Constructing $\mathbb{C}$ from $\mathbb{R}$

Question

I'm having difficulty grasping the notion that you can define the complex numbers as $\mathbb{C}=\mathbb{R}[t]/\langle t^2+1\rangle$.

As far as I understand, $\mathbb{R}[t]$ is the set of all polynomials in $t$ whose coefficients are real.

When we take the quotient, we are forming equivalence classes. But how does (some polynomial in $[t]$ with real coefficients) + $(t^2+1)$ translate into a complex number? Or should the operation be $\times$ instead of $+$?

At that point, do you get the set of complex numbers by considering the roots of these polynomials?

Answer 1

When taking quotient by $(t^2+1)$, the image of $t$ satisfies $t^2 + 1 = 0$, or, equivalently, $t^2 = -1$, so $t$ works as the imaginary unit.

Now, any element of the quotient can be uniquely represented by a polinomial with degree $\leq 1$ (using the fact that applying the identity $t^2 = -1$ reduces the degree). So, any element can be written as $a + tb$. Here, $a$ can be considered to be the real part and $b$ to be the imaginary part (just replace $t$ by $i$, if you feel better with that).

Addition of such polynomial is trivial (component-wise). Multiplication is a bit harder, but still very simple: $(a + bt)(c+dt) = ac + adt + bct + bdt^2 = (ac - db) + (ad + bc)t$ (since $t^2 = -1$). So, operations on this quotient ring are exactly those of complex numbers.

Answer 2

The complex number $a+bi$ corresponds to the equivalence class of polynomials that contains the polynomial $bt+a$.

For example $2+3i$ corresponds to the equivalence class $$ \{3t+2, t^2+3t+3, -t^2+3t+1, t^3+4t+2, \pi t^4+\pi t^2+3t+2,\ldots \} $$

You should be able to check that addition and multiplication of polynomials correspond to addition and multiplication of complex numbers -- that is, if $p(t)$ and $q(t)$ are polynomials that represent the complex numbers $z_1$ and $z_2$, then the polynomial $p(t)+q(t)$ represents the complex number $z_1+z_2$, and the polynomial $p(t)q(t)$ represents the complex number $z_1z_2$.

Answer 3

It is easy to see that this is a $2$-dimensional vector space over $\mathbb{R}$?

Now since in the quotient $t^2+1\cong 0$ then $t^2\cong -1$. By labeling $t$ as $i$ you are getting the "regular" complex numbers.

Answer 4

The solutions to the equation $$t^2+1=0$$ are the complex numbers commonly denoted $i$ and $-i$. (They are numbers which square to $-1$.)

In the ring $$R[t] / (t^2+1), $$ the polynomial $t^2+1$ is equivalent to $0$ (since it is certainly in the coset $(t^2+1)R[t]$. By my initial comment, another way of thinking about the fact that $t^2+1=0$ in this ring is to say that, in this ring, $t=i$ (or $t=-i$).

Since an arbitrary element of $R[t]$ looks like $$a_0 + a_1t + a_2t^2 + a_3t^3 + a_4t^4 + \cdots + a_nt^n, \quad a_i \in \mathbf{R}$$ in the quotient $R[t]/(t^2+1)$ this polynomial becomes $$f = a_0 + a_1 i + a_2 (i)^2 + a_3 (i)^3 + a_4 (i)^4 + \cdots + a_n (i)^n. $$

It's easy to check that since $i^2 = -1$, we have $i^3 = -i$ and $i^4 = 1$. Thus, collecting terms we can write $f$ in the form $$f = a + b i$$ for some real numbers $a, b \in \mathbf{R}$. This is the usual way of defining complex numbers.

Answer 5

There are two ways to see this:

1) The first is to give an isomorphism from $$\frac{\mathbb{R}[t]}{<t^2+1>} \rightarrow \mathbb{C}.$$ Define $\phi : \frac{\mathbb{R}[t]}{<t^2+1>} \rightarrow \mathbb{C}$ by $\phi(r) = r$ for $r \in \mathbb{R}$ and $\phi(t) = i.$ We can easily verify that this gives an isomorphism between the two required set.

2) Another way to see this fact is to understand the meaning of quotients. When we go quotient by an ideal, say $<t^2 + 1>$ then what we are doing is we are basically allowing the polynomial $t^2 +1 $ to have a solution in the quotient. This is essentially why we construct $\mathbb{C}$ in the first place. So the quotient is a ring which contains $\mathbb{R}$ and allows us to solve $t^2+1.$ And since $t^2+1$ is irreducible in $\mathbb{R}$ we see that $\frac{\mathbb{R}[t]}{<t^2+1>}$ is a field isomorphic to $\mathbb{C}.$

Answer 6

Pick any polynomial $P(t) \in \mathbb C[t]$. By long division you have

$$P(t)=Q(t)(t^2+1)+at+b $$

Now, in the factor you have $t^2+1=0$ thus $P(t)$ becomes "the same" as $at+b$. Here the same means, is in the same class.

Now, identifying each polynomial with its remainder when divided by $t^2+1$ is picking ONE single polynomial from each equivalence class.

If you think back about $\mathbb C$ as being complex numbers, this identification is basically replacing each polynomial $P(t)$ by the complex number $P(i)$. And the factorisation is basically identifying different polynomials for which $P(i)$ is the same number.