

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  4. Feb 21, 2013 #3

    micromass

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    No, it is not true.

    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).
     
  5. Feb 21, 2013 #4
    Thanks a lot. It was the non-commutative case I was worried about and it's good to see why it doesn't work.
     
  6. Feb 25, 2013 #5

    mathwonk

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    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.
     
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