How can we work with field extensions when our base fields aren't actually subfields?

Question

I've been wondering this for a little while. Say we are working with the rational numbers $\mathbb{Q}$, and then we wish to talk about the extension fields $E$ of $\mathbb{Q}$, by which we mean the fields $E$, such that $\mathbb{Q}\subseteq E$. We first give the example of $\mathbb{R}$ as an extension field, which I suppose is agreeable, given that we do not usually go into the formal constructions of the real numbers.

But then, for example, how to we justify saying then that $\mathbb{R}\subseteq \mathbb{C}$? If we have our initial construction of $\mathbb{R}$, and then decide to work with pairs in $\mathbb{R}\times\mathbb{R}$ defined in the usual way such that $(a,b)\cdot (c,d)=(ac-bd,bc+ad)$ and call that $\mathbb{C}$ then the fact is that $\mathbb{R}$ is $\textbf{not}$ a subfield of $\mathbb{C}$. The only solution I can see to this is that in reality a field extension $E$ is just a field for which there exists a subfield $F'$ which is isomorphic to our base field $F$...or else redefine "subfield".

At first I found the subobject classifier as introduced in Lawvere's "Sets for mathematics" so be unwieldy....but now I see that it in fact is a reflection of the fact that our "subsets" rarely actually appear to be subsets...

My question isn't the best phrased, I think, but that's more due to a general confusion about this.

Answer 1

In case you have objects $A,B$ with a canonical embedding $i\colon A\to B$, you can always view $A$ as a subobject of $B$ and, if those objects have something like "underlying sets", also $A\subseteq B$. As a last resort you may replace $B$ with someting like $(B\setminus i(A))\cup A$. (Actually, even more surgery has to be done if $A\cap B$ happens to be nonempty in an "unexpected" way). This allows us to consider $A$ as subset of (something just as good as) $B$ after all, thus allowing us to write $$\mathbb N \subset \mathbb Z\subset \mathbb Q\subset \mathbb R\subset \mathbb C$$ even though we construct these sets as

• pairs ($\mathbb C$ from $\mathbb R$)
• equivalence classes of pairs ($\mathbb Z$ from $\mathbb N$ or $\mathbb Q$ from $\mathbb Z$)
• equivalence classes of sequences ($\mathbb R$ from $\mathbb Q$)

Strictly speaking, $2\in \mathbb N$ is therefore not the same as $+2\in\mathbb Z$, or $\frac21\in\mathbb Q$ or $2.000\ldots\in\mathbb R$ or $2+0i\in\mathbb C$. The main point is however that once we have performed one construction or the other (e.g., for $\mathbb R$: Cauchy sequences modulo zero sequences or Dedekind cuts) we can forget the construction used and identify all those incarnations of the ultimately same number.

Answer 2

"Subfield" is a structure, not a property. That is, when you want to exhibit a field $K$ as a subfield of a field $L$, you need to specify some data, namely the data of a homomorphism $K \to L$. The way in which $K$ sits as a subfield of $L$ can vary a great deal depending on the choice of this homomorphism: for example, it turns out that $\mathbb{C}$ has a proper subfield isomorphic to itself.

When we say that $\mathbb{R}$ is a subfield of $\mathbb{C}$, we mean that we've picked a particular homomorphism $\mathbb{R} \to \mathbb{C}$, namely $a \mapsto a + 0i$, and that we want to think of real numbers as particular complex numbers via this homomorphism.

For the same reason, "field extension" is a structure, not a property.