0

This might be a silly question but i can't find a straight answer on this..

  • What's the domain and range of a polynomial in a polynomial ring R[x] (where R is a ring)?

is it all polynomials $p:R\to R$ (where the coefficients are from R) ? or can the domain/range be another ring S where R is a sub-ring of (but still, the coefficients are from R)?

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Ak2399
  • 217
  • 2
    Honestly, it's best not to think of elements of $R[x]$ as functions. Otherwise, $x^2+1$ and $x+1$ become equal polynomials in $\mathbb{Z}_2[x]$, which they are not. – Randall Jun 11 '24 at 14:10
  • it's a bit problematic because in textbooks I sometimes see them use $ p \in \mathbb{R}[x] $ then plug in an imaginary number in p and its confusing to say the least. – Ak2399 Jun 11 '24 at 14:14
  • Yes, understood. Polynomials in $R[x]$ may be evaluated (formal substitution) where $x$ is replaced by any element of $R$. In your example, they are considering $\mathbb{R}[x] \subset \mathbb{C}[x]$. – Randall Jun 11 '24 at 14:16
  • so I could say that the domain/range can be any ring S where R is a sub-ring of? – Ak2399 Jun 11 '24 at 14:20
  • 1
    That is one possible interpretation, yes. But of course, that isn't standard, so you must declare what this ring $S$ is in your exposition, for the reader's sake. – Randall Jun 11 '24 at 14:24
  • Rather than considering polynomials as functions I think of them as a direct sum $\bigoplus_{\mathbb{N}} R$ where the multiplication is defined in the usual way. Then you can define an evaluation map which is a ring homomorphism $R[x] \to R$ that works like a polynomial function. – CyclotomicField Jun 11 '24 at 15:14

1 Answers1

2

As Randall said in the comments, the elements of a polynomial ring are better viewed as formal polynomials. Meaning that they are not functions at all! See this old thread for a discussion.

In algebra we, of course, still want to evaluate polynomials. We want to plug in many possible entities in place of that $x$. For this to work well we usually restrict ourselves to commutative rings $R$. The reason is that we want evaluation to play nicely together with polynomial arithmetic in the sense that we expect the rules (here $\alpha$ is something we can plug in)

  • $f(x)+g(x)=h(x)\implies f(\alpha)+g(\alpha)=h(\alpha)$,
  • $f(x)g(x)=h(x)\implies f(\alpha)g(\alpha)=h(\alpha)$

to hold for all polynomials $f,g$.

The following is kind of an umbrella result on this theme:

Lemma. Assume that $R$ is a commutative ring, $S$ is another ring that contains $R$ as a subring, and $\alpha\in S$ commutes with all the elements of $R$. That is, $$r\alpha=\alpha r$$ for all $r\in R$. Then the evaluation mapping $ev_\alpha:R[x]\to S$, $$f(x)=\sum_{i=0}^na_ix^i\mapsto \sum_{i=0}^na_i\alpha^i$$ is a homomorphism of rings. In particular, the rules above work.

See this post for a quick explanation as to why commutativity is needed. And this answer for an example as to what goes spectacularly wrong, when we drop commutativity, explaining what goes wrong, when we substitute $x=j$ into the factorization $$x^2+1=(x-i)(x+i).$$ Here $i,j$ are quaternions.

The Lemma means that if $R$ is a commutative ring and $f(x)\in R[x]$, then we can, without much worries, evaluate $f(\alpha)$, when for example

  • $R$ is an integral domain, and $\alpha$ is an element of its field of fractions $K$.
  • Or, if $\alpha$ is an element of an extension field of $K$.
  • Of, if $\alpha$ is a square matrix with entries in a ring that contains $R$ in its center (here we need to adopt the usual convention of identifying $R$ with scalar matrices).
  • But, there is no maximal possible domain covering all the cases. In all cases range/domain need to match at least somewhat.
J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Jyrki Lahtonen
  • 140,891
  • Another comment as an answer. All three of the linked posts are, IMHO, needed to form a joint duplicate target. Everything here is well covered on the site, but many users frown at an explanation like this in comments. – Jyrki Lahtonen Jun 11 '24 at 16:01