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What are the basic differences between $\mathbb C$ and $\mathbb R^2$?

The points in these two sets are written as ordered pairs, I mean the structure looks similar to me. So what is the reason to denote these two sets differently?


marked as duplicate by Alex M., user147263, hardmath, user228113, Giovanni May 3 '16 at 0:27

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  • $\begingroup$ If you are familiar with the language of abstract algebra, $\mathbb{C}\cong \mathbb{R} \times\mathbb{R}.$ $\endgroup$ – Coffee_Table Apr 17 '13 at 4:00
  • $\begingroup$ That MUST be a duplicate. $\endgroup$ – Dominic Michaelis Apr 17 '13 at 4:08
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    $\begingroup$ As sets, they are identical. As groups, vector spaces, also. The point is that we denote and use $\mathbb{C}$ rather than $\mathbb{R}^2$ when we need or refer to its field structure. $\endgroup$ – Julien Apr 17 '13 at 4:54

The answer to this question depends on what you mean by $\mathbf{R}^2$. You can write $\mathbf{R}\times\mathbf{R}$, but the "$\times$" can have several different meanings depending on which category you are working in.

As sets:

If you view $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ as sets, this means you are ignoring any possible structure on these things besides their elements (so forget about multiplication, addition, anything.) From this perspective, $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ are the same object (in technical terms they are isomorphic in the category of sets) because there is a bijection $$(a,b)\ \leftrightarrow\ a+bi.$$ A key thing to note about this is that "$\times$" refers to a direct product of sets, i.e. the Cartesian product. This will change as we add more structure.

As real vector spaces:

You can give each of $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ the structure of a real vector space, meaning you can add vectors and multiply by real numbers. Then from the theory of linear algebra, we know that $\mathbf{R}\times\mathbf{R}$ has a basis of $\{(1,0), (0,1)\}$ and $\mathbf{C}$ has a (real) basis of $\{1, i\}$. Since these real vector spaces both have dimension 2, they are isomorphic (in the linear algebra sense, i.e. in the category of $\mathbf{R}$-modules). So from this perspective they are again the same object.

Note here that $\mathbf{R}\times\mathbf{R}$ can be interpreted as $\mathbf{R}\oplus \mathbf{R}$ which may be more familiar to linear algebra students. The point is that now we are requiring more from the operation (it has to preserve addition of vectors now).

As rings:

Here is where the difference comes in. We can think of $\mathbf{R}$ and $\mathbf{C}$ as rings, meaning we can add and multiply elements together according to some axioms. Then if you write $\mathbf{R}\times\mathbf{R}$, you mean a direct product in the category of rings, so now multiplication in $\mathbf{R}\times\mathbf{R}$ has to satisfy $$(a,b)\cdot (c,d)=(ac,bd).$$ But in particular this means things like $$(1,0)\cdot (0,1)=(0,0),$$ which means it is possible for two nonzero things to have a product zero. In contrast, if $z_1z_2=0$ in $\mathbf{C}$, then either $z_1=0$ or $z_2=0$. In this way, $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ have fundamentally different behavior as rings. Because of this, there is no isomorphism of rings between the two objects.

As fields:

A field is a commutative ring with more structure (we can invert multiplication for nonzero things). It turns out that $\mathbf{C}$ can be given the structure of a field because $z^{-1}$ exists for any nonzero $z\in\mathbf{C}$, but $\mathbf{R}\times\mathbf{R}$ cannot be a field because equations like $(1,0)\cdot (0,1)=(0,0)$ mess everything up (try to cancel something from the left side).

tl;dr -- You have to specify what you mean by "$\times$". $\mathbf{C}$ and $\mathbf{R}\times\mathbf{R}$ are exactly the same until you start saying you want to do things like multiply elements together.

  • $\begingroup$ That means the main difference is that , $\mathbb C$ is an integral domain but $\mathbb R \times \mathbb R$ is not an integral domain. $\endgroup$ – Germain Apr 17 '13 at 13:15
  • $\begingroup$ is there any difference between $\mathbb R^n$ and $\mathbb R \times \mathbb R \times....\times\mathbb R$ (n times) $\endgroup$ – Germain Apr 17 '13 at 13:20
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    $\begingroup$ @analysis89 Right, and because all fields are integral domains, this means it's impossible for something not an integral domain to be a field. $\mathbf{R}^n$ is just shorthand for $\mathbf{R}\times\cdots\times\mathbf{R}$. $\endgroup$ – Chris Brooks Apr 17 '13 at 16:35

Well $\mathbb{R}^2$ is (by default) interpreted as a vectorspace. So normally you only expect addition and scalar multiplication in it. Which give different algebraic structures.

And I want to add something. The points of $\mathbb{R}^2$ are written normally as tupels, namely $x\in \mathbb{R}^2$ then $ x=(x_1,x_2)$ with $x_1,x_2\in \mathbb{R}$. The elements of $\mathbb{C}$ have (at least) 3 common ways to denote. \begin{align*} z&=a+i\cdot b \quad a,b\in \mathbb{R}\\ z&=(a,b)\\ z&=|z|\cdot \exp(i\varphi) \end{align*} The first and the second look pretty much the same, but in $\mathbb{C}$ the multiplication is a function
$$ \mathbb{C} \times \mathbb{C} \to \mathbb{C}$$ while the scalar multiplication in $\mathbb{R}^2$ is a function $$ \mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}^2$$ You can copy the structure of $\mathbb{C}$ and put it on $\mathbb{R}^2$, to make it a field for example.

But for example the differentiability is different in $\mathbb{R}^2$ and in $\mathbb{C}$ namely a function which is differentiable on $\mathbb{R}^2$ need to fulfill the Cauchy Riemann differential equations to be complex differentiable.

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    $\begingroup$ To say it in an easy english, $\mathbb{R}^2$ and $\mathbb{C}$ are like twins, even though they are not the same they look pretty much the same $\endgroup$ – Dominic Michaelis Apr 17 '13 at 4:27

$\Bbb C$ is a field in its own right with an operation of multiplication. $\Bbb R^2$ does not have that.


To put it succinctly, $\mathbb{R}^2$ is a vector space while $\mathbb{C}$ is a field. One of the first difference to be noticed is that you can divide two complex numbers but you cannot divide two points in $\mathbb{R}^2$.

  • $\begingroup$ Some would argue that every field is a vector space over itself. $\endgroup$ – kahen Apr 17 '13 at 19:39
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    $\begingroup$ Yep every field is a VS over itself but every VS certainly isn't a field. So that's a (big) difference between $\mathbb{R}^2$ and $\mathbb{C}$. $\mathbb{C}$ is both but $\mathbb{R}^2$ is only a VS. $\endgroup$ – Fixed Point Apr 18 '13 at 8:03

If we desire that every integer has an inverse element, we have to invent rational numbers and many things become much simpler. If we desire every polynomial equation to have a root, we have to extend the real number field $\mathbb{R}$ to a larger field $\mathbb{C}$ of 'complex numbers', and many statements become more homogeneous. To construct a complex number, we associate with each real number $a$ second real number $b$. A complex number is then an ordered pair of real numbers (a,b). We write that new number as $a+bi$.

Complex numbers contain solutions to all polynomial equations and hence are an algebraically closed field unlike the real numbers. The complex numbers are not an ordered field. $\mathbb{C}$ is geometrically represented by identifying it with $\mathbb{R^2}$. (This is sometimes called the Argand diagram).


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