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This seems like a question that should be relatively easy to answer, but for the life of me I simply can't figure it out. My question is relatively simply put in the title, and the answer seems like it should be that $\mathbb{R}$ is a one dimensional space over itself whereas $\mathbb{C}$ is a two dimensional space over it. But this is unsatisfactory. Both of these fields are infinitely dimensional vector spaces over the rationals, but only $\mathbb{R}$ coexists in the same "number line". By that logic, $\mathbb{Q}(\sqrt{2})$ ought to be a plane, but since that's just a subfield of the reals, its on the number line as well. I suspect the answer to this has to do with the fact that $\mathbb{R}$ is totally ordered, but shouldn't it be possible to define a totally ordering on $\mathbb{C}$ that also respects the preexisting ordering? But then why is $\mathbb{H}$ 4-dimensional? Neither it nor $\mathbb{C}$ have apparent orderings, so what property of the former prevent is from residing within the complex plane?

Martin Sleziak
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tox123
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    While we can indeed put an ordering on $\mathbb{C}$, we cannot do so in a way compatible with the field structure - see the notion of ordered (or orderable) fields. – Noah Schweber May 23 '21 at 19:30
  • Just to clarify, $\mathbb{R}^2$ is a $2$ dimensional space, but $\mathbb{C}$ is a $1$ dimensional space. – copper.hat May 23 '21 at 19:34
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    @copper.hat $\mathbb{C}$ is a 2 dimensional space over $\mathbb{R}$. It's only one dimensional over itself. It satisfies all of the vector space axioms for real spaces and has a basis set of ${1,i}$ – tox123 May 23 '21 at 19:43
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    The concept of dimension depends on the base field. Sure, $\mathbb Q(\sqrt 2)$ is contained in the "real line", but the reals look like a line because you are seing it from the point of view of the reals.

    However $\mathbb Q(\sqrt 2)$ has dimension $2$ over $\mathbb Q$, so if you see it from $\mathbb Q$ you will see it as if it were a plane. For example, the same way you can reflect the complex plane around $\mathbb R$ with complex conjugation which sends $i\to -i$, you can reflect $\mathbb Q(\sqrt 2)$ around $\mathbb Q$ using the conjugation that sends $\sqrt 2 \mapsto -\sqrt 2$.

     – Jackozee Hakkiuz May 23 '21 at 19:57
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    I think @NoahSchweber has captured the essential thing here - the line provides a model for numbers in order. You can tell when a number is greater than zero, or between two others, by where the numbers fit on the line. We can fit integers or rational numbers or real numbers in a line according to an order which is compatible with the arithmetic involved. The key concept is order rather than dimension. – Mark Bennet May 23 '21 at 20:32

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Turning my comment into an answer:

I think the right way to make your question precise is to ask about the relationship between the algebraic and geometric natures of the real numbers. The key point in this context is not that there is an ordering of $\mathbb{R}$, but rather that there is an ordering of $\mathbb{R}$ which plays well with the algebraic structure. Formally, the ordering $\le$ satisfies the following algebraic properties:

  • $\le$ is a total ordering: that is, it is antisymmetric, transitive, and reflexive.

  • $a\le b$ implies $a+c\le b+c$ for all $c$.

  • $a\le b$ implies $ac\le bc$ for all $0\le c$.

These properties, together with some basic algebraic properties of addition and multiplication, constitute the ordered field axioms - so the point here is that $\mathbb{R}$ equipped with the usual ordering is an ordered field. Moreover, this ordering is unique in the sense that no other ordering on $\mathbb{R}$ turns it into an ordered field, and is algebraically definable since $a\le b$ iff there is some $x$ with $a+x^2=b$. All of this together says that we have a really nice way of ordering the real numbers.

By contrast, nothing of the sort will work for $\mathbb{C}$. Although we can put an ordering on $\mathbb{C}$, there is no way to do so which will satisfy the algebraic properties above. This is a good exercise (HINT: break into cases depending on whether $i$ is taken to be $\le0$ or $\ge0$).

Noah Schweber
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  • So in general can we consider the dimension of a ring is it's dimension as a vector space over the largest ordered subring? – tox123 May 25 '21 at 15:41