R02 Polynomial

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Polynomial Rings #

Polynomial rings #

Suppose $R$ be a ring. $R[x]$ is defined as the following set: $$ R[x] := \{r \in R^{\N} \mid \exists N \in \N, \forall n > N:r(n) = 0\} $$

  • With an indeterminate symbol $x$, we may write elements in $R[x]$ pretending they are polynomials in the following way: $$ f(x) = (a _ 0, \cdots, a _ n, 0, \cdots) = a _ 0 + a _ 1 x + \cdots + a _ n x^n \in R[x] $$
    • $f(x) \in R[x]$ is not a polynomial function. It is only a sequence $(a _ 0, a _ 1, \ldots)$. The resemblance is merely notational here.
    • The $a _ {i}$ are called coefficients of $f(x)$.
    • We often omit terms with zero coefficients.
    • If $R$ has unity $1 \neq 0$, we will write a term $1 x^{k}$ in such a sum as $x^{k}$.
    • Suppose $f(x) \neq 0$. The highest index nonzero coefficient is called the leading coefficient.
  • If $R$ has unity $1 \neq 0$, polynomials with leading coefficient $1$ are called monic.
  • The degree of $f(x) \in R[x]$ is a function $\deg: R[x] \mapsto \c{-\infty} \cup \N$.
    • Suppose $f(x) = 0$, $\deg f(x) := -\infty$.
    • Suppose $f(x) \neq 0$. $\deg f(x): =\max\c{k \mid a _ k \neq 0} \in \N$.

$R[x]$ is a ring when the binary operations on $R[x]$ are defined as if its elements are polynomials.

  • Addition is defined as $\sum _ {i}a _ i x^i + \sum _ ib _ i x^i := \sum _ i (a _ i + b _ i)x^i$.
  • Multiplication is defined as $\p{\sum _ i a _ i x^i}\p{\sum _ j b _ jx^j} := \sum _ i\sum _ ka _ {i - k}b _ kx^i$.
    • Also called the convolution of sequences $(a _ i)$ and $(b _ j)$.
  • $\a{R[x], +}$ is clearly an additive group.
  • Convolution of finite length sequences is associative, and distributes over addition.
    • Let's give a prove using the Einstein Summation notation.
    • Multiplication is associative: $$ (a _ i x^i)((b _ j x^j)(c _ k x^k)) = (a _ ix^i)(b _ {m - k}c _ k x^m) = (a _ {l - j}(b _ {j - k}c _ k))x^l = a _ {l - j}b _ {j - k} c _ k x^l $$
    • Multiplication distributes over addition: $$ c(x) (a(x) + b(x)) = c _ {i - j} (a _ j + b _ j) x^i = c _ {i - j} a _ j x^i + c _ {i - j} b _ j x^i = c(x) a(x) + c(x) b(x) $$

We have the following comments:

  • Suppose ring $F \le E$ then ring $F[x] \le E[x]$.
  • Suppose $R$ is commutative, then $R[x]$ is commutative.
  • We often identify $R$ with constant polynomials in $R[x]$.

Polynomial ring of $k$ variables is a subset of $R^{\N^k}$ with finitely many nonzero terms.

  • Under this definition $R[x _ 1, \cdots, x _ n], R[x _ 1, \cdots, x _ {n-1}][x _ n]$ and so on are clearly isomorphic.
Extending homomorphisms to polynomials #

Suppose $R$ and $S$ are rings.

Suppose $\sigma: R \to S$ is a homomorphism. We can extend the definition to $\sigma: R[x] \to S[x]$.

  • Let's identify $a \in R$ with $(a, 0, 0, \ldots) \in R[x]$. Similarly for $S$ and $S[x]$.
  • Extending $\sigma$ by applying $\sigma$ to each coordinate of $(a _ 0, a _ 1, \ldots) \in R[x]$.
  • The extended $\sigma$ is a ring homomorphism. (Einsum notation)
    • $\sigma((a _ ix^i)(b _ jx^j)) = \sigma(a _ {k-i}b _ ix^k) = \sigma(a _ {k-i}) \sigma(b _ i)x^k = \sigma(a _ ix^i) \sigma(b _ jx^j)$.
    • $\sigma((a _ i + b _ i)x^i) = (\sigma(a _ i) + \sigma(b _ i))x^i = \sigma(a _ i x^i) + \sigma(b _ i x^i)$.
  • The extended $\sigma$ is injective if and only if $\sigma$ is injective.
Evaluation homomorphism of polynomial rings #

Suppose $R$ is a NC ring. And $s \in R$.

For $f(x) \in R[x]$, and $f(x) = \sum _ i a _ i x^i$. The evaluation of $f(x)$ at $s \in R$ is denoted as $f(s):= \sum _ i a _ i s^i$.

Define the evaluation homomorphism $\theta _ s: R[x] \to S$ as $f(x) \mapsto f(s)$.

  • $\theta _ s$ is a ring homomorphism.
    • $\theta _ s((a _ ix^i)(b _ j x^j)) = \theta _ s(a _ {k - i}b _ i x^k) = a _ {k-i}b _ is^k = \theta _ s(a _ i x^i) \theta _ s(b _ j x^j)$.
    • $\theta _ s(a _ i x^i + b _ i x^i) = \theta _ s((a _ i + b _ i)x^i) = \theta _ s(a _ ix^i) + \theta _ s(b _ i x^i)$.
  • $\theta _ s$ is surjective.
  • Suppose $f(s) = 0$ for some $s \in S$, $s$ is called a zero of $f(x)$.
Polynomial ring of integral domains #

$R$ is an ID if and only if $R[x]$ is an ID.

  • $\to$ Suppose $R$ is an integral domain.

    • $R[x]$ is clearly commutative, since $R$ is commutative.
    • The element $1 \neq 0$ is the unity in $R[x]$.
    • Suppose $a(x), b(x) \neq 0$ but $a(x) b(x) = 0$. Two nonzero leading coefficient in $R$ must product to zero. Which is not possible.
    • So $R[x]$ is an integral domain.

  • $\from$ Suppose $R[x]$ is an ID, $R < R[x]$ so $R$ is an ID.

Immediately, when $F$ is a subdomain of $E$, $F[x]$ is a subdomain of $E[x]$.