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We know $\Bbb R$ is bigger than $\Bbb Q$ because its cardinality is bigger. We know that $\Bbb R^2$ is bigger than $\Bbb R$, which is bigger than $[0, 1]$ because the latter can be thought of as a subset of the former. However, the question is, how much bigger?

There is a notion of additive index, multiplicative index, and cardinality index. For example:

  • [0, 1] has additive index $2$ in [0, 2]. This is because [0, 2] is a union of 2 [0, 1]'s. Thanks to Captain Lama's comment, when we assign a measure on $\Bbb R$ and both sets are measurable, the additive index is simply the quotient of the measures of the two sets.
  • [0, 1] has additive index $\aleph_0$ in $\Bbb R$.
  • $\Bbb R$ has multiplicative index $\Bbb Z$/2 in $\Bbb R^2$, since $\Bbb R^2$ is a 2-dimensional vector space over $\Bbb R$. However, if we were looking at additive index, then $\Bbb R$ would be identified with $\Bbb R$ x {$0$}, and thus has additive index $\aleph_1$ in $\Bbb R^2$.
  • Similarly, $\Bbb R$ has multiplicative index $\Bbb Z$ in $\Bbb R^\Bbb Z$; however, the cardinality of both is the same.
  • Cardinality, then, is something to measure only the really really big differences between sets.

The most interesting index I've run into in math is, the multiplicative index. For example:

  • if $L$ is a finite-dimensional vector space over $K$ without additional structure, then we can define multiplicative index as simply $\Bbb Z/[L:K]$.

  • More generally, if $L$ is a Galois extension of $K$, then the multiplicative index can be defined to be the Galois group of $L/K$. In particular, this would allow differentiating things of countable multiplicative index, say those with Galois group $\Bbb Z$ x $\Bbb Z/2$ from those with Galois group $\Bbb Z$, or $\hat{\Bbb Z}$.

My questions is: how to define this multiplicative index when: 1) the extension is not Galois, and 2) cardinality is the same.

For example, $~~~~\bar{\!\!\!\!\!\!\!\!\!\rm if}$ denotes the algebraic closure, then we know $\bar{\Bbb Q}\cap\Bbb R/\Bbb Q$ is a countable extension. However, since the automorphisms of $\Bbb R$ are all trivial, this extension is not Galois. Is there a way to associate an index to this extension? What about to various extensions of $\bar{\Bbb Q}\cap\Bbb R$?

Alex
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  • Please: give us your definition of multiplicative index. It is not clear what it is just giving some examples. – Crostul Mar 02 '16 at 11:43
  • Have a look at http://math.stackexchange.com/q/243590/86776 – mvw Mar 02 '16 at 11:52
  • It seems very arbitrary to pick the cyclic group of a given order for the index, when you apparently want more structure once it happens to be a Galois extension. – Tobias Kildetoft Mar 02 '16 at 12:12
  • @mw This only addresses a bijection between two specific sets. I am asking how to compare cardinality of two given sets, something totally different.. – Alex Mar 02 '16 at 12:12
  • @TobiasKildetoft True, but that's the only choice that works for all extensions. I added "without additional structure" to reflect your point. – Alex Mar 02 '16 at 12:14
  • But everything here is dependent on extra structure anyway, as otherwise you would have nothing but cardinality. And you are definitely not talking about comparing cardinalities, but instead trying to get a better measure of the relative size of the sets, using the extra common structure. – Tobias Kildetoft Mar 02 '16 at 12:16
  • @TobiasKildetoft I mean, $\Bbb R^2$ / $\Bbb R$ is a vector space, but there is no notion of, say, $\Bbb R$-automorphisms of $\Bbb R^2$, because there's no reason to keep the first vs. the second coordinate fixed. – Alex Mar 02 '16 at 12:18
  • Any finite dimensional real vector space is also a real algebra, so you might as well use that structure (though this is of course also somewhat arbitrary). – Tobias Kildetoft Mar 02 '16 at 12:19
  • @Crostul The precise definition of the multiplicative index is the whole point of this question. How to define multiplicative index in a way to make, say, $\bar{\Bbb Q}∩\Bbb R$ have a meaningful multiplicative index over $\Bbb Q$. – Alex Mar 02 '16 at 12:20
  • @TobiasKildetoft Unless there is some a-priori algebra structure defined on the vector space, I believe the algebra structure we would impose would be based on some information we already know about the vector space... So I don't see how that would help defining a multiplicative index of any non-Galois extension. If you see something I don't, please post that as an answer. – Alex Mar 02 '16 at 12:28
  • Well, any finite dimensional vector space always has an algebra structure, it is just usually not very interesting. – Tobias Kildetoft Mar 02 '16 at 12:30
  • @TobiasKildetoft Once again, if that would somehow help define multiplicative index differently from just being a cyclic group, please post that as an answer. – Alex Mar 02 '16 at 12:35
  • Unless you put an extra structure on intervals (metric, or Lebesgue measure, perhaps ?) the fact that $[0;2]$ is the union of two copies of $[0;1]$ is meaningless. You need to specify what strucures you are working with. – Captain Lama Mar 02 '16 at 12:51
  • $\Bbb R^2$ is not bigger than $\Bbb R$. – BrianO Mar 03 '16 at 00:32
  • @BrianO Cardinality-wise, yes. But using the two other indices, it is bigger. $\Bbb R^2$ is the union of $\aleph_1$ copies of $\Bbb R$ (additive index), and is a $\Bbb Z/2$-vector space over $\Bbb R$ (multiplicative index). That being said, do you have an idea how to define multiplicative index to make it work for the question I asked? – Alex Mar 03 '16 at 01:43
  • But $\Bbb R$ is the union of $\aleph_1$ copies of $\Bbb R$. The "multiplicative index" is defined for vector spaces, not arbitrary sets. Via a bijection, we can transfer a one-dimensional real vector space structure to the set $\Bbb R^2$, so the notion collapses. – BrianO Mar 03 '16 at 02:41
  • @BrianO Good observation. I wish I could close my own question. – Alex Mar 03 '16 at 02:56
  • Too late now ;) And no problem, keep it. – BrianO Mar 03 '16 at 03:16

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