replace \fri

2021_Algebra_I.tex

@@ -363,13 +363,13 @@ Equivalent\footnote{used in a vague sense here} formulation:
 \begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
     $I \se \fm$ for some maximal ideal. $R / \fm$ is a field, since $\fm$ is maximal.
     $R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
-    There are thus a field extension $\fri / \mathfrak{l}$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\mathfrak{l}$-algebras.
-    By HNS2 (\ref{hns2}), $\fri / \mathfrak{l}$ is a finite field extension.
+    There are thus a field extension $\mathfrak{i} / \mathfrak{l}$ and an isomorphism $R / \fm \xrightarrow{\iota} \mathfrak{i}$ of $\mathfrak{l}$-algebras.
+    By HNS2 (\ref{hns2}), $\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
     Let $x_i \coloneqq \iota (X_i \mod \fm)$.
     \[
         P(x_1,\ldots,x_m) = \iota(P \mod \fm)
     \] 
-    Both sides are morphisms $R \to  \fri$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra.
+    Both sides are morphisms $R \to  \mathfrak{i}$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra.
 
     Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \fm = 0$ for  $P \in I \se \fm$).
     HNS1 (\ref{hns1}) can easily be derived from HNS1b.