## 9.13 Linear independence of characters

Here is the statement.

Lemma 9.13.1. Let $L$ be a field. Let $G$ be a monoid, for example a group. Let $\chi _1, \ldots , \chi _ n : G \to L$ be pairwise distinct homomorphisms of monoids where $L$ is regarded as a monoid by multiplication. Then $\chi _1, \ldots , \chi _ n$ are $L$-linearly independent: if $\lambda _1, \ldots , \lambda _ n \in L$ not all zero, then $\sum \lambda _ i\chi _ i(g) \not= 0$ for some $g \in G$.

Proof. If $n = 1$ this is true because $\chi _1(e) = 1$ if $e \in G$ is the neutral (identity) element. We prove the result by induction for $n > 1$. Suppose that $\lambda _1, \ldots , \lambda _ n \in L$ not all zero. If $\lambda _ i = 0$ for some, then we win by induction on $n$. Since we want to show that $\sum \lambda _ i\chi _ i(g) \not= 0$ for some $g \in G$ we may after dividing by $-\lambda _ n$ assume that $\lambda _ n = -1$. Then the only way we get in trouble is if

$\chi _ n(g) = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i\chi _ i(g)$

for all $g \in G$. Fix $h \in G$. Then we would also get

\begin{align*} \chi _ n(h)\chi _ n(g) & = \chi _ n(hg) \\ & = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i\chi _ i(hg) \\ & = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i\chi _ i(h) \chi _ i(g) \end{align*}

Multiplying the previous relation by $\chi _ n(h)$ and substracting we obtain

$0 = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i (\chi _ n(h) - \chi _ i(h)) \chi _ i(g)$

for all $g \in G$. Since $\lambda _ i \not= 0$ we conclude that $\chi _ n(h) = \chi _ i(h)$ for all $i$ by induction. The choice of $h$ above was arbitrary, so we conclude that $\chi _ i = \chi _ n$ for $i \leq n - 1$ which contradicts the assumption that our characters $\chi _ i$ are pairwise distinct. $\square$

Lemma 9.13.2. Let $L$ be a field. Let $n \geq 1$ and $\alpha _1, \ldots , \alpha _ n \in L$ pairwise distinct elements of $L$. Then there exists an $e \geq 0$ such that $\sum _{i = 1, \ldots , n} \alpha _ i^ e \not= 0$.

Proof. Apply linear independence of characters (Lemma 9.13.1) to the monoid homomorphisms $\mathbf{Z}_{\geq 0} \to L$, $e \mapsto \alpha _ i^ e$. $\square$

Lemma 9.13.3. Let $K/F$ and $L/F$ be field extensions. Let $\sigma _1, \ldots , \sigma _ n : K \to L$ be pairwise distinct morphisms of $F$-extensions. Then $\sigma _1, \ldots , \sigma _ n$ are $L$-linearly independent: if $\lambda _1, \ldots , \lambda _ n \in L$ not all zero, then $\sum \lambda _ i\sigma _ i(\alpha ) \not= 0$ for some $\alpha \in K$.

Proof. Apply Lemma 9.13.1 to the restrictions of $\sigma _ i$ to the groups of units. $\square$

Lemma 9.13.4. Let $K/F$ and $L/F$ be field extensions with $K/F$ finite separable and $L$ algebraically closed. Then the map

$K \otimes _ F L \longrightarrow \prod \nolimits _{\sigma \in \mathop{\mathrm{Hom}}\nolimits _ F(K, L)} L,\quad \alpha \otimes \beta \mapsto (\sigma (\alpha )\beta )_\sigma$

is an isomorphism of $L$-algebras.

Proof. Choose a basis $\alpha _1, \ldots , \alpha _ n$ of $K$ as a vector space over $F$. By Lemma 9.12.11 (and a tiny omitted argument) the set $\mathop{\mathrm{Hom}}\nolimits _ F(K, L)$ has $n$ elements, say $\sigma _1, \ldots , \sigma _ n$. In particular, the two sides have the same dimension $n$ as vector spaces over $L$. Thus if the map is not an isomorphism, then it has a kernel. In other words, there would exist $\mu _ j \in L$, $j = 1, \ldots , n$ not all zero, with $\sum \alpha _ j \otimes \mu _ j$ in the kernel. In other words, $\sum \sigma _ i(\alpha _ j)\mu _ j = 0$ for all $i$. This would mean the $n \times n$ matrix with entries $\sigma _ i(\alpha _ j)$ is not invertible. Thus we can find $\lambda _1, \ldots , \lambda _ n \in L$ not all zero, such that $\sum \lambda _ i\sigma _ i(\alpha _ j) = 0$ for all $j$. Now any element $\alpha \in K$ can be written as $\alpha = \sum \beta _ j \alpha _ j$ with $\beta _ j \in F$ and we would get

$\sum \lambda _ i\sigma _ i(\alpha ) = \sum \lambda _ i\sigma _ i(\sum \beta _ j \alpha _ j) = \sum \beta _ j \sum \lambda _ i\sigma _ i(\alpha _ j) = 0$

which contradicts Lemma 9.13.3. $\square$

## Comments (4)

Comment #5684 by Yiyang Wang on

A small typo: in the proof of 9.13.1,  "Since we want to show that... " then the sigma should be ka

Comment #7276 by peter a g on

Possible typo in " Multiplying the previous relation by $\chi_n(h)$ and substracting we obtain ",

where "substracting" should probably be "subtracting". Still, I looked it up: 'substract' is in fact a variant, although the OED says it's (now) non-standard. (The OED also teaches us that in Shakespeare a substractor is a slanderer.)

Comment #7335 by on

OK! Well, then I am going to leave it...

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