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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 $K[M]$.

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    $\begingroup$ 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, $\endgroup$ – Matt E Apr 15 '12 at 0:09
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    $\begingroup$ 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. $\endgroup$ – Makoto Kato Apr 15 '12 at 0:49
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    $\begingroup$ I'll learn it, Patrick. But it will take a while. $\endgroup$ – Makoto Kato Apr 15 '12 at 0:51
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    $\begingroup$ @MakotoKato I've seen an easier proof, don't know if it's the trick you are talking about. Would you accept it? $\endgroup$ – leo Dec 27 '13 at 3:35
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    $\begingroup$ @leo I would like to know the easy proof you have seen. $\endgroup$ – Makoto Kato Dec 27 '13 at 5:28
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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 $g$.

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  • $\begingroup$ why is $\sigma_n\ne\sigma_1$? Can you explain more? $\endgroup$ – ZHU Sep 19 '17 at 16:55

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