# A proof of Artin's linear independence of characters

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 $K[M]$.

• 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, Commented Apr 15, 2012 at 0:09
• 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. Commented Apr 15, 2012 at 0:49
• I'll learn it, Patrick. But it will take a while. Commented Apr 15, 2012 at 0:51
• @MakotoKato I've seen an easier proof, don't know if it's the trick you are talking about. Would you accept it?
– leo
Commented Dec 27, 2013 at 3:35
• @leo I would like to know the easy proof you have seen. Commented Dec 27, 2013 at 5:28

This is an approach different from yours. Let's precise some things.

Definition Let $G$ be a group and $F$ be a field.

1. A character from $G$ to $F$, it's a group homomorphism $\sigma:G\to F^\ast$, being $F^\ast$ the multiplicative group of units of $F$.
2. We say that a finite set of characters $\{\sigma_1,\ldots,\sigma_n\}$ is dependent, if there exist scalars $a_1,\ldots,a_n\in F$, not all $0$, such that $$\sum_{j=1}^n a_j \sigma_j(x) = 0\quad \forall x\in G.$$
3. A finite set of characters is independent if it is not dependent.

Theorem Let $G$ be a group and $F$ be a field. For any $n\in \Bbb N$, any set $\{\sigma_1,\ldots,\sigma_n\}$ of $n$ characters from $G$ to $F$ is independent.

Proof. Proceed by induction.

If $n=1$ and $a\in F$ then $$a\sigma(x) =0\quad \forall x\in G$$ implies $a=0$ because $\sigma(G)\subseteq F^\ast$.

Suppose that the theorem holds for any $k\in\{1,\ldots,n-1\}$, being this our induction hypothesis.

Arguing by contradiction, suppose that there is a set $\{\sigma_1,\ldots,\sigma_n\}$ of $n$ characters from $G$ to $F$ such that there exists $a_1,\ldots,a_n\in F$, not all $0$, such that $$\sum_{j=1}^n a_j\sigma_j(x) = 0\quad\forall x\in G. \tag{1}$$ Notice that if some $a_j$ is $0$ we'll have a dependent set of characters with less than $n$ elements. By our induction hypothesis, this can not be, so all the $a_j$ are not $0$.

Dividing in (1) by $a_n$, we can assume that $a_n=1$. So, we have $$0=a_1\sigma_1(x)+\cdots+a_{n-1}\sigma_{n-1}(x) + \sigma_n(x)\quad \forall x\in G.\tag{2}$$

Now, $\sigma_1\neq \sigma_n$ (otherwise $\{\sigma_1,\ldots,\sigma_n\}$ has not $n$ elements) and thus there is some $g\in G$ such that $\sigma_1(g)\neq \sigma_n(g)$. Equation (2) is valid for any element of $G$, particularly it is valid for elements of the from $gx$ with $x\in G$, then we get $$0=a_1\sigma_1(g)\sigma_1(x)+\cdots+a_{n-1}\sigma_{n-1}(g)\sigma_{n-1}(x)+\sigma_n(g)\sigma_n(x)\quad\forall x\in G.$$

Divide this last equation by $\sigma_n(g)$: $$0=a_1\frac{\sigma_1(g)}{\sigma_n(g)}\sigma_1(x)+\cdots+a_{n-1}\frac{\sigma_{n-1}(g)}{\sigma_n(g)}\sigma_{n-1}(x)+\sigma_n(x)\quad\forall x\in G.$$

Subtracting the equation (2) from this last one, we get $$0=a_1\left[\frac{\sigma_1(g)}{\sigma_n(g)}-1\right]\sigma_1(x)+\cdots+a_{n-1}\left[\frac{\sigma_{n-1}(g)}{\sigma_n(g)}-1\right]\sigma_{n-1}(x)\quad\forall x\in G.$$

Thanks to the independence of $\{\sigma_1,\ldots\sigma_{n-1}\}$ we obtain $$a_1\left[\frac{\sigma_1(g)}{\sigma_n(g)}-1\right]=0,$$ and since $a_1\neq 0$, this implies $\sigma_1(g)=\sigma_n(g)$, which is absurd due to the choose of $g$.

• why is $\sigma_n\ne\sigma_1$? Can you explain more?
– ZHU
Commented Sep 19, 2017 at 16:55