Polynomial ring

Revision as of 16:43, 26 March 2009 by Jam (talk | contribs) (Formal definition)

Given a (commutative) ring $R$, the polynomial ring $R[x]$ is, informally, "the ring of all polynomials in $x$ with coefficients in $R$." \[R[x]=\left\lbrace\sum_{i=0}^\infty a_ix^i\mid a_i\in R\right\rbrace\]

Formal Definition

We can rigorously define $R[x]$ to be the set of all sequences of elements of $R$ with only finitely many terms nonzero: \[R[x] = \{(a_0,a_1,a_2,\ldots)|\text{the set }\{i|a_i\neq 0\} \text{ is finite }\}\] The we call the elements of $R[x]$ polynomials (over $R$). For a polynomial $p=(a_0,a_1,a_2,\ldots)$, the terms $a_0,a_1,a_2,\ldots$ are called the coefficients of $p$.

For example, $(0,0,0,\ldots), (0,1,0,0,\ldots), (1,4,0,3,0,0,\ldots)$ would be considered polynomials, but $(1,1,1,1,\ldots)$ would not be.

At this point, our formal definition of a polynomial may seem unrelated to our intuitive notion of a polynomial. To relate these two concepts, we introduce the following notation:

We will denote the polynomial $(a_0,a_1,a_2,\ldots)$ by $a_0+a_1x+a_2x^2+\cdots$. For instance we would write:

$(0,0,0,)=0+0x+0x2+(0,1,0,0,)=0+1x+0x2+0x3+(1,4,0,3,0,0,)=1+4x+0x2+3x3+0x4+0x5+$ (Error compiling LaTeX. Unknown error_msg)

Typically, we repress the terms with coefficient $0$ and we do not write the coefficient on terms with coefficient $1$. We also do not care about the order in which the terms are written, and indeed often list them in descending order of power. So we would write:

$(0,0,0,)=0(0,1,0,0,)=x(1,4,0,3,0,0,)=3x3+4x+1$ (Error compiling LaTeX. Unknown error_msg)

It is important to note at this point that '$x$' is only a symbol, it has no independent meaning, and in particular it is not a variable, i.e. is does not represent an element of $R$. Furthermore, a polynomial is not a function.

One can now define addition and multiplication in $R[x]$ in the 'obvious' way:

  • \[\sum_i a_ix^i + \sum_i b_ix^x = \sum_i (a_i+b_i)x^i\]
  • \[\left(\sum_i a_ix^i\right)\cdot \left(\sum_j b_jx^j\right) = \sum_k\left(\sum_{i=0}^k a_ib_{k-i}\right)x^k\]

It is now a simple matter to verify that $R[x]$ indeed forms a commutative ring under these operations. This ring has additive identity $0=(0,0,0,\ldots)$ and multiplicative identity $1 = (1,0,0,\ldots)$.

$R$ can be thought of as a subring of $R[x]$ via the embedding $r\mapsto (r,0,0,\ldots)$.

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