\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
deepdream of a sidewalk
Show Question
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\groupAdd}[1] { +_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \require{mathtools} \)
Math and science::Algebra::Aluffi

Two new rings from polynomial ring quotients

Using the below theorem:

1. Polynomial evaluation

There is a \( f(x) \) so that \( \varphi : R[x] \to R[x]/(f(x)) \) sends any polynomial \( g(x) \) to \( g(a) \).

  1. What is \( f(x) \)? \( f(x) = x - a \)
  2. What is \( R[x]/(f(x)) \)? It is \( R \)
  3. Is \( \varphi \) an isomorphism of groups or rings? It's an isomorphism of rings
2. Complex ring, \( \mathbb{C} \)

There is a \( f(x) \) so that \( \varphi : R[x] \to R[x]/(f(x)) \) makes \( R[x]/(f(x)) \) isomorphic to \( \mathbb{C} \) as a ring.

  1. What is \( f(x) \)? \( f(x) = x^2 +1 \)
  2. How is the multiplication operation transferred to \( R[x]/(f(x)) \)? See below.

Let \( R \) be a commutative ring, \( R[x] \) a polynomial ring, and \( f(x) \in R[x] \) a polynomial. Let \( \varphi: R[x] \to R^{\bigoplus d} \) be a set function defined by by sending \( g(x) \in R[x] \) to the tuple representation of the remainder of \( g(x) \) when divided by \( f(x) \).

Then this function induces an isomorphism of abelian groups:

\[\frac{R[x]}{(f(x))} \cong R^{\bigoplus d} \]

Polynomial evaluation

Let \( f(x) = x - a \) for some \( a \in R \). And let \( \varphi : R[x] \to R[x]/(x-a) \) be the function that sends a polynomial in \( R[x] \) to its remainder when divided by \( x-a \). Then for some \( g(x) \in R[x] \), we can write:

\[ g(x) = q(x)(x-a) + r \]

where \( r \in R \) is the value of \( \varphi(g(x)) \). \( r \) is also the value of \( g(x) \) when evaluated at \( a \).

Of note is that fact that we have the equivalence:

\[ g(a) = 0 \iff g(x) \in (x-a) \]

What have we done? We have "projected" all polynomials in \( R[x] \) to the much smaller \( R \). Polynomials which experience this mapping as an identity are those which have no contribution from \( x-a \).

Creating \( \mathbb{C} \) from \( R[x] \)

Let \( f(x) = x^2 + 1 \). Then \( \varphi : R[x] \to R[x]/(x^2 + 1) \) sends a polynomial \( g(x) \) to its remainder when divided by \( x^2 + 1 \). As per the theorem above, this induces an isomorphism of abelian groups:

\[ \frac{R[x]}{(x^2 + 1)} \cong R \bigoplus R \]

But we can go further by transferring the multiplication operation. Let \( a, b \in \frac{R[x]}{(x^2 + 1)} \). We must have both:

\[ \begin{align} a &= a_1x + a_0 \\ b &= b_1x + b_0 \end{align} \]

as they are remainders with degree less than 2. We can try to multiply them as if they were polynomials in \( R[x] \):

\[ \begin{align} ab &= (a_1x + a_0)(b_1x + b_0) \\ &= a_1b_1x^2 + (a_1b_0 + a_0b_1)x + a_0b_0 \\ \end{align} \]

We can rewrite the last statement:

\[ \begin{align} ab &= a_1b_1x^2 + (a_1b_0 + a_0b_1)x + a_0b_0 \\ &= (a_1b_1)(x^2 + 1) + (a_1b_0 + a_0b_1)x + a_0b_0 - a_1b_1 \\ \end{align} \]

And taking the remainder after the quotient by \( x^2 + 1 \), forces us to equate:

\[ (a_1, a_0) \cdot (b_1, b_0) = (a_1b_0 + a_0b_1)x + a_0b_0 - a_1b_1 \]

If these pairs are from \(\mathbb{R} \bigoplus \mathbb{R} \), then this is the multiplication operation of the complex numbers. And so:

\[ \frac{R[x]}{(x^2 + 1)} \cong \mathbb{C} \]

From criteria to \( f(x) \)

As an example, consider the criteria: a polynomial ring \( R[x] \) and a value for \( a_0 = x \) such that \( a_0^3 = -1 \). For every polynomial in \( g(x) \in R[x] \) when evaluated with \( x = a_0 \), if we were to divide it by \( x^3 + 1 \), the remainder would be \( g(a_0) \), as any multiple of \( a_0^3 + 1 \) would be zero, and so \( g(a_0) \) has no contribution from the polynomial \( (x^3 + 1) \).

In the "evaluation" example, the criteria was \( x = a_0 \), and when this criteria is met, then the polynomial \( x - a_0 \) has no contribution to \( g(x) \), and so the remainder of \( g(x) \) when divided by \( x - a_0 \) is just \( g(a_0) \).

What is amazing, and something I don't understand yet, is how the sets of polynomials (or their evaluations?) form a group and possibly a ring, by transferring the operations from \( R[x] \) through the quotient map \( \varphi \).

Computation perspective

Remember that \( R[x] \) expresses all possible \( (+, \cdot ) \) computation networks where there is 1 "wire" whose value has not been set yet, and this wire can be connected to an arbitrary number of \( + \) or \( \cdot \) bivariate operations or "gates". \( R[x, y] \) would have two free wires.

With just one free wire, \( x \), any computation can be evaluated, like lambda terms are evaluated, in order to reach a reduced form which is a polynomial like \( 4x^5 + 3x^2 + 1 \), which can be achieved with one 3-input add-gate, one 6 input mult-gate and one 3-input mult-gate. Expressing \( g(x) \) as \( f(x)q(x) + r \) is a way of factorizing the computation network.

Source

Aluffi p149