replace fm

2021_Algebra_I.tex

@@ -361,17 +361,17 @@ Equivalent\footnote{used in a vague sense here} formulation:
 \end{theorem}
 
 \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.
+    $I \se \mathfrak{m}$ for some maximal ideal. $R / \mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal.
-    $R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
+    $R / \mathfrak{m}$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
-    There are thus a field extension $\mathfrak{i} / \mathfrak{l}$ and an isomorphism $R / \fm \xrightarrow{\iota} \mathfrak{i}$ of $\mathfrak{l}$-algebras.
+    There are thus a field extension $\mathfrak{i} / \mathfrak{l}$ and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of $\mathfrak{l}$-algebras.
     By HNS2 (\ref{hns2}), $\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
-    Let $x_i \coloneqq \iota (X_i \mod \fm)$.
+    Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
     \[
-        P(x_1,\ldots,x_m) = \iota(P \mod \fm)
+        P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
     \] 
     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  $P \in I \se \fm$).
+    Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m} = 0$ for  $P \in I \se \mathfrak{m}$).
     HNS1 (\ref{hns1}) can easily be derived from HNS1b.
 \end{proof}
 
@@ -1165,17 +1165,17 @@ Then \[
         \item $\#\mSpec R = 1$
         \item  $R \sm R^{\times }$ is an ideal.
     \end{itemize}
-    If this holds, $\fm_R \coloneqq R \sm R^{\times }$ is the unique maximal ideal of $R$.
+    If this holds, $\mathfrak{m}_R \coloneqq R \sm R^{\times }$ is the unique maximal ideal of $R$.
 \end{definition}
 \begin{proof}
-    Suppose $\mSpec R = \{\fm\}$. If $x \in  \fm$, then $x \not\in R^{\times }$ as otherwise $xR = R \implies \fm = R$.
+    Suppose $\mSpec R = \{\mathfrak{m}\}$. If $x \in  \mathfrak{m}$, then $x \not\in R^{\times }$ as otherwise $xR = R \implies \mathfrak{m} = R$.
-    If $x \not\in  R^{\times }$ then $xR$ is a proper ideal, hence contained in some maximal ideal. Thus $x \in \fm$.
+    If $x \not\in  R^{\times }$ then $xR$ is a proper ideal, hence contained in some maximal ideal. Thus $x \in \mathfrak{m}$.
 
-    Assume that $\fm = R \sm R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \fm$. Hence $R = xR \se I \se \fm$. It follows that $\fm$ is the only maximal ideal of $R$.
+    Assume that $\mathfrak{m} = R \sm R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$. Hence $R = xR \se I \se \mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
 \end{proof}
 \begin{remark}
     \begin{itemize}
-        \item Any field is a local ring ($\fm_K = \{0\}$).
+        \item Any field is a local ring ($\mathfrak{m}_K = \{0\}$).
         \item The null ring is not local as it has no maximal ideals.
     \end{itemize}
 \end{remark}
@@ -1183,7 +1183,7 @@ Then \[
 Many questons of commutative algebra are easier in the case of local rings. Localization at a prime ideal is a technique to reduce a problem to this case.
 
 \begin{proposition}[Localization at a prime ideal]\label{locatprime}
-    Let $A$ be a ring and $\fp \in  \Spec A$. Then $S \coloneqq A \sm \fp$ is a multiplicative subset, $A_S$ is a local ring with maximal ideal $\fm = \fp_S =\{\frac{p}{s}| p \in  \fp, s \in S\} $.
+    Let $A$ be a ring and $\fp \in  \Spec A$. Then $S \coloneqq A \sm \fp$ is a multiplicative subset, $A_S$ is a local ring with maximal ideal $\mathfrak{m} = \fp_S =\{\frac{p}{s}| p \in  \fp, s \in S\} $.
 
     We have a bijection
     \begin{align}
@@ -1207,8 +1207,8 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
 % More remarks on localization at a prime ideal
 
 \begin{remark}
-    Let $B = \mathfrak{k}[X_1,\ldots,X_n]$, $x \in \mathfrak{k}^n$ and $\fm$ the maximal ideal such that $V(\fm) = \{x\}$.
+    Let $B = \mathfrak{k}[X_1,\ldots,X_n]$, $x \in \mathfrak{k}^n$ and $\mathfrak{m}$ the maximal ideal such that $V(\mathfrak{m}) = \{x\}$.
-    The elements of $B_\fm$ are the fractions $\frac{b}{s}, b \in B, s \in B \sm \fm$, i.e. $s(x) \neq 0$.
+    The elements of $B_\mathfrak{m}$ are the fractions $\frac{b}{s}, b \in B, s \in B \sm \mathfrak{m}$, i.e. $s(x) \neq 0$.
     These are precisely the rational functions which are well-defined in some neighbourhood of $x$.
     This will be rigorously formulated in \ref{proplocalring}.
     %Hence the name localization.
@@ -1291,10 +1291,10 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
                An element $x \in  A_\fp$ has the form $x = \frac{a}{s}$ for some $s \in  R \sm \fp$ and where $a \in  A$ is integral over $R$.
                Hence $a^n = \sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$. Thus $x^n = \sum_{i=0}^{n-1} \rho_i x^i$ with $\rho_i \coloneqq s^{i-n} r_i \in R_\fp$.
            \end{subproof}
-           As $i$ is injective and $R_\fp \neq  \{0\} $ ($R_\fp$ is local!) $A_\fp \neq  \{0\}$, there is $\fm \in \mSpec A_\fp$.
+           As $i$ is injective and $R_\fp \neq  \{0\} $ ($R_\fp$ is local!) $A_\fp \neq  \{0\}$, there is $\mathfrak{m} \in \mSpec A_\fp$.
-           D has already been shown and applies to $A_\fp / R_\fp$, hence $i\inv(\fm) = \fp_\fp$ is the only maximal ideal of the local ring $R_\fp$. Hence $\fq = \alpha\inv(\fm)$ satisfies
+           D has already been shown and applies to $A_\fp / R_\fp$, hence $i\inv(\mathfrak{m}) = \fp_\fp$ is the only maximal ideal of the local ring $R_\fp$. Hence $\fq = \alpha\inv(\mathfrak{m})$ satisfies
            \[
-               \fq \cap R = \alpha\inv(\fm) \cap R = \rho\inv(i\inv (\fm)) = \rho\inv(\fp_\fp) = \fp
+               \fq \cap R = \alpha\inv(\mathfrak{m}) \cap R = \rho\inv(i\inv (\mathfrak{m})) = \rho\inv(\fp_\fp) = \fp
            \] 
        \item[B] The map $\Spec A_\fp \xrightarrow{\alpha\inv}  \Spec A$ is injective with image equal to $\{\fq \in  \Spec A | \fq \cap R \se \fp\}$. In particular, it contains the set of all $\fq$ lying over $\fp$.
            If $\fq = \alpha\inv(\fr)$ lies over $\fp$, then \[\rho\inv(i\inv(\fr)) = (\alpha\inv(\fr)) \cap R = \fq \cap R = \fp = \rho\inv(\fp_\fp)\]
@@ -1651,25 +1651,25 @@ and it remains to show the opposite inequality.
 
 \begin{claim}
     For any maximal ideal $\fp \in \mSpec A$ \[
-        \hght(\fm) \ge \trdeg(K / \mathfrak{l})
+        \hght(\mathfrak{m}) \ge \trdeg(K / \mathfrak{l})
     \] 
 \end{claim}
 \begin{subproof}
     By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in  A^d$ which are algebraically independent over $\mathfrak{l}$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base of $K / \mathfrak{l}$.
 \begin{claim}
-    We can choose $x_i \in \fm$
+    We can choose $x_i \in \mathfrak{m}$
 \end{claim}
 \begin{subproof}
-     By the Nullstellensatz (\ref{hns2}), $\mathfrak{k}(\fm) = A / \fm$ is a finite field extension of $\mathfrak{l}$. Hence there exists a normed polynomial $P_i \in  \mathfrak{l}[T]$  with $P_i(x_i \mod \fm) = 0$ in $\mathfrak{k}(\fm)$.
+     By the Nullstellensatz (\ref{hns2}), $\mathfrak{k}(\mathfrak{m}) = A / \mathfrak{m}$ is a finite field extension of $\mathfrak{l}$. Hence there exists a normed polynomial $P_i \in  \mathfrak{l}[T]$  with $P_i(x_i \mod \mathfrak{m}) = 0$ in $\mathfrak{k}(\mathfrak{m})$.
-    Let $\tilde x_i  \coloneqq P_i(x_i) \in  \fm$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S / \tilde S$. It follows that $A / \tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in \fm$.
+    Let $\tilde x_i  \coloneqq P_i(x_i) \in  \mathfrak{m}$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S / \tilde S$. It follows that $A / \tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in \mathfrak{m}$.
 
     
 \end{subproof}
 % TODO: fix names A_1 = A_S, k_1 = R_S 
    The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$.
-    For $0 \le  i \le  d$, let $\fp_i \se R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\fm \sqcap R = \fp_0$ as all $X_i \in \fm$, hence $X_i \in \fm \sqcap R$ and $\fp_0$ is a maximal ideal.
+    For $0 \le  i \le  d$, let $\fp_i \se R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in \mathfrak{m}$, hence $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal.
-    By applying going-down and induction on $i$, there is a chain $\fm = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$.
+    By applying going-down and induction on $i$, there is a chain $\mathfrak{m} = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$.
-    It follows that $\hght(\fm) \ge  d$.
+    It follows that $\hght(\mathfrak{m}) \ge  d$.
 \end{subproof}
 This finishes the proof in the case of $\fp \in \mSpec A$.
 
@@ -1686,7 +1686,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
 \end{proof}
 
 \begin{remark}
-    As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\fm \in \mSpec A} \hght(\fm)$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite.
+    As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\mathfrak{m} \in \mSpec A} \hght(\mathfrak{m})$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite.
 \end{remark}
 \begin{dexample}[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}}
     Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$.
@@ -1750,7 +1750,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
         A &\longmapsto \bigcap_{\fp \in A}  \fp\\
         \Vspec(I) &\longmapsfrom I
     \end{align}
-    Under this bijection, the irreducible subsets correspond to the prime ideals and the closed points $\{\fm\}, \fm \in \Spec A$ to the maximal ideals.
+    Under this bijection, the irreducible subsets correspond to the prime ideals and the closed points $\{\mathfrak{m}\}, \mathfrak{m} \in \Spec A$ to the maximal ideals.
 \end{proposition}
 \begin{proof}
     If $A = \Vspec(I)$, then by \ref{sqandvspec} $\sqrt{I}  = \bigcap_{\fp \in A}  \fp$. Thus, an ideal with $\sqrt{I}  = I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I}  = I$ to closed subsets is surjective.
@@ -1887,18 +1887,18 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal.
 
 \subsection{The Jacobson radical}\limrel
 \begin{proposition}
-    For a ring $A, \bigcap_{\fm \in \mSpec A} \fm = \{a \in A | \A x \in  A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical}  of $A$.
+    For a ring $A, \bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m} = \{a \in A | \A x \in  A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical}  of $A$.
 \end{proposition}
 \begin{proof}
-    Suppose $\fm \in \mSpec A$ and $a \in  A \sm \fm$. Then $a \mod \fm \neq 0$ and $A / \fm$ is a field. Hence $a \mod \fm$ has an inverse $x \mod \fm$.
+    Suppose $\mathfrak{m} \in \mSpec A$ and $a \in  A \sm \mathfrak{m}$. Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field. Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$.
-    $1 - ax \in \fm$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS.
+    $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS.
 
-    Conversely, let $a \in A$ belong to all $\fm \in \mSpec A$. If there exists $x \in  A$ such that $1 - ax \not\in  A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\fm$. As $a \in \fm, 1 = (1-ax) + ax \in  \fm$, a  contradiction.
+    Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \mSpec A$. If there exists $x \in  A$ such that $1 - ax \not\in  A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\mathfrak{m}$. As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in  \mathfrak{m}$, a  contradiction.
-    Hence every element of $\bigcap_{\fm \in \mSpec A} \fm$ belongs to the right hand side.
+    Hence every element of $\bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m}$ belongs to the right hand side.
 \end{proof}
 
 \begin{example}
-    If $A$ is a local ring, then $\rad(A) = \fm_A$.
+    If $A$ is a local ring, then $\rad(A) = \mathfrak{m}_A$.
 \end{example}
 \begin{example}
     If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g.  $\Z$ of $\mathfrak{k}[X]$), then $\rad(A) = \{0\}$:
@@ -2748,10 +2748,10 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X
 \subsubsection{The local ring of an affine variety}
 \begin{definition}
     If $X$ is a variety, the stalk $\cO_{X,x}$ of the structure sheaf at $x$ is called the \vocab{local ring}  of $X$ at $x$.
-    This is indeed a local ring, with maximal ideal $\fm_x = \{f \in \cO_{X,x} | f(x) = 0\}$.
+    This is indeed a local ring, with maximal ideal $\mathfrak{m}_x = \{f \in \cO_{X,x} | f(x) = 0\}$.
 \end{definition}
 \begin{proof}
-    By \ref{localring} it suffices to show that $\fm_x$ is a proper ideal, which is trivial, and that the elements of $\cO_{X,x} \sm \fm_x$ are units in $\cO_{X,x}$.
+    By \ref{localring} it suffices to show that $\mathfrak{m}_x$ is a proper ideal, which is trivial, and that the elements of $\cO_{X,x} \sm \mathfrak{m}_x$ are units in $\cO_{X,x}$.
     Let $g = (U, \gamma)/\sim \in \cO_{X,x}$ and $g(x) \neq 0$.
     $\gamma$ is  Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \sm V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$.
     By the third point of \ref{localinverse} we have $\gamma \in \cO_X(U)^{\times}$.
@@ -2760,34 +2760,34 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X
 
 \begin{proposition}\label{proplocalring}
     Let $X = \Va(I) \se \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \se R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$.
-    $\{P \in  R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \se R$ is maximal, $I \se \fn_x$ and  $\fm_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$.
+    $\{P \in  R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \se R$ is maximal, $I \se \fn_x$ and  $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$.
-    If $\lambda \in  A \sm \fm_x$, we have $\lambda_x \in  \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to  \cO_{X,x}$.
+    If $\lambda \in  A \sm \mathfrak{m}_x$, we have $\lambda_x \in  \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to  \cO_{X,x}$.
-    By the universal property of the localization, there exists a unique ring homomorphism $A_{\fm_x} \xrightarrow{\iota} \cO_{X,x}$ 
+    By the universal property of the localization, there exists a unique ring homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} \cO_{X,x}$ 
     such that 
     \begin{figure}[H]
         \centering
         \begin{tikzcd}
-            A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\fm_x} \arrow[dotted, bend left]{ld}{\Eone \iota} \\
+            A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\Eone \iota} \\
             \cO_{X,x}
         \end{tikzcd}
     \end{figure}
     commutes.
    
-    The morphism $A_{\fm_x}\xrightarrow{\iota} \cO_{X,x}$ is an isomorphism.
+    The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} \cO_{X,x}$ is an isomorphism.
 
 \end{proposition}
 \begin{proof}
     To show surjectivity, let $\ell = (U, \lambda) / \sim  \in  \cO_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. 
     We have $X \sm U = V(J)$ where $J \se A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \sm V(f)$ we may assume $U = X \sm V(f)$.
     By \ref{oxulocaf}, $\cO_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$.
-    Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\fm_x}$.
+    Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\mathfrak{m}_x}$.
 
 
-    Let $\lambda = \frac{\vartheta}{g} \in  A_{\fm_x}$ with $\iota(\lambda) = 0$.
+    Let $\lambda = \frac{\vartheta}{g} \in  A_{\mathfrak{m}_x}$ with $\iota(\lambda) = 0$.
     It is easy to see that $\iota(\lambda) = (X \sm V(g), \frac{\vartheta}{g}) / \sim $.
     Thus there is an open neighbourhood $U$ of $x$ in $X \sm V(g)$ such that $\vartheta$ vanishes on $U$.
     Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \sm V(h) \se U$.
-    By the isomorphism $\cO_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \fm_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\fm_x}$ vanishes, implying $\lambda = 0$.
+    By the isomorphism $\cO_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$.
 \end{proof}
 \subsubsection{Intersection multiplicities and Bezout's theorem}
 \begin{definition}
@@ -2834,7 +2834,7 @@ Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\
 
 
 \begin{itemize}
-    \item[HNS2 $\implies$ HNS1b] Let $I \se \mathfrak{l}[X_1,\ldots,X_n]$. $I \se \fm$ maximal. $R / \fm$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2.
+    \item[HNS2 $\implies$ HNS1b] Let $I \se \mathfrak{l}[X_1,\ldots,X_n]$. $I \se \mathfrak{m}$ maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2.
     \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$).
         $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and  $L / K$ is finite.
     \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge  \aleph_1$.

algebra.sty

@@ -36,7 +36,6 @@
 \DeclareMathOperator{\hght}{ht}
 \newcommand{\Wlog}{W.l.o.g. }
 
-\newcommand{\fri}{\ensuremath\mathfrak{i}}
 \newcommand{\fm}{\ensuremath\mathfrak{m}}
 \newcommand{\Vspec}{\ensuremath V_{\mathbb{S}}}%\Spec}}
 \newcommand{\Vs}{\ensuremath V_{\mathbb{S}}}%\Spec}}