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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.

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  • $\begingroup$ 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. $\endgroup$ – Mathematician 42 Oct 20 at 8:40
  • $\begingroup$ Yes...Thank you $\endgroup$ – Ppp Oct 20 at 8:41
  • $\begingroup$ I know the answer is almost the same as the comment, but it became too long for a comment :p $\endgroup$ – Mathematician 42 Oct 20 at 8:43
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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 $i^2=-1$.

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

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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 $$i=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$

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