# Can the complex numbers be realized as a quotient ring?

Can the complex numbers be realised as some $R/M$ where $R$ a ring and $M$ a maximal ideal like the integers modulo some prime? I understand that unlike the latter case, such a maximal ideal would need to partition the ring into infinitely many cosets.

• $\mathbb{C} \cong \mathbb{C} / (0)$. The zero ideal is maximal in any field. Dec 28, 2014 at 13:16

Yes, $\mathbb{C} = \mathbb{R}[x] / \langle x^2 + 1 \rangle$ is the standard contruction, where $\mathbb{R}[x]$ is the ring of polynomials in one variable with real coefficients; and $\langle x^2 + 1 \rangle$ is the ideal generated by the polynomial $x^2 + 1$.

If you think about it, since $x^2 + 1$ generates the ideal, it's true that

$$x^2 + 1 \equiv 0 \pmod{x^2 + 1}$$

and therefore $x^2 \equiv-1$ so that the equivalence class of the polynomial $x$ is a square root of $-1$.

By mapping $1$ to $1$ and $i$ to $x$ you can get an isomorphism.

[Markup question, Why do I have an extra space in my mod expression above?]

• Thank you. I am now curious if the rationals can be created by a similar construction, do you know about this? Dec 28, 2014 at 5:01
• You should ask that as a separate question Dec 28, 2014 at 5:08
• I would use \pmod {x^2 + 1} which does its own spacing Dec 28, 2014 at 5:13
• Compare \pmod and \mod.
– Pedro
Dec 28, 2014 at 5:23
• You might also be interested in \bmod.
– Hoot
Dec 28, 2014 at 5:42

The $\mathbb R[x] / (x^2 + 1)$ solution is what leapt to everyone's mind, but there is an even simpler solution:

$\Bbb C[x]/(x)\cong \Bbb C$.

In both cases, the polynomial whose ideal is being modded out is a maximal ideal of the ring (it would have to be maximal, after all), so it is very much like the integers modulo a prime, as you wished.

Or, for that matter, you could even just say $\Bbb C/\{0\}\cong \Bbb C$, if you don't mind a trivial solution.

One possible definition of $$\mathbf C$$ is that it is the splitting field of the polynomial $$X^2 +1$$. As such, it is isomorphic to the quotient ring $$\mathbf R[X]/(X^2+1)$$ (polynomials with real coefficients, modulo the ideal generated by $$X^2+1$$). The imaginary number i is then simply the congruence class of X.

I would like to add that the standard construction $$\mathbb{R}[x]/(x^2+1)$$ mentioned by several of the answers here can be thought of as a private case of a more general theorem regarding field extensions:

Theorem: If $$E/F$$ is a field extension, and some element $$a\in E$$ is algebraic over $$F$$, then $$F(a)$$ (the field generated by $$a$$ over $$F$$) is isomorphic to $$F[x]/(m_a)$$ (where $$m_a$$ is the minimal polynomial of $$a$$).

Proof Idea: We can define a homomorphism from $$F[x]$$ to $$F(a)$$ given by $$f(x)\mapsto f(a)$$, and then use the first isomorphism theorem for rings.

In our case, it is obvious that $$\mathbb{C}=\mathbb{R}(i)$$, and $$m_i=x^2+1$$, and we immediately get: $$\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$$