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I'm trying to understand the difference between polynomial functions and the evaluation homomorphisms. I noticed that I don't know what's the exact definition of a polynomial function, although I've used it since I was in school.

I'm confused.

Thanks in advance

user42912
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1 Answers1

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  1. A polynomial with coefficients in a commutative ring $R$ is a formal sum $r_n x^n + \ldots + r_1 x + r_0$. It is not a function. It is merely a finite sum of terms of the form $r_i x^i$, where $x$ is some abstract symbol. The ring structure of $R$ lets us make these symbols into a ring, that we denote $R[x]$.

  2. For each $r \in R$, there is a ring homomorphism $e_r:R[x] \rightarrow R$. These are the evaluation maps, defined by replacing each instance of $x$ in the formal sum of a polynomial with $r$ and considering this as an element of $R$. We sometimes denote $e_r(f(x))$ as $f(r)$, but this is not the same as the formal polynomial $f(x)$. $f(r)$ for $r \in R$ is actually an element of the ring, while the formal polynomial $f(x)$ is not.

  3. For each polynomial $f(x) \in R[x]$, there is a map $R \rightarrow R$ defined by $r \mapsto e_r(f(x)) = f(r)$. This is not a ring homomorphism! (For instance, the polynomial function on $\mathbb{R}$ defined by $f(x) = x^2$ is not a homomorphism of rings.) It is a function determined by a polynomial, and so we call such a function a polynomial function.

All of these things can be generalized to polynomials in multiple variables, as well.

A note: Why the fuss about formal polynomials and the functions they define? Typically, you're safe not making a distinction between the formal sum and the function it defines. However, some issues can arise and you need to be aware of them. For instance, consider the polynomial $x^2 - x = (x-1)x$ with coefficients in the ring $\mathbb{F}_2$, the field with two elements. This polynomial, formally, is not zero... it has nonzero coefficients! However, the function it defines on $\mathbb{F}_2$ is the zero function, since there are only two points in $\mathbb{F}_2$ and the evaluation homomorphism for each point sends the function to zero.

Zach L.
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    Tangentially, an oft-overlooked point is that the evaluation map is not a ring homomorphism if $R$ is noncommutative. – anon May 25 '13 at 02:40
  • Added "commutative" to cover my bases! – Zach L. May 25 '13 at 02:41
  • thanks a lot, great answer, that's what I needed. – user42912 May 25 '13 at 03:22
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    The fuss about formal polynomials and functions can be completely resolved if you recognize that evaluation maps are naturally defined on all $R$-algebras: if $S$ is any $R$-algebra, then any element $s \in S$ defines an evaluation map $e_r : R[x] \to S$. A polynomial over a finite field is completely determined by its behavior as a function provided that you work over at least every finite extension of the field. – Qiaochu Yuan May 25 '13 at 04:22