The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 $\sum \lambda _ i\sigma _ i(g) \not= 0$ 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CKL. Beware of the difference between the letter 'O' and the digit '0'.