Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite
1

I'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.

In the world of vector spaces and fields, I keep coming across the definition of $\mathbb R^2$ as a vector space ontop of the field $\mathbb R$.

This makes me think, Why can't $\mathbb R^2$ be a field of its own? Would that make $\mathbb R^2$ a field and a vector space?

Thanks

share|improve this question
1  
A field has multiplication. How would you define multiplication on $\mathbb R^2$ so that it is a field? (There is a way to do so, but it isn't "obvious" until you realize that the resulting field is the complex numbers...) – Thomas Andrews Dec 4 '12 at 16:07
5  
$\mathbb{R}^2$ can be a field but with multiplication defined as follows: $(a,b)(c,d) = (ac - bd, ad + bc)$. Indeed, this is one way of defining the complex numbers. – Rankeya Dec 4 '12 at 16:08
1  
But, if you want to try to do this for $\mathbb{R}^n$, for $n \geq 3$ such that $\mathbb{R}$ is naturally embedded in $\mathbb{R}^n$ as a subfield, then it is not possible to do so, and this is a harder fact to prove. – Rankeya Dec 4 '12 at 16:11
ok this just answered my follow-up question. Is there any proof of that being true? – vondip Dec 4 '12 at 16:11
1  
Vondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is "larger", 4 + 7i or 6 + 5i ? (Answer: well, defining how "large" or "the length" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about. – Adam Rubinson Dec 4 '12 at 16:37
show 2 more comments

2 Answers

up vote 4 down vote accepted

If you define:

$$(a,b)+(x,y):=(a+x,b+y)$$

$$(a,b)\cdot (x,y):=(ax-by,ay+bx)$$

then the set $\,\Bbb R^2=\Bbb R\times\Bbb R\,$ turns into a field, and a rather well known and important one. Can you identify it?

share|improve this answer
Spoiler: take a look at one of the comments given. $\^\smile\^$ – FrenzY DT. Dec 4 '12 at 16:10
   
complex numbers indeed! fantastic how it all connects! Would that make R^2 a field and a vector space? – vondip Dec 4 '12 at 16:11
Yes, it would, because addition in $\mathbb{R}^2$, as @DonAntonio defines it, is component wise (the usual way). Remember, the vector space structure depends only on the underlying abelian group. – Rankeya Dec 4 '12 at 16:15
What do you mean "vector space"? Any field is a vector space over any subfield, so the field $\,\Bbb R^2\cong\Bbb C\,$ is a vector field ove an infinite number of subfields, say $\,\Bbb C\,,\,\Bbb R\,,\,\Bbb Q\,,\,\Bbb Q(i)\ldots\,$ , etc. – DonAntonio Dec 4 '12 at 16:15
So would that mean that I any vector space could be defined when F -being the field, as : F^n ? – vondip Dec 4 '12 at 16:17
show 2 more comments
up vote 5 down vote

It is important to understand that a set on its own has no algebraic structure. By defining operators on $\mathbb{R}^2$ you could turn it into (almost) anything you like.

The natural operators on $\mathbb{R}^2$, namely $(x, y) + (a, b) \mapsto (x+a, y+b)$ and $(x, y) \cdot (a, b) \mapsto (x\cdot a, y\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.