# Subrings generated by a set

I have a misunderstanding with subrings generated by a set. (The intersection of all subrings of R containing a set X and a subring R0 of R is the subring generated by X over R0 in the ring R). Firstly, this definition seems not intuitive. Also I have a more specific issue: on the one hand, for the definition to work the generating set must be contained in the ring $R$. On the other hand, when talking about polynomials there is a statement asserting that $$R[x]= \{ \text{formal } a_0 + \ldots + a_n x^n \text{ where } a_0, \ldots, a_n \in R\}$$ is the ring generated over $R$ by $x$. The problem I have with the definition is that $x$ isn't in $R$. (Moreover, what is $a_n x^n$ formally if $x$ isn't in $R$?)

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$R[x]$ isn’t a subring of $R$; it’s the ring of polynomials over $R$. – Brian M. Scott Oct 21 '12 at 19:44
If you're going to complain about a definition not being intuitive, you should at least say what definition you're talking about. – Chris Eagle Oct 21 '12 at 20:13
@Chris added the definition in the beginning – Idan Oct 21 '12 at 20:55

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.

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And what about R[x] being the generated ring by x over R? – Idan Oct 21 '12 at 20:56
@Idan: Are you asking why it’s sometimes called the ring generated by $x$ over $R$? That’s simply a name, and not one that I’ve seen very often, either; at any rate, it has nothing to do with the the subring of $R$ generated by a subset $X\subseteq R$. These are two unrelated ways of generating new rings from $R$; one way gives a subring, and the other gives something completely new. Any similarity in terminology simply reflects the fact that we are somehow building a new ring out of $R$. – Brian M. Scott Oct 21 '12 at 21:02
Though, it is worth noting that the polynomial algebra $R[x]$ is indeed generated by $x$ as a subring of itself, which is important when describing the $R$-algebra homomorphisms $R[x] \to A$. – Zhen Lin Oct 21 '12 at 21:34
@ZhenLin: True, but only, I think, after the basic confusion has been cleared up. (Thanks for catching the typo.) – Brian M. Scott Oct 21 '12 at 21:36