Construct the field of complex numbers as the quotient ring of real polynomials

Question

I am trying to construct the field of complex numbers as the quotient ring of real polynomials.

Suppose that

1. $\mathbb C, \mathbb R$ are the fields of complex and real numbers respectively.

2. $\mathbb R [X]$ is the ring of polynomials over $\mathbb R$.

3. $\left \langle X^{2} + 1 \right \rangle = \left \{p(X^{2}+1) \mid p \in \mathbb R [X] \right \}$ is the ideal generated by $(X^{2}+1)$.

4. $D = \mathbb R [X] / \left \langle X^{2} + 1 \right \rangle$ 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.

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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:

1. 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 $(1)$.

2. 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.

Answer 1

Good job! I would just add to that a proof of the fact that $\phi\bigl([1]\bigr)=1$ (that's trivial, of course).

Answer 2

Well, the quotient ring $D = {\Bbb R}[x]/\langle x^2+1\rangle$ is a field since $x^2+1$ is irreducible over $\Bbb R$.

Moreover, the quotient ring $D$ contains a zero of $x^2+1$, namely $\bar x = x+\langle x^2+1\rangle$. This zero satisfies $\bar x^2+1=0$. The other zero is $-\bar x$.

This shows that $D = {\Bbb R}[\bar x]$ (ring adjoint) and since $D$ is a field, $D={\Bbb R}(\bar x)$ (field adjoint).

The mapping

${\Bbb C}=\{a+ ib\mid a,b\in{\Bbb R}\}\rightarrow {\Bbb R}(\bar x) = \{a+b\bar x\mid a,b\in{\Bbb R}\}: a+bi \mapsto a+b\bar x$

is an isomorphism of fields.