$\mathbb R^2$ is not a field, but $2$-tuple arithmetic rules like $(a,b)(c,d)=(ac-bd,ad+bc)$ coupled with $\mathbb R^2$ make it a field, but are there other rules than $(a,b)(c,d)=(ac-db,ad+bc)$ combined with $\mathbb R^2$ that make it a field?

Is $\mathbb C$ the only field that contains $\mathbb R$ as a subfield or are there other fields that contain $\mathbb R$ as subfields?

Is $\mathbb C$ the only that can be made by coupling $\mathbb R^2$ and rules like $(a,b)(c,d)$ $=$ $(ac-db,ad+bc)$? Are there other rules that can be coupled with $\mathbb R^2$ to make it a field?

PS: Just trying to pin down some properties that make $\mathbb C$ special/unique from other fields that can/may be constructed using some specific algebraic structure of 2 tuples in $\mathbb R$ e.g. $(\sqrt 2,\pi)$. At this point this is just a wide net as I lack the terminology to make this question specific. But any hints on how to make this question specific to listing properties that make $\mathbb C$ unique or gives strong characterisation of relationship between $\mathbb C$ and $\mathbb R$ are welcome.

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    $\begingroup$ Could you elaborate on what you mean by "field in $\mathbb R^2$"? I suppose I want to know what structure you are putting on $\mathbb R^2$ already. $\endgroup$ – user714630 May 17 '13 at 6:56
  • $\begingroup$ @user1 : if you have seen the 2 tuple approach to complex numbers, where there is no mention of $i$ but instead (0,1), or the matrix form is used, then it seemed natural to think as (0,1) as a point in $\mathbb R^2$ with some funny rules for addition (a,b)+(b,c)=(a+b,c+d) and multiplication (a,b)(b,c)=(ab-bc,ac+bd) , {(a,b),(c,d)} being points in $\mathbb R^2$ with strange arithmetic, where R just becomes (a,0),(b,0), etc. $\endgroup$ – Arjang May 17 '13 at 9:14
  • $\begingroup$ Assuming you mean $(a,b)(c,d)=(ac-bd,bc+ad)$, then you cannot distinguish this from $\mathbb C$ as a ring; hence, they will agree as fields too. $\endgroup$ – user714630 May 17 '13 at 9:17
  • $\begingroup$ @user1 : Edited to clarify more, hope this helps in making what I am trying to say clearer. $\endgroup$ – Arjang May 17 '13 at 9:24
  • $\begingroup$ In sum, I would guess you mean that $\mathbb R^2$ should be taken as the vector space (no multiplication given besides scalar multiplication) and you want to see what fields agree with that under addition. Both answerers seem to have guessed this anyways. :) $\endgroup$ – user714630 May 17 '13 at 9:26

For any field $F$, the field $F(t)$ of rational functions in the variable $t$ is a field that contains $F$. The degree $[F(t):F]$ is infinite, and $F(t)$ is a transcendental (i.e., non-algebraic) extension of $F$. Either of these observations suffices to check that $\mathbb{R}(t)\not\cong\mathbb{C}$ (as extensions of $\mathbb{R}$), because $[\mathbb{C}:\mathbb{R}]=2$ and $\mathbb{C}/\mathbb{R}$ is algebraic. Moreover, there is not even an abstract isomorphism between $\mathbb{R}(t)$ and $\mathbb{C}$ (i.e., one which does not necessarily respect the way $\mathbb{R}$ sits inside each of them) because $\mathbb{R}(t)$, and indeed $F(t)$ for any field $F$, is not algebraically closed, whereas $\mathbb{C}$ is.

Also, this is up to isomorphism; technically, if you've chosen what you mean by "$\mathbb{C}$" and I take your $\mathbb{C}$ and paint the elements of $\mathbb{C}\setminus\mathbb{R}$ red, that is a field different from $\mathbb{C}$ that contains $\mathbb{R}$. Similarly, $\mathbb{C}$ and $\mathbb{R}^2$ are different things; yes, $\mathbb{C}$ is a two-dimensional real vector space, and so we often think of it as a plane, but this is an identification we are making - they are not literally equal.

Yes, if $F$ is a field that contains $\mathbb{R}$ and $[F:\mathbb{R}]=2$ (i.e., $F\cong \mathbb{R}^2$ as real vector spaces), then $F\cong \mathbb{C}$. This is because $[F:\mathbb{R}]$ finite $\implies$ $F$ is an algebraic extension of $\mathbb{R}$, so $\overline{F}$ (the algebraic closure of $F$) is an algebraically closed field containing $\mathbb{R}$ and which is an algebraic extension of $\mathbb{R}$, so that $\overline{F}$ is also an algebraic closure of $\mathbb{R}$, and therefore $\overline{F}\cong\mathbb{C}$.

The most straightforward characterization of $\mathbb{C}$ is just that it is an algebraic closure of $\mathbb{R}$. It is also true that $\mathbb{C}$ is (up to isomorphism) the unique algebraically closed field of characteristic $0$ and cardinality $\mathfrak{c}$.

  • $\begingroup$ Maybe this is a bit to chatty but is $\mathbb{C}$ really a two dimensional vectorspace? For sure it is obviously isomorphic to one, but that doesn't mean it is already one, does it ? $\endgroup$ – Dominic Michaelis May 17 '13 at 7:17
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    $\begingroup$ @Dominic Given a field extension $L/K$, $L$ is a vector space over $K$. (Unless you really want to distinguish them and think of them as in different types of categories connected by a faithful functor, or constructed with different nuts-and-bolts set-theoretically, or something else at least arguably pedantic.) $\endgroup$ – anon May 17 '13 at 7:21
  • $\begingroup$ @anon I am more on the pedantic side of different types of catgories :) I mean for sure, as we have a faithful functor there is no reason to distinguish them at all. $\endgroup$ – Dominic Michaelis May 17 '13 at 7:23

First off, ${\bf R}^2$ is not a field. Let me build up the idea you want. If $L/K$ is a field extension, then in particular $L$ is a vector space over $K$, and the degree of the extension $[L:K]$ is defined to be the dimension $\dim_KL$ of $L$ over $K$. So perhaps you want to ask

Question: What kinds of degree $2$ extensions of the reals are there?

Answer: The complex numbers are the only such extension.

This requires only the most basic field theory; if $K/{\bf R}$ is a degree two extension, then $K={\bf R}(a)$ for any $a\in K\setminus{\bf R}$, and such an $a$ has minimal polynomial of degree two over $\bf R$, the only irreducibles over $\bf R$ are of the form $(x+s)^2+b$ with $b>0$, so ${\bf R}(a)\cong {\bf R}(\sqrt{-b})={\bf R}(i)={\bf C}$.

Note: the notation $F(a)$, where $F$ is some field, means "the smallest field containing $F$ and $a$," where $a$ lives in some field containing $F$.

There are many fields containing $\bf R$ other than just $\bf C$. Field theory gives ways to construct field extensions of any field, in fact. Here is one way: if $F$ is a field that is not algebraically closed, then there is some polynomial $p(x)\in F[x]$ that does not have a root in $F$, and we can adjoin a root of the polynomial to $F$ via the quotient ring $F[x]/(p(x))$.

(However, it does make sense to adjoin different roots and obtain distinct extensions, though the extensions are isomorphic as fields, from the perspective of some overlying field $L/F$. For instance, adjoining the real cube root of $2$ to $\bf Q$ and adjoining the complex cube root of $2$ to $\bf Q$ yield two distinct number fields, finite field extensions of $\bf Q$.)

Another way is to adjoin a transcendental element. If $F$ is a field and we want $T$ to be a transcendental element over $F$, then $F(T)$ can be thought of as the field of all rational functions in the variable $T$ with coefficients from $F$. (In the case of positive characteristic, we should be quick to note that two rational functions, in fact two polynomials, can act as the same function even though they are different as abstract expressions.)

This makes sense: if there were two distinct rational functions $a(\cdot)/b(\cdot)$ and $c(\cdot)/d(\cdot)$ such that $a(T)/b(T)=c(T)/d(T)$ in $F(T)$, then $a(T)d(T)-b(T)c(T)=0$ would make $T$ algebraic, a contradiction. Thus $F(T)$ contains all rational functions in $T$ as distinct elements, and by definition of minimality this means $F(T)$ need contain no further elements.

There is an analytically inspired way of creating fields that I will also note. If a field $F$ is also a metric space, then the metric space completion of the field $F$ will also be a field. This can be constructed abstractly as the ring of all sequences of elements of $F$ that are Cauchy with respect to the metric, modulo the maximal ideal comprised of null sequences (those that converge to $0$ in the metric).

Question: What kinds of ways are there of constructing new fields from old ones?

Answer: Adjoining algebraic elements, adjoining transcendental elements, completing with respect to a metric, taking certain types of closures.

An example of a closure is an algebraic closure of a field. We say $\bar{F}$ is an algebraic closure of $F$ is it is (i) algebraically closed, (ii) contains $F$, and (iii) there is no field lying between $\bar{F}$ and $F$ that is also algebraically closed. In fact any two algebraic closures of a field $F$ will be isomorphic (a fact that requires some amount of choice.) We can restrict ourselves to other closures too, such as maximal abelian extension, maximal (totally/tamely) ramified extensions, etc.

There is more to say about uniqueness of field extensions and its relation to $\bf C$ than in the first section above. In fact with the axiom of choice,

Question How many algebraically closed fields of cardinal size $\kappa\ge{\frak c}:=|{\bf R}|$ are there?

Answer: There is always only one, up to isomorphism.

I don't actually know how to prove such a fact.

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    $\begingroup$ I know $\mathbb R^2$ is not a field, but 2 tuple arithmetic rules like (a,b)(c,d)=(ac-db,ad+bc) etc. coupled with $\mathbb R^2$ make a field, but are there rules other than (a,b)(c,d)=(ac-db,ad+bc) combined with $\mathbb R^2$ that make a field? $\endgroup$ – Arjang May 17 '13 at 9:20

Just because I don't think the very first question is explicitly addressed in other answers:

Are there other rules than $(a,b)\times(c,d)=(ac−db,ad+bc)$ that make (the $\Bbb R$-vector space) $\Bbb R^2$ a field?

Yes there are, but the field defined is always isomorphic to $\Bbb C$ (by transfer through some linear isomorphism of $\Bbb R^2$, usually different from the identity). To limit trivial variations a bit we may insist that the element $1$ of the field will be $(1,0)$, so that in any case $(a,0)\times(c,0)=(ac,0)$. If you call $x$ the element $(0,1)$, then $(a,b)=a+bx$, so the multiplication is completely determined by knowing $x^2=(s,t)$, namely $(a,b)\times(c,d)=(ac+sbd,ad+bc+tbd)$. Not all choices will result in a field though. The element $x$ will satisfy the equation $x^2=s+tx$, so if we put $P=X^2-tX-s$ then the structure will be that of the quotient ring $\Bbb R[X]/(P)$. This will be a field if and only if $P$ has no real roots, which happens whenever it has negative discriminant: $D=t^2+4s<0$. In this case the linear map to the complex numbers, i.e., to $\Bbb R^2$ with the usual product, which sends $(1,0)$ to $(1,0)$, and $x=(0,1)$ to the complex root $(\frac t2,\frac12\sqrt{-D})$ of the polynomial $P$, is easily shown to be an isomorphism of fields.


This is not really an answer to your question, but you seem to be interested in the extent to which the complex structure on $\mathbb{C}$ is 'special' or 'unique'. I think the Gelfand-Mazur theorem might be of interest to you. It comes up in the study of spectral theory of operators on Banach spaces:

If $A$ is a commutative complex Banach algebra in which every nonzero element is invertible, then $A$ is isometric and isomorphic to $\mathbb{C}$.

In other words, if I give you some Banach algebra $A$ over $\mathbb{C}$ which satisfies the multiplicative and additive inequalities $\|xy\| \leq \|x\|\|y\|$ and $\|x+y\| \leq \|x\|+\|y\|$, so long as the space is i) complete and ii) a ''field'', that space must be the complex numbers.

The proof is also very simple if you assume the (non-trivial) result that the spectrum of any element $x \in A$ non-empty. To say that $\lambda \in \mathrm{sp}(x)$ is to say that $x-\lambda e$ is not invertible. The only non-invertible element in $A$ is 0, so $x-\lambda e = 0$. The isomorphism $\phi(x) = \lambda$ is also an isometry.

This theorem becomes useful in defining the Gelfand transform (see Rudin's functional analysis, chapter 11).


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