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.
9.13 Linear independence of characters
Here is the statement.
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 subtracting 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
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