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.
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ $\mathbb{C} \cong \mathbb{C} / (0)$. The zero ideal is maximal in any field. $\endgroup$– Konstantin ArdakovDec 28, 2014 at 13:16
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
6
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?]
-
$\begingroup$ Thank you. I am now curious if the rationals can be created by a similar construction, do you know about this? $\endgroup$ Dec 28, 2014 at 5:01
-
-
1$\begingroup$ I would use \pmod {x^2 + 1} which does its own spacing $\endgroup$ Dec 28, 2014 at 5:13
-
-
$\begingroup$
$\endgroup$
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.
answered Dec 28, 2014 at 12:15
$\begingroup$
$\endgroup$
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.
$\begingroup$
$\endgroup$
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: $$\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$$
