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. 
 A: 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.
A: 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.
A: 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?]
A: 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:
$$\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$$
