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

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.

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 $$(1)$$.

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.

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

• Thank you so much for your verification ;) Jun 29 '19 at 9:15
• I'm glad I could help. Jun 29 '19 at 9:35

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.