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author committer Tim Hosgood 2021-02-03 17:46:11 +0000 Tim Hosgood 2021-02-03 17:46:11 +0000 cd090d047414587cfa1103862a139d1279b66d7e (patch) 2e932dd4a2f2d1df1787222890d74cf7e17effd7 68b053959cae6459b4e9f5b100c78822ab307791 (diff) ega-cd090d047414587cfa1103862a139d1279b66d7e.tar.gzega-cd090d047414587cfa1103862a139d1279b66d7e.zip
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 diff --git a/ega1/ega1-1.tex b/ega1/ega1-1.texindex 63f7506..eab90c4 100644--- a/ega1/ega1-1.tex+++ b/ega1/ega1-1.tex@@ -269,7 +269,8 @@ for each $f'\in A'$, we therefore have, by definition, \begin{proof} The relation ${}^a\vphi(x)\in V(E')$ is, by definition, equivalent to $E'\subset\vphi^{-1}(\mathfrak{j}_x)$, so $\vphi(E')\subset\mathfrak{j}_x$, and finally $x\in V(\vphi(E'))$, hence (i).-To prove (ii), we can suppose that $\mathfrak{a}$ is equal to its radical, since $V(\rad(\mathfrak{a}))=V(\mathfrak{a})$ \sref{I.1.1.2}[v] and $\vphi^{-1}(\rad(\mathfrak{a}))=\rad(\vphi^{-1}(\mathfrak{a}))$;+To prove (ii), we can suppose that $\mathfrak{a}$ is equal to its radical, since $V(\rad(\mathfrak{a}))=V(\mathfrak{a})$ \sref{I.1.1.2}[(v)] and $\vphi^{-1}(\rad(\mathfrak{a}))=\rad(\vphi^{-1}(\mathfrak{a}))$;+if we set $Y=V(\mathfrak{a})$, and $\mathfrak{a}'=\mathfrak{j}({}^a\varphi(Y))$, then we have $\overline{{}^a(Y)=V(\mathfrak{a}')}$ (\sref{I.1.1.4}[(ii)]) the relation $f'\in\mathfrak{a}'$ is, by definition, equivalent to $f'(x')=0$ for each $x\in{{}^a\vphi(Y)}$, so, by Equation~\hyperref[1.1.2.1]{(1.2.1.1)}, it is also equivalent to $\vphi(f')(x)=0$ for each $x\in Y$, or to $\vphi(f')\in\mathfrak{j}(Y)=\mathfrak{a}$, since $\mathfrak{a}$ is equal to its radical; hence (ii). \end{proof}