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First Ring Isomorphism Theorem

Example

Define \rho : \mathbb{R}[x] \to \mathbb{C} such that \rho(f(x)) = f(i) for f(x) \in \mathbb{R}[x]

Lemma

ker(\rho) = (x^2 + 1) \mathbb{R}[x]
Proof
\begin{align} f(x) &= (x^2 + 1) \mathbb{R}[x] + ax + b \in ker(\rho) \\ f(i) &= ai + b = 0 \end{align}
  1. if a \ne 0, i = \frac{-b}{a} ( contradiction )
  2. if a = 0, b = 0, then f(x) = (x^2 + 1) \mathbb{R}[x]

By First Ring Isomorphism Theorem

\mathbb{R}[x] / (x^2 + 1) \mathbb{R}[x] \cong \rho(\mathbb{R}[x])

  • \mathbb{R}[x] / (x^2 + 1) \mathbb{R}[x] 可以看成 \{ax + b | a, b \in \mathbb{R}\} 做加減乘除運算模 (x^2 + 1)
  • \rho(\mathbb{R}[x]) = \{ai + b | a, b \in \mathbb{R}\} = \mathbb{C}

結論是我們可以把複數上面的運算轉換成實數多項式上的運算