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Subsections

Rings of polynomials

Rings

Despite their usefulness in a variety of areas, groups are still limited because they have only one operation associated with them. The next algebraic category to work with is a ring.
$\begin{definition} A {\bf ring} $\la R,+,\cdot\ra$ is a set $R$ with two bina... ...isplaymath} (a+b)c = (ac)+(bc) \end{displaymath} \end{enumerate}\end{definition}$
Notice that we do not require that the multiplication operation form a group: there may not be multiplicative inverses in a ring.
$\begin{example} The set of $\matsize{2}{2}$ matrices under usual definitions of addition and multiplication form a ring. \end{example}$

$\begin{example} $\la \Zbb_6,+,\cdot\ra$ forms a ring. Recall that multiplica... ...roup. But $\Zbb_6$ still satisfies the requirements to be a ring. \end{example}$

$\begin{definition} In a ring let $a \in R$, $R$ a ring and let $na$ denote $... ... such positive integer exists, the $R$ is of characteristic 0. \end{definition}$

$\begin{example} In the ring $\Zbb_6$, the characteristic is 6. In general, in t... ...aracteristic is $n$. In the ring $\Qbb$, the characteristic is 0. \end{example}$

Rings of polynomials

Let be a ring. A polynomial of degree with coefficients in is

$\begin{displaymath}f(x) = \sum_{i=0}^{n} a_i x^i \end{displaymath}$

where $a_i \neq 0$ . The symbol

is said to be an indeterminate. If the coefficient of the highest power of

is equal to 1, the polynomial is said to be monic. The set of all polynomials with an indeterminate

with coefficients in a ring

is denoted as

.
$\begin{example} Let $R = \la \Zbb_6,+,\cdot\ra$, and let $S = R[x] = \Zbb_6[x]$... ...}\begin{displaymath}(4+2x)(5+4x) = 2 + 2x + 2x^2. \end{displaymath}\end{example}$

$\begin{example} $\Zbb_2[x]$ is the ring of polynomials with coefficients that ... ..., note that \begin{displaymath}(1+x)(1+x) = 1+x^2 \end{displaymath}\end{example}$
It is clear that polynomial multiplication does not, in general, have an inverse. For example, in the ring of polynomials with real coefficients $\Rbb[x]$ , there is no polynomial solution

$\begin{displaymath}f(x)(x^2+3x+1) = x^3 + 2x+1 \end{displaymath}$

One reason polynomials are of interest in signal processing is that polynomial multiplication is equivalent to convolution. The convolution of the sequence

$\begin{displaymath}\abf = \{a_0,a_1,a_2,\ldots,a_n\} \end{displaymath}$

with the sequence

$\begin{displaymath}\bbf = \{ b_0, b_1,b_2, \ldots, b_m\} \end{displaymath}$

can be accomplished by forming the polynomials

$\begin{displaymath}a(x) = a_0 + a_1x + a_2x^2 + \cdots +a_n x^n \end{displaymath}$

$\begin{displaymath}b(x) = b_0 + b_1x + b_2x^2 + \cdots +b_m x^m \end{displaymath}$

and multiplying them

$\begin{displaymath}c(x) = a(x)b(x). \end{displaymath}$

Then the coefficients of

$\begin{displaymath}c(x) = c_0 +c_1x + c_2 x^2 + \cdots + c_{n+m} x^{n+m} \end{displaymath}$

are equal to the values obtained by convolving $\abf * \bbf$ .

In addition to the representing the arithmetic operations on sequences, polynomials can be used to represent a shift data. For the sequence

$\begin{displaymath}\abf = \{a_0,a_1,\ldots,a_n\} \end{displaymath}$

a shifted version of the data, represented by the operator $\sigma \abf$ is

$\begin{displaymath}\sigma \abf = \{0,a_0,a_1,\ldots,a_n\}. \end{displaymath}$

This shift can be represented using polynomials as a multiplication by

. If

is the polynomial representing $\abf$ , then

is the polynomial representing $\sigma \abf$ .

Just as it is possible to define addition and multiplication modulo a number (as in $\Zbb_5$ or $\Zbb_2$ ), it is also possible to define multiplication modulo a polynomial. To clarify, if we write $p(x) \pmod{q(x)}$ , what is commonly meant is to divide by and to take the remainder using conventional polynomial long division. Thus in $\Rbb[x]$ ,

$\begin{displaymath}x^2+3x+2 \pmod{x+2} = 0 \end{displaymath}$

since there is no remainder, and

$\begin{displaymath}x^2 + 3x+4 \pmod{x+4} = -x+4. \end{displaymath}$

Cyclic convolution can also be represented using polynomial multiplication. Cyclic convolution on points is equivalent to multiplication of polynomials modulo . We will denote the -point cyclic convolution of the sequence $\abf$ with the sequence $\bbf$ as $\abf \cycc \bbf$ or, to emphasize the length, $\abf \cycc_n \bbf$ .
$\begin{example} The 5-point cyclic convolution of $\abf = \{1,1,2\}$ and $\bb... ...oefficients are in correct order, highest degree to lowest degree. \end{example}$
It may be observed that a cyclic shift (wrap around shift) on the data can be accomplished using multiplication modulo a polynomial. Let the -cyclic shift on the sequence

$\begin{displaymath}\{a_0,a_1,\ldots,a_{n-1}\} \end{displaymath}$

be defined by

$\begin{displaymath}\sigma_n \{a_0,a_1,\ldots,a_{n-1}\} = \{a_{n-1},a_0,a_1,\ldots,a_{n-2}\} \end{displaymath}$

This can be represented using polynomials as $x a(x) \pmod(x^n-1)$ .
$\begin{example} Let $\abf = \{1,2,3,4,5\}$, and do a 5-cyclic shift on it: \beg... ...+4x^3+5x^4$, and $x a(x) \pmod{x^n-1} = 5 + x+2x^2 + 3x^3 + 4x^4$. \end{example}$

Next: Fields Up: lecture3 Previous: Groups

Todd Moon 2009-11-06