Same symbol, different meanings? Subgroup index & field extension

1. Jul 26, 2011

nonequilibrium

Hello,

Say we have field (F,+,.) and field extension (E,+,.), then the degree of the field extension (i.e. the dimension of the vector field E across the field F) is given the symbol $[E:F]$.

But we can also see F and E as the groups (F,+) and (E,+), and then the same symbol denotes the subgroup index (in relation to Lagrange's theorem).

Is there any relation between these two cases, or is it just that the same symbol is used for two totally different things and it's assumed that the context makes clear what is meant?

In any case they're not equal (doesn't mean there can't be some meaningful connection, however), because for example as fields $[ \mathbb C : \mathbb R ] = 2$ while as groups $[ \mathbb C : \mathbb R ] = 2^\aleph_0$ (anyway I think so, because $\mathbb C = \cup_{r \in \mathbb R} (r.i + \mathbb R)$)

Then again for the rationals and irrationals they seem to coincide, but that might well be a coincidence.

Last edited: Jul 26, 2011
2. Jul 26, 2011

micromass

Are you certain of this?? Doesn't this need to be

$$[\mathbb{C}:\mathbb{R}]=2^{\aleph_0}$$

since the reals have cardinality $2^{\aleph_0}$.

In general if F is a field extension of K, then we can write $F=K^n$ as vector space (and where n is not necessarily finite).

So, as groups, we have

$$[F:K]=|F/K|=|K^n/K|=|K^{n-1}|$$

while as fields (or vector spaces), we have

$$[F:K]=\dim_K{F^n}=n$$

In general, it seems obvious that [F:K] as fields is always smaller than [F:K] as groups. But I cannot quite see any other interesting relation.

3. Jul 26, 2011

spamiam

Well, there are several similarities. For a group, $[G]$ is the number of distinct left cosets of H in G, or equivalently, the number of elements in a set of coset representatives for left cosets of H in G. Similarly, $[F:E]$ is the number of elements in a basis for F as a vector space over E. Other results hold in both cases as well. For instance, if $F \subseteq K \subseteq L$ are fields or groups, then
$$[L:F] = [L:K][K:F] \; .$$
I think there are other results that are true for both interpretations of the symbol [F:E], often involving divisibility.

Also micromass is right: as a group under addition, $\mathbb{C} \cong \mathbb{R}^2$ so $[\mathbb{C}: \mathbb{R}] = |\mathbb{R}| = 2^{\aleph_0}$.

4. Jul 26, 2011

nonequilibrium

Indeed the cardinality was a typo (fixed it in my OP)

thanks for the rest