# Algebraic structure built from the set of complex numbers

Is it correct to say that $\mathbb{C}$ is both a field, and a dimension 2 vector space? The complex numbers can be the elements of the field, or they can be considered as 2 elements of $\mathbb{R}$, in which case $\mathbb{C}$ exhibits the structure of a vector space.

To generalize, what algebraic structures can we attribute to $\mathbb{C}^{n}$?

So that we are all on the same page (if it was not clear), let's adopt the following standard definitions for field and vector space (there shouldn't be any surprise here, IMHO):

field: a set F together with two operations, usually called addition and multiplication, and denoted by + and ·, respectively, such that the following axioms hold: closure of F under + and . ; associativity of + and . ; commutativity of + and . ; existence of identity elements for + and . ; existence of inverses for + and . ; distributivity of . over +.

vector space: A vector space over a field F is a set V together with two binary operations (+ from VxV to V, * from FxV to F) that satisfy the axioms: associativity, commutativity, existence of identity and inverses for + ; a(u+v) = a u + b v ; (a+b)u = a u + b v ; a(b v) = (ab) v ; 1 v = v ; where a,b,c are in F and u, v are in V.

• The answer depends on exactly what you mean with what you're saying. What is a field to you? And what about a vector space? Aug 19, 2013 at 15:07
• I thought "field" and "vector space" where fairly unambiguous, except maybe for detail points. I'll specify/recall the usual definitions in my post. Aug 19, 2013 at 15:09
• Can you please define what 'set together with two operations' mean? Aug 19, 2013 at 15:22
• The complex numbers form a field. They form a one-dimensional complex vector space, a two-dimensional real vector space, a continuum-dimensional rational vector space, etc. The dimension of a vector space depends on what you take its scalar field to be. Up to isomorphism any finite-dimensional $\Bbb C$-algebra is a quotient of a multivariable polynomial ring ${\Bbb C}[x,y,\cdots,z]$.
– anon
Aug 19, 2013 at 15:23
• @GitGud Your questions seem inane and potentially derailing. There was no need for Frank to define what he means by "field," and you very well know what a binary operation is.
– anon
Aug 19, 2013 at 15:27

Frank - you are correct. $\mathbb{C}$ is a 2-dimensional vector space over $\mathbb{R}$. Similarly $\mathbb{C}^n$ can be considered an $n$-dimensional vector space over $\mathbb{C}$ and also as a $2n$-dimensional vector space over $\mathbb{R}$.
You can also think of $\mathbb C^n$ as the complexification of $\mathbb R^n$.
• Isn't it $\mathbb{C}^{n}$ and $\mathbb{R}^{2n}$? I will check how "complexification" is actually defined. Thanks! Aug 19, 2013 at 18:28
• @Frank It's $\mathbb R^n$. By the way, to answer your general question, the Wikipedia article does a pretty good job of describing the structure of a complexification of a real vector space. It's basically equivalent to a complex vector space together with a complex conjugation map. I haven't personally worked out the details, though, so I didn't want to include that in my answer. Aug 19, 2013 at 18:43