18
$\begingroup$

I came up with a proof of Artin's linear independence of characters in field theory. The usual proof uses a clever trick devised by Artin. Since I'm not as clever as him, I prefer a proof which doesn't use a clever trick. Is this proof well-known? The proof consists of a few easy steps.

Step 1.

Let $K$ be a field. Let $A \neq 0$ be a not-necessarily-commutative associative unital $K$-algebra. Let $f_1,\dotsc,f_n$ be distinct $K$-algebra homomorphisms from $A$ to $K$. Let $\phi:A \to K^n$ be the map defined by $\phi(x) = (f_1(x),\dotsc,f_n(x))$. Then $\phi$ is surjective.

The proof is an easy consequence of Chinese remainder theorem.

Step 2.

Let $f_1,\dotsc,f_n$ be as above. There are elements $x_1,\dotsc,x_n$ of $A$ such that $f_j(x_i) = \delta(i, j)$ where $\delta(i, j)$ is Kronecker's delta.

The proof is an easy consequence of Step 1.

Step 3

Let $K$ and $A$ be as above. Let $\text{Homalg}(A, K)$ be the set of $K$-algebra homomorphisms from $A$ to $K$. Let $\text{Hom}(A, K)$ be the set of $K$-linear maps from $A$ to $K$. Then $\text{Homalg}(A, K)$ is a linearly independent subset of $\text{Hom}(A, K)$.

The proof is an easy consequence of Step 2.

Step 4 (Artin's linear independence of characters)

Let $K$ be a field. $K$ is regarded as a monoid by multiplication. Let $M$ be a not-necessarily-commutative monoid. Let $\text{Hom}(M, K)$ be the set of monoid homomorphisms. Let $K^M$ be the set of maps from $M$ to $K$. $K^M$ is regarded as a vector space over $K$. Then $\text{Hom}(M, K)$ is a linearly independent subset of $K^M$.

The proof is an easy consequence of Step 3 if one considers the monoid algebra $K[M]$.

  • 0
    May I suggest learning $\LaTeX$? Your questions will immediately be more readable, and it will save others from having to re-typeset your questions. See http://www.math.harvard.edu/texman/2012-04-14
  • 3
    Dear Makoto, This is a nice argument, which I haven't seen written explicitly in this manner before. Have you looked in *Bourbaki* to see how they argue? They often have conceptual arguments of this nature. Regards,2012-04-15
  • 1
    Thanks, Matt. I'm a big fan of Bourbaki and the style of the proof was influenced by them. However, they did use the Artin's trick to prove this theorem.2012-04-15
  • 1
    I'll learn it, Patrick. But it will take a while.2012-04-15
  • 1
    @MakotoKato I've seen an easier proof, don't know if it's the trick you are talking about. Would you accept it?2013-12-27
  • 1
    @leo I would like to know the easy proof you have seen.2013-12-27
  • 0
    Okay. I'll definitions, so you can see if we are talking about the same stuff.2013-12-27
  • 0
    Oh well, in my notes a character is a group homomorphism $\sigma:G\to F^\ast$, where $F^\ast$ is the group of units of the field $F$. Do you want to see the proof anyway?2013-12-27
  • 0
    @leo It's almost the same thing. Yes, I would like to see the proof.2013-12-27

1 Answers 1