Dans le diagramme commutatif

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toutes les flèches sont plates. Il s’ensuit que l’on a pour tout entier $n$ et tout $\mathscr{A}$-module de type fini des isomorphismes naturels

$$((\mathscr{I}^{n}\mathscr{F})/(\mathscr{I}^{n+1}\mathscr{F}))_{\mathbf{x}}\simeq(\mathscr{I}^{n}\mathscr{F}_{\mathbf{x}})/(\mathscr{I}^{n+1}\mathscr{F}_{\mathbf{x}})$$

et

$$((\mathscr{I}^{n}\mathscr{F})/(\mathscr{I}^{n+1}\mathscr{F}))_{x}\simeq(\mathscr{I}^{n}\mathscr{F}_{x})/(\mathscr{I}^{n+1}\mathscr{F}_{x}).$$

Par conséquent,

$$\dim_{\mathscr{O}_{X,x}/\mathscr{I}}(\mathscr{I}^{n}\mathscr{F}_{x})/(\mathscr{I}^{n+1}\mathscr{F}_{x})=\dim_{\mathscr{O}_{\mathscr{X},\mathbf{x}}/\mathscr{I}}(\mathscr{I}^{n}\mathscr{F}_{\mathbf{x}})/(\mathscr{I}^{n+1}\mathscr{F}_{\mathbf{x}}).$$

Le lemme en résulte par dévissage. ∎