# 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 $(x^2 + 1)$
• $\rho(\mathbb{R}[x]) = \{ai + b | a, b \in \mathbb{R}\} = \mathbb{C}$