02.3.環の準同型定理

Proposition

剰余環

Statement

$R$:Ring
$I$:Ideal
define $x \sim y$ if $x - y \in I$
then $\sim$ is 同値関係.
剰余類 $R/\sim$ を $R/I$ で表す.
$x + I, y + I \in R/I$に対して
$(x + I) + (y + I) = (x + y) + I$
$(x + I)(y + I) = xy + I$
と定義するとこの演算はWell-Definedになり $R/I$ は環になる.

Proof

(同値関係)
(1) $x \sim x \because (x - x) = 0 \in I$
(2) $x \sim y \Rightarrow y \sim x \because (x - y) \in I \Rightarrow (y - x) = -(x - y) \in I$
(3) $x \sim y, y \sim z \Rightarrow x \sim z \because (x - y), (y - z) \in I \Rightarrow (x - z) = (x - y) + (y - z) \in I$
(Well-Defined)
$x \sim x',y \sim y'$
then
$x - x' \in I, y - y' \in I$
$(x + y) - (x' + y') = (x - x') + (y - y') \in I$
$\therefore x + y \sim x' + y'$
$x = x' + i_x, y = y' + i_y$
$xy = (x' + i_x)(y' + i_y) = x'y' + i_x y' + x' i_y + i_x i_y$
$xy - x'y' = i_x y' + x' i_y + i_x i_y \in I$
$\therefore xy \sim x'y'$

Proposition

準同型定理

Statement

$R,R'$:Ring
$f:R \mapsto R'$ is a homomorphism.
then
$R/\text{Ker }f \simeq \text{Img }f$

Proof

$\text{Ker }f$ is an ideal
$\because \forall x \in R, i \in \text{Ker }f \Rightarrow f(xi) = f(x)f(i) = f(r)0 = 0 \therefore xi \in \text{Ker }f$
Let
$I = \text{Ker }f$
$\phi:\bar x = x + I \in R/I \mapsto f(x) \in R'$
then
$\phi$ is well-defined
$\because$
$\bar x, \bar x' \in R/I$
$x - x' \in I = \text{Ker }f \therefore 0 = f(x - x') = f(x) - f(x')$
$\therefore f(x) = f(x')$
$\phi$ is a homomorphism.
$\phi(\bar x + \bar x') = f(x + x') = f(x) + f(x') = \phi(x) + \phi(x')$ $\phi(\bar x \bar x') = f(x x') = f(x) f(x') = \phi(x) \phi(x')$ $\phi$ is injective
$\phi(\bar x) = \phi(\bar x')$
$\Leftrightarrow \phi(\bar x) - \phi(\bar x') = 0$
$\Leftrightarrow f(x) - f(x') = 0$
$\Leftrightarrow f(x - x') = 0$
$\Leftrightarrow x - x' \in \text{Ker }f$
$\Leftrightarrow \bar x = \bar x'$
$\phi$ is surjective
$\because \forall y \in \text{Img} f,\exists x \in R,y = f(x) = \phi(\bar x) \quad \bar x \in R/I$ QED

Proposition

体の準同型写像は単射

Statement

$F,F'$:Field
$f:F \mapsto F'$ homomorphism
then
$f$ is injective.

Proof

$f(1) = f(1\cdot 1) = f(1)f(1)$
$\therefore 1 = f(1)f(1)^{-1} = f(1)f(1)f(1)^{-1} = f(1)$ -----(1)
$\text{Ker}f = \{0\}$ -----(2)
$\because$(
Suppose
$x \in \text{Ker}f, x\neq 0$
then
$f(1) = f(xx^{-1}) = f(x)f(x^{-1}) = 0$
contradict (1)
) $f(x) = f(x')$
Then
$0 = f(x) - f(x') = f(x-x')$
$x - x' = 0 \because$ (2)
$x = x'$
QED