Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $R$?)

share|improve this question
3  
$R[x]$ 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

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.