Polynomial u-ring

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Not my strong suit but I've done the proofs for this


Let [ilmath]R[/ilmath] be a u-ring with the standard operations of [ilmath]+[/ilmath] and [ilmath]*[/ilmath]. The set (or further-more: u-ring) of polynomials over [ilmath]R[/ilmath] (in the indeterminate [ilmath]X[/ilmath]), denoted universally as [ilmath]R[X][/ilmath], is defined as follows[1]:

  • Let [ilmath]M:\eq\{e,X,X^2,X^3,\ldots,X^n,\ldots\} [/ilmath] be the free monoid generated by [ilmath]\{X\} [/ilmath].
  • A polynomial, [ilmath]P\in R[X][/ilmath], over [ilmath]R[/ilmath] is a mapping: [ilmath]P:M\rightarrow R[/ilmath] by [ilmath]P:X^n\rightarrow a_n[/ilmath] such that [ilmath]\vert\{a_n\in P(M)\ \vert\ a_n\neq 0\}\vert\in\mathbb{N} [/ilmath][Note 1]
  • The set of such polynomials is [ilmath]R[X][/ilmath] - the set of all polynomials over [ilmath]R[/ilmath]


  • It is normal to identify [ilmath]r\in R[/ilmath] with the constant polynomial[1]: [ilmath]a_n:\eq\left\{\begin{array}{lr}0 & \text{if }n>0\\r&\text{otherwise}\end{array}\right.[/ilmath], which we can write more simply as [ilmath]r[/ilmath]
    • Later we shall write this as [ilmath]r+0X+0X^2+\cdots 0X^n+\cdots[/ilmath] or more simply: [ilmath]r[/ilmath] (hence the "naturalness" of the identification)
  • It is also normal to identify [ilmath]X[/ilmath] with the polynomial [ilmath]a_n:\eq\left\{\begin{array}{lr}0 & \text{if }n\neq 1\\1&\text{otherwise}\end{array}\right.[/ilmath]
    • Later we shall write this as [ilmath]0+1X+0X^2+\cdots+0X^n+\cdots[/ilmath] or simply [ilmath]X[/ilmath] (hence the "naturalness" of the identification)


Claim 1: we may write [ilmath]P\in R[X][/ilmath] as:

  1. A summation: [ilmath]\sum_{n\in\mathbb{N} }a_nX^n[/ilmath] or more commonly: [ilmath]\sum^\infty_{n\eq 0}a_nX^n[/ilmath]; or
  2. [ilmath]P:\eq a_0+a_1X+a_2X^2+\cdots+a_nX^n[/ilmath] for some [ilmath]n\in\mathbb{N} [/ilmath].
    • Such an [ilmath]n[/ilmath] is actually called the degree of the polynomial Caveat:and can be defined only by convention if [ilmath]P[/ilmath] is the zero polynomial

U-ring operations

Claim 2: [ilmath]R[X][/ilmath] is a u-ring itself. The operations are:

  1. [ilmath]P+Q[/ilmath]
  2. [ilmath]PQ[/ilmath]
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The terms are:
  1. [ilmath](P+Q)_n\eq (P)_n+(Q)_n[/ilmath] and
  2. [ilmath](PQ)_n\eq \sum_{i+j\eq n}(P)_i(Q)_j[/ilmath] - meaning the sum over terms whose indices add to [ilmath]n[/ilmath]
Don't forget to flesh out the identity and unity!

Warning:That grade doesn't exist!

Identity (of the abelian additive group) and unity of the ring:

  1. The zero polynomial: [ilmath]0+0X+0X^2+\cdots+0X^n+\cdots[/ilmath] which by the notation section we can simply write as: [ilmath]0[/ilmath]
  2. The unity of the ring is the polynomial [ilmath]1[/ilmath], (the thing we identify the unity of [ilmath]R[/ilmath] to).
    • More explicitly, the polynomial: [ilmath]1+0X+0X^2+\cdots+0X^n+\cdots[/ilmath]

Proof of claims

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Done on paper:

See also


  1. This may be said as:
    1. [ilmath]a_n\eq 0[/ilmath] for almost all [ilmath]n\in\mathbb{N}_0[/ilmath] (as per one of the references - Abstract Algebra: Grillet),
    2. [ilmath]a_n\eq 0[/ilmath] almost always, or
    3. [ilmath]\vert\{a_n\in P(M)\ \vert\ a_n\eq 0\}\vert\eq\aleph_0[/ilmath]
    I have written: [ilmath]\vert\{a_n\in P(M)\ \vert\ a_n\neq 0\}\vert\in\mathbb{N} [/ilmath] - which is quite literally "there are finitely many elements that are the additive identity of [ilmath]R[/ilmath]"


  1. 1.0 1.1 Abstract Algebra - Pierre Antoine Grillet