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 $φ : R [x] \to R [x] / (f (x))$ sends any polynomial $g (x)$ to $g (a)$ .

What is $f (x)$ ? $f (x) = x - a$
What is $R [x] / (f (x))$ ? It is $R$
Is $φ$ an isomorphism of groups or rings? It's an isomorphism of rings

2. Complex ring, $C$

There is a $f (x)$ so that $φ : R [x] \to R [x] / (f (x))$ makes $R [x] / (f (x))$ isomorphic to $C$ as a ring.

What is $f (x)$ ? $f (x) = x^{2} + 1$
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 $φ : R [x] \to R^{⨁ d}$ 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))} ≅ R^{⨁ d}

Polynomial evaluation

Let $f (x) = x - a$ for some $a \in R$ . And let $φ : R [x] \to R [x] / (x - a)$ 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 $φ (g (x))$ . $r$ is also the value of $g (x)$ when evaluated at $a$ .

Of note is that fact that we have the equivalence:

g (a) = 0 ⟺ 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 $C$ from $R [x]$

Let $f (x) = x^{2} + 1$ . Then $φ : R [x] \to R [x] / (x^{2} + 1)$ 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)} ≅ R ⨁ 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{aligned} a & = a_{1} x + a_{0} \\ b & = b_{1} x + b_{0} \end{aligned}

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

\begin{aligned} a b & = (a_{1} x + a_{0}) (b_{1} x + b_{0}) \\ = a_{1} b_{1} x^{2} + (a_{1} b_{0} + a_{0} b_{1}) x + a_{0} b_{0} \end{aligned}

We can rewrite the last statement:

\begin{aligned} a b & = a_{1} b_{1} x^{2} + (a_{1} b_{0} + a_{0} b_{1}) x + a_{0} b_{0} \\ = (a_{1} b_{1}) (x^{2} + 1) + (a_{1} b_{0} + a_{0} b_{1}) x + a_{0} b_{0} - a_{1} b_{1} \end{aligned}

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_{1} b_{0} + a_{0} b_{1}) x + a_{0} b_{0} - a_{1} b_{1}

If these pairs are from $R ⨁ R$ , then this is the multiplication operation of the complex numbers. And so:

\frac{R [x]}{(x^{2} + 1)} ≅ 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 $φ$ .

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 $4 x^{5} + 3 x^{2} + 1$ , 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