Let $R$ be a ring, then define the polynomial space $R[x]$ as

$$R[x]=\{ (a_0, a_1, \ldots) : a_j \neq 0 \text{ for a finite number of } a_j \in R \}.$$

So formally, a polynomial is just a tuple. But how can I imagine it as function? What is $f(c)$ formally? It's clear that $f(c)$ is just the value of the polynomial applied on $c$ but how does this harmonize with the definition of $f$ not as function but as tuple of coefficients?

And what is $x$ in $R[x]$? It doesn't appear in the definition so what is the difference between $R[x]$ and $R[y]$? Why does one specifies the polynomial space on $x$?


1 Answer 1


Of course, a tuple is not a function. Formally, you have a natural morphism $\ell: R[x] \rightarrow \{f:R \rightarrow R\}$ given by $\sum_n{a_nx^n} \longmapsto (c \longmapsto \sum_n{a_nc^n})$.

So, if $f$ is a polynomial and $c \in R$, then what is denoted as $f(c)$ is actually $(\ell(f))(c)$.

In $R[x]$, $x$ is formally how you name the sequence $(0,1,0,\ldots)$. You specify it because it’s nicer to write expressions as sums of powers of $x$ than as stationary sequences.

Moreover, there are polynomials in several variables, and it’s then nice to be able to differentiate between several variables. It is also useful to specify $R[x]$ when $R$ is already a ring of polynomials (or, eg, formal series), with respect to another variable, so that you don’t mix them up.

  • $\begingroup$ But why is - notation wise - $\sum_n a_n x^n \in R[x]$? How and especially why can I interpret a tuple as sum? $\endgroup$
    – ATW
    May 2, 2020 at 11:06
  • 1
    $\begingroup$ Actually, what happens is that you can add and multiply sequences between themselves (addition is pointwise, product follows the rule for polynomials) and multiply a given sequence by an element of $R$ (pointwise). So, for each $n$, $a_nx^n$, in the definition of my post for $x$, is the sequence that is $0$ everywhere outside $n$ and is $a_n$ at $n$. So $\sum_n{a_nx^n}$ is the sequence that is at any $n$, $a_n$. $\endgroup$
    – Aphelli
    May 2, 2020 at 15:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .