I am trying to construct the field of complex numbers as the quotient ring of real polynomials.
Suppose that
$\mathbb C, \mathbb R$ are the fields of complex and real numbers respectively.
$\mathbb R [X]$ is the ring of polynomials over $\mathbb R$.
$\left \langle X^{2} + 1 \right \rangle = \left \{p(X^{2}+1) \mid p \in \mathbb R [X] \right \}$ is the ideal generated by $(X^{2}+1)$.
$D = \mathbb R [X] / \left \langle X^{2} + 1 \right \rangle$ is the quotient ring of $\mathbb R [X]$ modulo $\left \langle X^{2} + 1 \right \rangle$.
Then $$(\mathbb C, +, \cdot) \cong (D, +, \cdot)$$
Could you please verify if my attempt contains logical gaps/errors? Any suggestion is greatly appreciated.
My attempt:
Lemma (Long Division of Polynomial): Let $K$ be a field and $p, q \in K[X]$ with $q \neq 0 .$ Then there are unique polynomials $r, s \in K[X]$ such that $$p=s q+r \quad \text {and} \quad \operatorname{deg}(r)<\operatorname{deg}(q) \tag 1$$
Proof:
- Existence
Define $\mathcal Q: K[X] \times (K[X] – \{0\}) \to K[X]$ by $$\mathcal Q(p,q) = \begin{cases} \bar p / \bar q X^{\operatorname{deg} (p)-\operatorname{deg} (q)} & \text{if } \operatorname{deg} (p) \ge \operatorname{deg} (q) \\ 0 & \text{otherwise}\end{cases}$$
where $\bar p, \bar q$ are the coefficients corresponding to $\operatorname{deg} (p), \operatorname{deg} (q)$ respectively.
Define $\langle p_n, s_n \rangle_{n \in \mathbb N}$ recursively by
$$\begin{aligned}\langle p_0, s_0 \rangle &= \langle p, \mathcal Q(p,q) \rangle \\ \langle p_{n+1}, s_{n+1} \rangle &= \langle p_n – s_n q, \mathcal Q (p_{n+1},q) \rangle \end{aligned}$$
Let $n' = \min \{ n \in \mathbb N \mid s_n = 0\}$. It is easy to verify that $s = \sum_{i = 0}^{n'} s_n$ and $r = p_{n'}$ satisfy $(1)$.
- Uniqueness
Suppose that $s’$ and $r’$ are other polynomials such that $p=s’ q+r’$ and $\operatorname{deg} (r’) < \operatorname{deg}(q)$. Then $(s’-s) q=r-r’$. If $s’-s \neq 0$ then, from $\operatorname{deg} (p q) = \operatorname{deg}(p) + \operatorname{deg}(q)$, we would get
$$\operatorname{deg} (r-r’) = \operatorname{deg} ((s’-s) q) = \operatorname{deg} (s’-s) + \operatorname{deg}(q)>\operatorname{deg}(q)$$
which, because $\operatorname{deg} (r-r’) \leq \max \{\operatorname{deg} (r), \operatorname{deg} (r’)\} <\operatorname{deg} (q)$, is not possible. Thus $s’=s$ and also $r’=r$.
By lemma, each set in $D$ has at least one element of the form $a+bX$. Assume that both $a + b X$ and $a' +b' X$ belong to the same set in $D$. Then $(a + b X) \sim (a' + b' X)$ and thus $(a + b X) – (a' + b' X) = p (X^2 + 1)$ for some $p \in K[X]$. It follows that $a = a'$, $b = b'$, and $p = 0$. As such, each set in $D$ has exactly one element of the form $a+bX$.
Consider $$\phi : D \rightarrow \mathbb{C}, \quad [a+bX] \mapsto a+b i$$
Clearly, $\phi$ is surjective.
Assume $[a+bX], [a'+b'X] \in D$ such that $\phi ([a+bX]) = \phi ([a'+b'X])$. Then $a+b i = a'+b' i$, and thus $a=a'$ and $b = b'$. Hence $\phi$ is injective.
Next we show that $\phi$ is a homomorphism w.r.t $+$ and $\cdot$.
$$\begin{aligned} \phi([a+b X]+[a'+b' X]) &= \phi([(a+a')+(b+b')X ]) \\
&= (a+a')+(b+b') i \\
&= (a+b i)+(a'+b' i) \\
&= \phi([a+b X])+\phi([ a'+b' X])\end{aligned}$$
$$\begin{aligned}\phi([a+b X]\cdot[a'+b' X]) &=\phi([(a+b X)\cdot (a'+b' X)])\\
&=\phi ([aa'+(ab'+ba') X+bb' X^2]) \\
& = \phi ([aa'-bb'+ (ab'+ba') X+bb' (X^2 + 1)]) \\
&=\phi([aa'-bb'+ (ab'+ba') X] )\\
&=(aa'-bb')+ (ab'+ba') i\\
&=(a+b i) \cdot (a'+b' i) \\
&=\phi([a+b X]) \cdot \phi([a'+b' X]) \end{aligned}$$
This completes the proof.
Best Answer
Good job! I would just add to that a proof of the fact that $\phi\bigl([1]\bigr)=1$ (that's trivial, of course).