# Definition of the complex numbers

I know that a complex number is an ordered pair $$(x,y) \in \mathbb{R} × \mathbb{R}$$ which can be written as $$z=x+iy$$, where $$i^2=-1$$. I want some abstract definition of complex number, I searched in google but didn't get any. Kindly help me with this.

• I'm not sure what you are looking for, but we can realize $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$. In other words, if we formally add a solution of the polynomial $P(X)=X^2+1$ to the field $\mathbb{R}$, we can endow the larger set with a natural field structure extending the field structure of $\mathbb{R}$. The resulting field is $\mathbb{C}$ and can be described as you say. – Mathematician 42 Oct 20 at 8:40
• Yes...Thank you – Ppp Oct 20 at 8:41
• I know the answer is almost the same as the comment, but it became too long for a comment :p – Mathematician 42 Oct 20 at 8:43

You wrote that you know that “a complex number is an ordered pair $$(x,y)\in\mathbb R\times\mathbb R$$ which can be written as $$z=x+iy$$, where $$i^2=−1$$.” You cannot possibly know that since that makes no sense.

You can define (as Hamilton did) a complex number as an ordered pair $$(x,y)\in\mathbb R\times\mathbb R$$ and then you define addition and multiplication:

• $$(a,b)+(c,d)=(a+c,b+d)$$;
• $$(a,b)\times(c,d)=(ac-bd,ad+bc)$$.

After that, you can define $$i=(0,1)$$. Then if you identify each real number $$a$$ with the complex number $$(a,0)$$, it will be true that $$i^2=-1$$.

• This answer IMHO it is very nice because is the same that I use. – Sebastiano Oct 20 at 8:50
• Thanks for the clarification. I understood my mistake now. – Ppp Oct 20 at 8:53

I'm not sure what you are looking for, but we can realize $$\mathbb{C}$$ is the algebraic closure of $$\mathbb{R}$$. In other words, if we formally add a solution of the polynomial $$P(X)=X^2+1$$ to the field $$\mathbb{R}$$, we can endow the larger set with a natural field structure extending the field structure of $$\mathbb{R}$$. The resulting field is $$\mathbb{C}$$ and can be described as you say.

I'm lying a bit. If I say I want to introduce $$\mathbb{C}$$ as the algebraic closure of $$\mathbb{R}$$, we should formally add all solutions of all real polynomials to $$\mathbb{R}$$. It turns out adjoining a solution of $$X^2+1$$ is sufficient to get the algebraic closure of $$\mathbb{R}$$.

This procedure works for any field.

See Kronecker's theorem on field extensions for a detailed general discussion on how to formally adjoin roots of a polynomial to a field.

A simple algebraic definition is that $$\mathbf C$$ is the quotient ring $$\mathbf R[X]/(X^2+1),$$ which happens to be a field as $$(X^2+1)$$ is a maximal ideal in the polynomial ring $$\mathbf R[X]$$. The square root of $$-1$$ is the congruence class $$X+(X^2+1)$$.

Another algebraic definition is as a set of $$2{\times}2$$ matrices: $$\mathbf C=\biggl\{\begin{pmatrix}a&b\\-b&a\end{pmatrix}\;\bigg\vert\;a,b\in\mathbf R\biggr\},$$ endowed with the usual addition and multiplication. In this definition, the unit element is … the unit matrix, as you may have guessed, and $$\sqrt{-1}$$ is $$i=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$