This tower of fields is being ridiculous

Question

Suppose $K\subseteq F\subseteq L$ as fields. Then it is a fact that $[L:K]=[L:F][F:K]$. No other hypotheses are needed (I'm looking at you, Hungerford V.1.2).

Now obviously $[\mathbf{C}:\mathbf{R}]=2$. But consider the fact that the algebraic closure of $\mathbf{R}(t)$ has cardinality $2^{\aleph_0}$---this implies that $\overline{\mathbf{R}(t)}\cong\mathbf{C}$, so in particular we can embed $\mathbf{R}(t)$ into $\mathbf{C}$.

If we embed $\mathbf{R}$ into $\mathbf{R}(t)$ in the natural way, we get $$\mathbf{R}\subset\mathbf{R}(t)\subset\mathbf{C}.$$

So our good fact at the beginning would have us believe

$$2=[\mathbf{C}:\mathbf{R}(t)][\mathbf{R}(t):\mathbf{R}].$$

What is the meaning of this? Either these two degrees really are both finite or (more likely) I've made a huge mistake. Perhaps it would all be clear if I were more precise about "embedding" $\mathbf{R}(t)$ in $\mathbf{C}$.

Answer 1

Your mistake is that $[\mathbb{C} : \mathbb{R}] \neq 2$!

To define the degree of a field extension is not enough to know the two fields involved (except in special cases): you actually have to know what the field extension is. In this case, the field extension $\mathbb{R} \to \mathbb{C}$ you constructed is not the field extension that comes from the inclusion $\mathbb{R} \subseteq \mathbb{C}$, and therefore it can, and does, have different degree.

A simpler (and more dramatic!) example of this phenomenon is the field extension $F(x) / F(x)$ given by the embedding $F(x) \to F(x)$ that sends $x \to x^2$. In this case, we have

$$ [F(x) : F(x)] = 2 $$

The supposition "$K \subseteq F \subseteq L$ as fields" usually implies more than it says: it also implies that the letter $F$ will sometimes be used not for a field, but for the field extension defined by the inclusion $K \to F$. Occasionaly we might disambiguate by writing $F/K$ rather than $F$. Similarly, $L$ will sometimes mean a field, and it will sometimes mean $L/F$ and it will sometimes mean $L/K$.

These sort of technicalities is the price we pay for the greater flexibility of allowing extensions to be any injective map $F \to E$, rather than requiring field extensions to come from actual subset relationships $|F| \subseteq |E|$ among sets. (In that last expression, $|F|$ means the underlying set, and $\subseteq$ has its usual set-theoretic meaning)

P.S. if you're ever in the situation where you consider the field extension $F(x) \to F(x)$ above, do yourself a favor and rename the indeterminate variable in one of the two copies of $F(x)$ rather than blithely forge ahead as I did above for dramatic effect. Similarly, it's probably wise to add a decoration to $\mathbb{R}$ to indicate when you are using it in a way inconsistent with its canonical inclusion into $\mathbb{C}$. (or decorate $\mathbb{C}$)

Answer 2

Note that $\mathbb R(t)\supset\mathbb R[t]$ and $\mathbb R[t]/(x^2+1)\sim\mathbb C$. Note that $[\mathbb R(t):\mathbb R]>2$. Note that all inequalities and inclusions in this answer are strict. Finally, note that if $R$ is a ring, $I$ is an ideal, and R is a vector space over a field $F$, then $[R:F]\le[R/I:F]$ if $R/I$ is still a vector space over $F$.

To see that $\mathbb R[t]/(x^2+1)\sim\mathbb C$, consider the homomorphism $f:\mathbb R[x] \to\mathbb C$ defined by $f(p(x)) = p(i)$.

This is injective, because $f(a + bx) = a + bi$ for any $z = a + bi$ in $\mathbb C$, and $(a + bx)$ is in $\mathbb R[x]$.

Next, $\ker (f) = \{p(x) \in\mathbb R[x]: p(i) = 0\} = \{p(x) \in\mathbb R[x]: p(i) = 0 \land p(-i) = 0\}, \text{since p has real coefficients} = \{p(x) \in\mathbb R[x]: (x - i)|p(x) \land (x + i) | p(x)\} = \{p(x) \in\mathbb R[x]: (x - i)(x + i) | p(x)\} = \{(x^2+1) q(x) : q(x) \in R[x]\} = (x^2 + 1)$.

Now from the homomorphism theorem: $\mathbb R[x]/\ker (f) = \mathbb R[x]/(x^2 + 1) \sim C$.

I hope this helps!