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I have trouble with understanding this sentence in the context of polynomial rings: "We will assume that polynomials satisfy right evaluation. That is, a polynomial can only be evaluated once it is written so that powers of $x$ appear to the right of their coefficients".

I don't underatand this statement. I really appreciate if someone could explain it to me. Thanks in advance.

Mahtab
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  • It seems that they allow "scalars" from a noncommutative ring, but some more context would be really helpful. – Amateur_Algebraist Feb 28 '24 at 12:51
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    @Amateur_Algebraist yes, coefficients come from a noncommutative ring. – Mahtab Feb 28 '24 at 13:00
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    In fact, the author of the paper (I'm reading) works with polynomials with non-commuting coefficients (Null ideals of subsets of matrix rings over fields by Werner). – Mahtab Feb 28 '24 at 13:03

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(clarification from comments: this is from "Null ideals of subsets of matrix rings over fields" by N.J.Werner, and the polynomials are over a noncommutative ring)

They define the evaluation map $ev_r$: $\ldots \to R$ only on polynomials of form $f(x) = \sum\limits_{i=0}^n a_i x^i$ with $f \mapsto \sum\limits_{i=0}^n a_i r^i$. Unlike the commutative case, this definition need not extend to a homomorphism $R[x] \to R$, hence the need for a convention.

Edit: Noncommutative rings and the evaluation homomorphism

Amateur_Algebraist
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  • Dear @Amateur_Algebraist , Thank you so much for the answer. I got your explanations but could you please explain me these sententes ""We will assume that polynomials satisfy right evaluation. That is, a polynomial can only be evaluated once it is written so that powers of $x$ appear to the right of their coefficients"? – Mahtab Feb 29 '24 at 12:55
  • It means that $x^i a$ $(a\in R)$ can not be evalutaed. Am I right? – Mahtab Feb 29 '24 at 12:59
  • Indeed ,why we can't evaluate polynomials of the form $\sum_i x^i \alpha_i$? – Mahtab Feb 29 '24 at 13:04
  • We can, but the author wants to avoid confusion. Note that just prior they assume that "$x$ commutes with all elements of $R$", so e.g. $ax$ and $xa$ are considered to be the same polynomial. (In other words, the polynomials are uniquely defined by their coefficient lists, as in the commutative case.) But if $a$ is noncentral, these two forms will give different results upon evaluation. – Amateur_Algebraist Feb 29 '24 at 20:15
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    I got it now. I really appreciate you for your nice explanation. – Mahtab Mar 01 '24 at 14:44