second half of lec 2

algebra1.sty

@@ -9,7 +9,7 @@
 \usepackage{mkessler-math}
 
 % Fancy theorem environments
-\usepackage[number in = section]{fancythm}
+\usepackage[number in = section, cache]{fancythm}
 
 % Set up of different types of todonotes
 \usepackage{mkessler-todo}

inputs/lectures/lec_01.tex

@@ -229,6 +229,7 @@ The lecture notes are available on eCampus.
 The following is a very important corollary:
 
 \begin{corollary}
+  \label{cor:preimage-of-prime-ideal-is-prime}
   Let $f\colon A \to  B$ be a ring homomorphism and $\mathfrak{p} \subset B$
   a prime ideal.
   Then $f^{-1}(\mathfrak{p}) \subset A$ is a prime ideal.

inputs/lectures/lec_02.tex

@@ -153,3 +153,163 @@ In particular, $k[x_1, \dotsc, x_n]$ is a unique factorization domain.
 \end{proof}
 
 % end of first half
+
+\begin{goal}
+  We would like to show that if $k$ is algebraically closed,
+  then every maximal ideal in $k[x_1, \dotsc, x_n]$ is of the form $\mathfrak{m}_P$
+  for some point $P \in k^n$.
+
+  We have just seen this for $n = 1$, but not generally.
+\end{goal}
+
+\begin{definition}
+  Let $k$ be a field and $A$ a $k$-algebra.
+  \begin{enumerate}[h]
+    \item An element $a\in A$ is called \vocab{algebraic over $k$}
+      if there exists $f\neq 0 \in k[x]$ with $f(a) = 0$.
+    \item $A$ is called \vocab{algebraic over $k$}
+      if all $a\in A$ are algebraic over $k$.
+    \item A finite subset $\set{ a_1,\dotsc,a_n  }\subset A $ is called
+      \vocab{algebraically independent} if $\ev_{a_1, \dotsc, a_n} \colon k[x_1, \dotsc, x_n] \to A$ is injective.
+
+      A set $M\subset A$ is called algebraically independent if all finite subsets
+      of $M$ are algebraically independent.
+    \item If $A$ is a field, a \vocab{transcendence basis} for $A$ over $k$
+      is an algebraically independent subset $M\subset A$ such that $A$ is algebraic
+      over $k(M)$.
+  \end{enumerate}
+\end{definition}
+
+\begin{example}+
+  $x^n\in k(x)$ is a transcendence basis of $k(x)$ over $k$ for all $n\geq 1$.
+\end{example}
+
+\begin{theorem}[Existence of transcendence basis]
+  \label{thm:transcendence-basis-completion}
+  Let $L / K$ be a field extension.
+  \begin{enumerate}[h]
+    \item $M \subset L$ is a transcendence basis for $L / K$
+      iff $M$ is maximally algebraically independent.
+    \item If $M' \subset L$ is algebraically independent and $M'' \subset L$ is such that
+      $L / K(M'')$ is algebraic,
+      then there exists a transcendence basis with $M' \subset M \subset M' \cup M''$.
+  \end{enumerate}
+  In particular, transcendence bases exist.
+\end{theorem}
+
+\begin{recall}
+  If $A$ is an integral domain,
+  the \vocab{field of fractions} $\Frac(A)$ of $A$ is the set of
+  equivalence classes of $\frac{a}{b}$ with $a \in A, b\neq 0 \in A$
+  under the relation $\frac { ac } { bc } = \frac{a}{b}$.
+\end{recall}
+
+\begin{lemma}[Universal property of the field of fractions]
+  \label{lm:universal-prop-of-field-of-fractions}
+  Let $A$ be an integral domain and $f\colon A \to  B$ a ring homomorphism.
+  Then, $f$ factors through 
+    \begin{equation*}
+    \begin{array}{c c l} 
+    A & \longrightarrow & \Frac(A) \\
+    a & \longmapsto &  \frac{a}{1}
+    \end{array}
+  \end{equation*}
+  if and only if $f(A \setminus \set{ 0 } ) \subset B^{\times}$.
+\end{lemma}
+
+\begin{lemma}
+  \label{lm:integral-domain-algebraic-over-k-is-field}
+  Let $A$ be an algebra over a field $k$.
+  \begin{enumerate}[h]
+    \item If $A$ is an integral domain and algebraic over $k$,
+      then $A$ is a field.
+    \item If $A$ is a field and contained in a finitely generated $k$-algebra,
+      then $A$ is algebraic over $k$.
+  \end{enumerate}
+\end{lemma}
+
+\begin{oral}
+  One might be tempted to deduce 2) from 1) by arguing that $A$
+  is itself finitely generated over $k$ since it is contained in a finitely generated
+  $k$-algebra.
+  However, this is not true in general, as the following example shows.
+\end{oral}
+
+\begin{example}+
+  Let $k$ be a field and consider the finitely generated $k$-algebra.
+  However, the subalgebra $B = k[x, xy, xy^2, xy^3, \dotsc]$ is not finitely generated.
+\end{example}
+
+\begin{corollary}[Zariski's Lemma]
+  \label{cor:zariski-lemma}
+  Let $L/K$ be a field extension. If $L$ is finitely generated as a $K$-algebra,
+  then $L / K$ is finite.
+\end{corollary}
+
+\begin{note}
+  The notions of being finitely generated as a $k$-algebra and being finitely generated
+  as a field extension over $k$ are not the same.
+
+  For example, $k(x) / k$ is finitely generated as a field extension,
+  but not as a $k$-algebra.
+\end{note}
+
+\begin{corollary}
+  Let $k$ be a field and let $A \to  B$ be a morphism of $k$-algebras.
+  Let $\mathfrak{m} \subset B$ be a maximal ideal.
+
+  If $B$ is finitely generated as a $k$-algebra, then $f^{-1}(\mathfrak{m})$ is maximal.
+\end{corollary}
+
+\begin{proof}
+  As in \autoref{cor:preimage-of-prime-ideal-is-prime}, by the homomorphism
+  theorem we get a commutative diagram
+  \[
+    \begin{tikzcd}
+      &
+      A
+      \ar[swap]{d}{}
+      \ar{r}{}
+      &
+      B
+      \ar{d}{}
+      \\
+      k
+      \ar{r}
+      &
+      \faktor{A}{f^{-1}(\mathfrak{m})}
+      \ar[swap,hook]{r}{}
+      &
+      \faktor{B}{\mathfrak{m}}
+    \end{tikzcd}
+  \]
+  Since $B$ is finitely generated as a $k$-algebra, so is $B / \mathfrak{m}$.
+  Since $B / \mathfrak{m}$ is also a field, by \autoref{cor:zariski-lemma}
+  it is algebraic over $k$.
+
+  Hence, also $A / f^{-1}(\mathfrak{m})$ is algebraic over $k$.
+  Since $A / f^{-1}(\mathfrak{m})$ is an integral domain,
+  by \autoref{lm:integral-domain-algebraic-over-k-is-field},
+  $f^{-1}(\mathfrak{m})$ is in fact maximal.
+\end{proof}
+
+\begin{corollary}
+  \label{cor:classification-of-max-ideals-in-polynomial-ring-over-alg-closed-field}
+  Let $k$ be an algebraically closed field
+  and $\mathfrak{m} \subset k [x_1, \dotsc, x_n]$ a maximal ideal.
+  Then, there exists a point $P \in k^n$ such that $\mathfrak{m} = \mathfrak{m}_P$.
+\end{corollary}
+
+\begin{corollary}[Hilbert's Nullstellensatz v1]
+  Let $k$ be an algebraically closed field
+  and $I \subsetneq k[x_1, \dotsc, x_n]$ a proper ideal.
+  Then, there exists $P \in k^n$ such that $f(P) = 0$ for all $f\in I$.
+\end{corollary}
+
+\begin{proof}
+  By Zorn's lemma, there is some maximal ideal $\mathfrak{m}$ with
+  $I \subset \mathfrak{m}$.
+  By \autoref{cor:classification-of-max-ideals-in-polynomial-ring-over-alg-closed-field},
+  $\mathfrak{m} = \mathfrak{m}_P = \ker(\ev_P)$ for some point $P \in k^n$.
+  Hence, for all $f\in I$, $f(P) = \ev_P(f) = 0$.
+\end{proof}