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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 $\mathbf{C}$.

Watson
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Chris Brooks
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  • I'm certainly open to title changes, it's just very late and I'm tired. – Chris Brooks Jun 05 '13 at 07:36
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    You can't embed $\mathbb{R}(t)$ in $\mathbb{C}$ as an $\mathbb{R}$-algebra. – Cocopuffs Jun 05 '13 at 07:40
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    @Joseph What would $t$ map to? Elements in $\mathbb{C}$ are roots of degree-two polynomials over $\mathbb{R}$, and if you've got an (injective) $\mathbb{R}$-algebra homomorphism from $\mathbb{R}(t)$ to $\mathbb{C}$, that would apply to $t$ as well. – Cocopuffs Jun 05 '13 at 07:46
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    Joseph: To your comment to @Cocopuffs, because then $2=\infty\cdot\infty$... :-) – Asaf Karagila Jun 05 '13 at 07:47
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    @Joseph It's not enough that $\mathbb{R}(t)$ and $\mathbb{C}$ have some arbitrary structure as fields - it needs to be compatible in the sense that $\mathbb{R}(t)$ is a subalgebra of $\mathbb{C}$. Maybe this isn't mentioned explicitly in your reference. – Cocopuffs Jun 05 '13 at 07:48
  • @Cocopuffs Hmm, this must be implicit somewhere in the exposition. Though I'm surprised that I missed this point in class and in the books I've read (or maybe it's just that I need to get some sleep) – Chris Brooks Jun 05 '13 at 07:59
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    @YACP: I don't understand the edit. While I don't conform to the provably mistaken tradition of using $\bf N,Z,Q,R,C$ instead of their blackboard bold counterparts, it's still a legitimate and common notational choice. Editing your preference over a common one seems like a wrong thing to do. See also this meta thread. – Asaf Karagila Jun 05 '13 at 08:00
  • @AsafKaragila What makes the tradition "provably mistaken"? In the past I've searched for the merits of one way vs. the other but come up empty (and no one I know cares enough to have an opinion) – Chris Brooks Jun 05 '13 at 08:05
  • @YACP: I have no intention to get into rollback wars, or edit wars, or any other wars of this kind. I'm merely pointing out that while editing out peculiar and unusual notations, where common ones exist is one thing; to edit out common notational choices is another. – Asaf Karagila Jun 05 '13 at 08:06
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    @anon, Joseph: I use the blackboard bold notation, and I'm always right... So it is provable that using $\bf R$ is wrong! :-) – Asaf Karagila Jun 05 '13 at 08:07

2 Answers2

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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 $\mathbb{C}$)

  • As far as I know the degree of a field, extension is the dimension of the extension as vector field over the smaller field. By this definition $[\mathbb C:\mathbb R]=2$ – superAnnoyingUser Jun 05 '13 at 08:10
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    @Ivan: Only if you choose the usual $\mathbb{R}$-vector space structure on $\mathbb{C}$, which we are not doing here. –  Jun 05 '13 at 08:19
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    @Ivan: A field can be contained in a bigger field in more than one way, allowing different degrees depending on which embedding you pick. Obviously the symbols $\bf C$ and $\bf R$ are supposed to come automatically equipped with a canonical inclusion, but we are forfeiting that for OPs purposes with a different embedding $\bf R\hookrightarrow C$. – anon Jun 05 '13 at 08:22
  • @Ivan: Note that $\Bbb C$ is isomorphic, as an additive group, to any $\Bbb Q$-vector space of dimension $2^{\aleph_0}$. So we can endow $\Bbb C$ with a different structure of scalar multiplication, in which case we end up with it being infinitely dimensional vector space over $\Bbb R$. – Asaf Karagila Jun 05 '13 at 08:26
  • The point is not to look at $\bf C$ itself any differently but rather to look at how $\bf R$ sits inside $\bf C$ differently, though. – anon Jun 05 '13 at 08:30
  • But endowing different scalar multiplication changes the nature of $\mathbb C$ and turns in into something else, does it not? I mean that we are considering $\mathbb C$ as a ring, with the usual multiplication. Otherwise we are just talking about $\mathbb R^2$, aren't we? – superAnnoyingUser Jun 05 '13 at 08:44
  • @Ivan: Yes, if by "$\mathbb{C}$" you mean "all of the usual implied relationships between $\mathbb{R}$ and $\mathbb{C}$". We're still considering $\mathbb{C}$ as a field in the usual way. However, we are considering the field extension defined by the field homomorphism $\mathbb{R} \to \mathbb{R}(t) \to \overline{\mathbb{R}(t)} \to \mathbb{C}$ the OP constructed rather than the canonical field extension defined by the canonical inclusion $\mathbb{R} \to \mathbb{C}$. –  Jun 05 '13 at 08:48
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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!

superAnnoyingUser
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  • That can't be right. I'm fairly certain that $\Bbb R(t)$ is isomorphic to an ordered field, whereas $\Bbb C$ is not. – Asaf Karagila Jun 05 '13 at 08:16
  • If $R$ is a ring and $I$ an ideal, the quotient $R/I$ is not canonically a subring of $R$, nor need it be isomorphic to any subring at all. In particular ${\bf R}(t)$ has no square roots of $-1$. – anon Jun 05 '13 at 08:18
  • Couldn't I impose the lexicographical order on $\mathbb C$, @AsafKaragila – superAnnoyingUser Jun 05 '13 at 08:37
  • Now it does not state that quotients of rings are subrings. It is strange that so much is devoted to the obvious isomorphism ${\bf R}[t]/(t^2+1)\cong\bf C$, and also I'm not sure what it has to do with OP's question directly, though it is useful to see that what constitutes an extension versus a subobject depends on more than just an object's abstract structure, but also on how things sit inside other things. – anon Jun 05 '13 at 09:03
  • Ivan, just to make it an ordered set? Sure. But if you want the order to be compatible with the structure of the field, then no. – Asaf Karagila Jun 05 '13 at 12:28