Math and science::Algebra::Aluffi

Group isomorphism: quotient of R[x] ≅ direct sum of R

In what way can we construct an isomorphism using $R [x]$ and $R$ ?

If we restrict our aim to just a group isomorphism then we can achieve the following:

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 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}

Can you construct the proof?

Extra:

How to construct the complex ring $C$ ?
How to construct a ring of polynomial evaluations?

$R [x] / (f (x)) ≅ R^{⨁ d}$ . Proof idea.

If $f (x)$ is a monic polynomial of degree $d$ , then mapping $g (x)$ to $r (x)$ in $g (x) = q (x) f (x) + r (x)$ is a is a function from elements of $R [x]$ to elements of $R^{⨁ d}$ . We will call this mapping $φ$ . It turns out that this mapping meets the requirements to be a homomorphism of abelian groups (for the addition operation). A taster of this is that if $g_{3} (x) = g_{1} (x) + g_{2} (x)$ then $r_{3} (x) = r_{1} (x) + r_{2} (x)$ , and so the mapping preserves addition. What is left is to show preservation of the identity and inverses.

We then turn to the homomorphism decomposition theorem: $φ$ should be decomposable into a surjective homomorphism followed by an injective homomorphism to the image. The surjective part is from $R [x]$ to $R [x] / \ker φ$ . The injective part is from this set to $im φ$ . It turns out that $\ker φ = (f (x))$ , and so we have $R [x] / (f (x)) ≅ some subset of R^{⨁ d}$ . But $φ$ is itself surjective so the codomain is all of $R^{⨁ d}$ . The function is surjective as every polynomial of degree less than $d$ is its own remainder, and thus fills up the entire codomain of d-tuples. And so we end up with $R [x] / (f (x)) ≅ R^{⨁ d}$ .

The actual proof is best broken down into some lemmas.

Lemma 1

$φ$ is a surjective function.

Proof. $φ$ is a well-defined function because we can always calculate a unique remainder of $g (x)$ when divided by $f (x)$ (shown previously). $φ$ is surjective, as every polynomial of degree less than $d$ is its own remainder when divided by $f (x)$ , and so for any element in $R^{⨁ d}$ we can find a polynomial of degree less than $d$ that maps to it. (Remember: $f (x)$ itself would require $d + 1$ coordinates to represent.)

Lemma 2

$φ$ is a abelian group homomorphism.

Proof. Let $g_{1} (x), g_{2} (x) \in R [x]$ . We must show that:

$φ (g_{1} (x) + g_{2} (x)) = φ (g_{1} (x)) + φ (g_{2} (x))$
$φ (0) = 0$
$φ (- g (x)) = - φ (g (x))$

Let $g_{1} (x), g_{2} (x) \in R [x]$ . We can write:

$\begin{aligned} g_{1} (x) & = q_{1} (x) f (x) + r_{1} (x) \\ g_{2} (x) & = q_{2} (x) f (x) + r_{2} (x) \end{aligned}$

for some $q_{1} (x), q_{2} (x), r_{1} (x), r_{2} (x) \in R [x]$ , unique to $g_{1} (x)$ and $g_{2} (x)$ . Then:

$g_{1} (x) + g_{2} (x) = (q_{1} (x) + q_{2} (x)) f (x) + r_{1} (x) + r_{2} (x)$

As $r_{1} (x) = φ (g_{1} (x))$ and $r_{2} (x) = φ (g_{2} (x))$ , we have:

$\begin{aligned} φ (g_{1} (x) + g_{2} (x)) & = r_{1} (x) + r_{2} (x) \\ = φ (g_{1} (x)) + φ (g_{2} (x)) \end{aligned}$

as desired.
$0 = 0 f (x) + 0$ , and so $φ (0) = 0$ .
TODO!

Lemma 3

$\ker φ = (f (x))$ .

Proof. Let $g (x) \in \ker φ$ . Then $φ (g (x)) = 0$ . We know that:

φ (g (x)) = 0 ⟺ g (x) = q (x) f (x) + 0

and so

g (x)

is in the ideal generated by

f (x)

. In other notation,

g (x) \in (f (x))

). Thus,

\ker φ = (f (x))

$R [x] / (f (x)) ≅ R^{⨁ d}$ . Proof.

$φ$ is a surjective group homomorphism, by Lemma 1 and Lemma 2. By the first isomorphism theorem for groups, we have:

R [x] / \ker φ ≅ R^{⨁ d}

By Lemma 3, $\ker φ = (f (x))$ , and so:

R [x] / (f (x)) ≅ R^{⨁ d}

as desired.

As a recap, here is the first isomorphism theorem for groups:

First isomorphism theorem for groups

Suppose $φ : G \to H$ is a surjective group homomorphism. Then:

H ≅ G / \ker φ

$d$ and not $d + 1$

Why is the dimension of the vector space $R^{⨁ d}$ and not $R^{⨁ d + 1}$ ? We need $d$ coordinates to represent a polynomial of degree $d - 1$ because we also need 1 coordinate for the 0th term.

$R [x] / (f (x)) ≅ R^{⨁ d}$ . Motivation and proof idea.

One way to construct an isomorphism between $R [x]$ and $R^{⨁ d}$ is to restrict the polynomial ring to only contain polynomials of degree less than $d$ and then only consider the addition operation. This is a bit pointless, as the now reduced $R [x]$ group is practically identical to $R^{⨁ d)}$ —it just uses $x, x^{1}, x^{2}$ etc. as notation instead of commas that might be used for writing down elements of $R^{⨁ d)}$ .

Another way to restrict ourselves to polynomials of $\deg \leq d$ is to "project" all polynomials onto the smaller set of polynomials of degree less than or equal to $d$ . One way to project $g (x) \in R [x]$ onto the set of polynomials of degree less than or equal to $d$ is first fix a polynomial $f (x)$ of degree $d$ and then map $g (x)$ to the remainder of $g (x)$ when divided by $f (x)$ . The uniqueness of the remainder given $g (x)$ and $f (x)$ insures that this projection is a set function. What is left to do is show that the function maps the $+$ operation in $R [x]$ to the $+$ operation in $R^{⨁ d}$ . This is analogous to showing that for integers $(a + b) \mod c$ is the same as $(a \mod c) + (b \mod c)$ .

Example

TODO: add the complex number example

Source

Aluffi p148