This introduction to polynomial rings is very brief, but it does make the important points. Here’s a slightly different way to think about it that may help to clear up some of the confusion.
Given a ring $R$, we can form a new ring, which for a moment I’ll call $R^*$, whose elements are infinite sequences of elements of $R$ that are have only finitely many non-zero terms:
$$R^*=\left\{\langle r_k:k\in\Bbb N\rangle\in{^{\Bbb N}R}:\exists m\in\Bbb N~\forall k\ge m(r_k=0_R)\right\}\;.$$
Addition in $R^*$ is component-wise: $$\langle r_k:k\in\Bbb N\rangle+\langle s_k:k\in\Bbb N\rangle=\langle r_k+s_k:k\in\Bbb N\rangle\;.$$
Multiplication is the Cauchy product: if $\bar r=\langle r_k:k\in\Bbb N\rangle$ and $\bar s=\langle s_k:k\in\Bbb N\rangle$, then $$\bar r\bar s=\langle t_k:k\in\Bbb N\rangle\;,\text{ where }t_k=\sum_{i=0}^kr_is_{k-i}\;.$$
It’s not hard to verify that these really are operations on $R^*$. In particular, if $r_k=0_R$ for $k\ge m$, and $s_k=0_R$ for $k\ge n$, then $t_k=0_R$ for $k\ge m+n-1$.
$R^*$ also contains a nice embedded copy of $R$:
$$R\hookrightarrow R^*:r\mapsto\langle r,0_R,0_R,0_R,\dots\rangle\;.$$
The polynomial ring $R[x]$ is just this $R^*$ in disguise. For $\bar r=\langle r_k:k\in\Bbb N\rangle$ define $\deg\bar r$, the degree of $\bar r$, to be $-1$ if $\bar r=0_{R^*}=\langle 0_R,0_R,0_R,\dots\rangle$, and otherwise to be the least $m\in\Bbb N$ such that $a_k=0_R$ for all $k>m$. If $0_{R^*}\ne\bar r\in R^*$, and $\deg\bar r=m$, then all of the information about $\bar r$ is contained in the finite sequence $\langle r_0,r_1,\dots,r_m\rangle$. We can just as well present this information in the form $$r_0+r_1x+r_2x^2+\ldots+r_mx^m\;,\tag{1}$$ where $x$ is a new symbol not in $R$ whose rôle is to carry the exponent that tells which term of the sequence $\bar r$ is which. Instead of writing $r_0,\dots,r_m$ as a sequence, and using the position in the sequence to keep the terms straight, we write the polynomial $(1)$ and use the symbols $x^k$ to keep the terms straight.
It’s routine to check that if you endow this family $R[x]$ of polynomials with the usual operations of polynomial addition and multiplication, the map that sends $\bar r\in R^*$ of degree $m\ge 0$ to $r_0+r_1x+r_2x^2+\ldots+r_mx^m\in R[x]$ and $0_{R^*}$ to the zero polynomial is an isomorphism of $R^*$ and $R[x]$.
The thing to remember is that these polynomials in $R[x]$ are just a way to line up finite sequences of elements of $R$; they should not be thought of as functions. This is made very clear by the $R^*$ representation of them, but the $R[x]$ representation has the great advantage that the manipulations, including multiplication, are already familiar.
Best Answer
A sub-ring of a ring $(R,+,\times)$ is a non-empty subset $R'$ that is stable under + and x and satisfies the axioms for a ring with the binary operations + restricted to $R'$ and x restricted to $R'.$ Note that for any $g \in R$ we can define $zg$ for any integer $z$ and $g^m$ for any positive integer $m$ whether or not $R$ contains an identity element. Thus if $f \in R$ we define $$\langle f\rangle=\left\{\sum_{i=1}^na_if^{n+1-i}\>\Biggm|\> n \in \mathbb Z,n>0;a_i \in \mathbb Z \text { for } 1 \le i \le n\right\}.$$ Then $\langle f\rangle$ is a subset of $R$ that contains $f$ and is a sub-ring of $R.$ Indeed, $\langle f\rangle$ is the smallest subset of $R$ that contains $f$ and is a sub-ring of $R$. We call $\langle f\rangle$ the sub-ring of $R$ generated by $f$.
If $R$ is a ring with identity 1, these ideas must be modified in several ways. First, we can, for any integer $z$, write $z$ as a symbol for a member of $R$ that is the sum of a finite number of $1$'s or of $-1$'s depending on whether $z$ is positive or negative as an integer. But we must remember that $z$ as an integer is not the same as $z$ as a member of $R.$ Then for $f \in R$ we define $$\langle f\rangle=\left\{\sum_{i=0}^na_if^{n-i}\>\Biggm|\> n \in \mathbb Z,n \ge 0;a_i \in \mathbb Z \text { for } 0 \le i \le n\right\}.$$ Then $\langle f\rangle$ is a ring-with-identity 1 and is the smallest ring-with-identity that contains $f$.