Often in mathematics I find we can ask "why" something is true. Of course, this is not a well defined question. However, it usually prompts answers that includes words like "intuitively..." or "morally...". These answers often can provide useful heuristics/analogies/intuition for reasoning about mathematics. Such questions I believe prompt useful discussion and can generate insights, and this is the sort of answer I'd like here.


Define a character $\chi$ of a group with values in the field $L$ to be a homomorphism

$$\chi\colon G \longrightarrow L^{\times}$$

We call the collection $\chi_1, \chi_2, \dots, \chi_n$ linearly independent (as functions) over $L$ if

$$a_1\chi_1 + a_2\chi_2 + \dots + a_n\chi_n = 0$$ for $a_i \in L$ implies that $a_1 = \dots = a_n = 0$.

There is a result that states that if $\chi_1, \chi_2, \dots, \chi_n$ are distinct characters of $G$ with values in $L$ then they are linearly independent over $L$. Of course there are formal proofs for this fact, one of which can be found in Dummit and Foote, Sec 14.2. However I feel that, given the amount of structure on the algebraic objects in this picture ($L$ a field, $G$ a group) there is a "reason" why this is true.

This result is important because it is used in the proof of the following theorem (which is a precursor to the fundamental theorem of Galois Theory):

Theorem: Let $G = \{\sigma_1 = \mathrm{id}, \sigma_2, \dots,\sigma_n\}$ be a subgroup of automorphisms of a field $K$ and let $F$ be the fixed field. Then $[K\colon F] = n = |G|$.

To prove this theorem, Dummit and Foote use the fact that distinct embeddings $\sigma_1, \dots, \sigma_n$ of a field $K$ into a field $L$ are linearly independent as functions on $K$. In particular, distinct automorphisms of $K$ have this property.

The main thing:

It seems there is some "thing" about the structure of fields that prevents characters and field isomorphisms/embeddings from being linear combinations of one another. Why is this so? Is this the case with weaker structures than a field as the image set? Can I represent field isomorphisms as combinations of higher degrees of one another? I.e is it possible for there to be some polynomial $f(x_1, \dots, x_n) \in F[x_1, \dots, x_n]$ such that

$$f(\chi_1, \chi_2, \dots, \chi_n) = 0$$

In summary, it seems that the structures surrounding Galois theory are somehow inter-related in some way I am yet to perceive, and I would like more insight on this matter.

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    $\begingroup$ I'm not sure if there's a simple intuition for why this is. If so, I suspect it will rely on some pretty heavy commutative algebra. Are you familiar with that type of language? $\endgroup$ – Alexander Gruber Mar 27 '14 at 6:28
  • $\begingroup$ @AlexanderGruber To some extent. Additionally, I will be getting a lot more familiar with commutative algebra in the coming weeks. Please don't let heavy machinery get in the way of an answer, I'm fascinated by this subject and will be reading much more about it in an attempt to understand it better. If there is anything I don't understand immediately, I will be able to revisit it later. $\endgroup$ – providence Mar 27 '14 at 6:30
  • $\begingroup$ Higher degrees of one another...? What does that mean? If $\phi:A\to B$ is a ring homomorphism, powers of $\phi$ may not be homomorphisms. If you only wanted to talk about multiplicative characters, if $\chi$ is a multiplicative character then $\chi^n$ for $n>1$ will still be, and so polynomials in multiplicative characters are just linear combinations of them. $\endgroup$ – blue Mar 28 '14 at 6:27
  • $\begingroup$ @seaturtles The definition given above singles out linear characters. If you can say something interesting using multiplicative characters I'd be interested to hear. Also note that I've specified linearly dependence/independence as functions. $\endgroup$ – providence Mar 28 '14 at 22:22
  • $\begingroup$ I know you're talking about linear in/dependence as functions. Linear characters and multiplicative characters are the same thing; group homomorphisms. The adjectives are used to distinguish from more general characters, which are traces of representations. Do you understand the points I made in my original comment? $\endgroup$ – blue Mar 28 '14 at 23:04

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