add exercises ch6 & polish Zariski's lemma, Weak Nullstellensatz, (strong) Nullstellensatz theorems (#8)

* add exercises 6.3, 6.4

* small typos

* small polishing

* polish Zariski's lemma & weak Nullstellensatz proofs

.github/workflows/typos.toml

@@ -8,6 +8,8 @@ iddeal = "ideal"
 iddeals = "ideals"
 allpha = "alpha"
 fieldd = "field"
+kernetl = "kernel"
+extenaion = "extension"
 
 # strings that are not a typo:
 thm = "thm"

commutative-algebra-notes.tex

@@ -1009,12 +1009,12 @@ As in with rings, it is equivalent to say that
 
 \subsection{A-algebras and integral domains}
 
-\begin{defn}{}[A-algebra]
+\begin{defn}{}[A-algebra / k-algebra]
   An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
 
   $B$ is an $A$-module with multiplication defined by $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
 
-  When $A \subset B$, $B$ is an extenaion ring of $A$; denoted $\psi(A) = A' \subset B$.
+  When $A \subset B$, $B$ is an extension ring of $A$; denoted $\psi(A) = A' \subset B$.
 \end{defn}
 
 \begin{defn}{R.4.1}\label{R.4.1}
@@ -1347,6 +1347,31 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
   thus there exists inverse in $A$, so $A$ is a field too.
 \end{proof}
 
+\begin{thm}{4.10.prev}[Zariski's lemma] \label{zariski}
+  $k$ a field, $L$ fingen $k$-algebra and a field. Then $L$ is a finite algebraic extension of $k$.
+
+  If $k$ is algebraically closed, then $L=k$.
+\end{thm}
+
+\vspace{0.4cm}
+From Zariski's lemma we can see that:\\
+let maximal ideal $m \subset k[X_1, \ldots, X_n]$, then $k[X_1, \ldots, X_n]/m$ is a (quotient) field.
+
+It's a fingen $k$-algebra, so by Zariski's lemma, $k[X_1, \ldots, X_n]/m$ is algebraic over $k$.
+
+Since $k[X_1, \ldots, X_n]/m$ is algebraically closed, it must equal $k$; ie.
+$$\frac{k[X_1, \ldots, X_n]}{m}=k$$
+Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown at \ref{5.2}).
+
+\vspace{0.4cm}
+Then:
+For $k$ algebraically closed, every maximal ideal of $k[X_1, \ldots, X_n]$ has the form 
+$(X_1 - a_1, \ldots, X_n - a_n)$ for some $a_i \in k$.
+
+Equivalently, $V(I) = \emptyset ~~\Longleftrightarrow  1 \in I$.
+
+\vspace{0.4cm}
+
 \begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski}
   let $k$ a field, $K$ a $k$-algebra which
   \begin{enumerate}
@@ -1420,7 +1445,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   The ideal generated by these terms is a subset of $m$:
   $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
 
-  Since $J$ is the kernetl of the evaluation map at point $(a_1, \ldots, a_n)$,
+  Since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$,
   then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have
   $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
 
@@ -1544,23 +1569,32 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   \begin{enumerate}[a.]
     \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\
   Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\
-  Then $L=k[X_1, \ldots, X_n]/m$ is a field (by TODO ref).
 
-  By Zariski's lemma (\ref{zariski}), since $L$ is generated as a $k$-algebra by the images of the variables $x_i$, and $k$ is algebraically closed.
+  Now,
+  \begin{itemize}
+    \item[-] since $m$ maximal, $L= k[X_1, \ldots, X_n]/m$ is a field.
+    \item[-] since $k[X_1, \ldots, X_n]$ is a fingen $k$-algebra, $L$ is a fingen $k$-algebra.
+  \end{itemize}
+  $\Longrightarrow~$ thus $L$ is a finite field extension of $k$.
 
-  Then the only algebraic extension of $k$ is $k$ itself. Thus $L \cong k$.
+  (Recall: if $k$ algebraically closed and $L$ a fingen field extension of $k$,
+  then $L \cong k$.)
 
-  \vspace{0.3cm}
+  Therefore,
-  Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$.
+  $$L=\frac{k[X_1, \ldots, X_n]}{m} \cong k$$
 
-  Let $a_i = \psi(x_i)$. Then $x_i - a \in ker(\psi) = m ~\forall~ i$.
+  Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, $\exists$ a surjective homomorphism
+  $$\psi: k[X_1, \ldots, X_n] \longrightarrow k$$
 
-  Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$.
+  Let $a_i = \psi(x_i)$. Then $x_i - a_i \in ker(\psi) = m ~\forall~ i \in [n]$.
 
-  Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$.
+  Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$. (as in \ref{5.2})
 
-  Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$.\\
+  Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$,
-  $\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$.
+  $$f \in m ~\Longleftrightarrow~ f(P)=0$$
+
+  Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$, ie. every element of $J$ vanishes at $P$.\\
+  $\Longrightarrow~$ therefore $P \in V(J)$, and thus $V(J) \neq \emptyset$.
 
 \item $I(V(J)) = rad J$:\\
   \begin{align*}
@@ -1742,7 +1776,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
 \end{eg}
 
 \begin{lemma}{6.2} \label{6.2}
-  For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong A[X]/(Xf-1)$.
+  For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$.
 
   Then
   $$A_f \cong \frac{A[X]}{(Xf-1)}$$
@@ -1775,7 +1809,7 @@ By the 1st isomorphism theorem:
 
   Now, we want to prove that $ker(\psi) \subseteq (Xf-1)$.
 
-  Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \foracll~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$.
+  Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \forall~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$.
 
   \vspace{0.3cm}
   Want to prove that $h(X)$ is a multiple of $(Xf-1)$.
@@ -1797,7 +1831,8 @@ By the 1st isomorphism theorem:
   $$\Longrightarrow~~ f^n \cdot h(X)=C \pmod{Xf-1}$$
   $$\Longleftrightarrow~~ f^n \cdot h(X)=Q(X) \cdot (Xf-1) + C$$
 
-  Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of fractions), $a\f^n = 0$ iff $\exist~ k$ such that $f^k \cdot a = 0$ in $A$.
+  Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of
+  fractions), $\frac{a}{f^n} = 0$ iff $\exists~ k$ such that $f^k \cdot a = 0$ in $A$.
 
   So, multiply both sides by $f^k$:
   $$f^k \cdot f^n \cdot h(X)=f^k \cdot (Q(X) \cdot (Xf-1) + C)$$
@@ -2531,6 +2566,148 @@ $$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
 
 \end{proof}
 
+
+\subsection{Exercises Chapter 6}
+
+\begin{ex}{R.6.3.a}
+  Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.
+\end{ex}
+\begin{proof}
+  ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$.
+
+  Localization map
+  \begin{align*}
+    \psi: A &\longrightarrow A'\\
+    (a', a'') &\longmapsto a'
+  \end{align*}
+
+  note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$.
+
+  Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$.
+
+  Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$.
+
+  In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$.
+  
+  Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$,
+
+  with $s \in S$, so $s=(1,0)$, hence
+  $$(1,0)\cdot(a',a'')=(0,0)$$
+  $$\Longrightarrow~ (a', 0) = (0, 0)$$
+  \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$.
+
+  $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$
+
+  $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$
+
+  By the 1st isomorphism theorem, 
+
+  \begin{tikzpicture}[node distance=1.5cm, auto]
+    \node (G)    {$A$};
+    \node (H) [right of=G] {$A'$};
+    \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$};
+
+    \draw[->] (G) to node {$\psi$} (H);
+    \draw[->] (G) to node [swap] {$\phi$} (GmodK);
+    \draw[<->] (GmodK) to node [swap] {$\eta$} (H);
+  \end{tikzpicture}
+  
+  $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$
+
+
+  \vspace{0.5cm}
+  Take the localization map
+  \begin{align*}
+    \phi: A &\longrightarrow S^{-1}A\\
+    a &\longmapsto a/1
+  \end{align*}
+
+  $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$
+
+  and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$
+
+  $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$
+
+  which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$.
+
+  \vspace{0.2cm}
+  $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel,
+  thus $A' \cong S^{-1}A$.
+\end{proof}
+
+\begin{ex}{R.6.4}
+  \begin{enumerate}[a.]
+  \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$.
+  \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by
+    $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$
+  \end{enumerate}
+\end{ex}
+\begin{proof}
+  \begin{enumerate}[a.]
+    \item
+  (\emph{saturated sets})\\
+
+  General rule of saturation:\\
+  two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
+
+  Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
+  $$\hat{S} = \{ a \in A ~|~ \exist~ b \in A ~\text{s.th.}~ ab \in S \}$$
+
+  \vspace{0.4cm}
+  Example 1: $A=\mathbb{Z}$,\\
+
+  $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$
+  localization:
+  $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$
+
+  $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$
+
+  Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\
+  since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$.
+
+  \vspace{0.4cm}
+  Example 2: $A= k[x]$,
+  $$S = \{ x^n ~|~ n \geq 0 \}$$
+  $$T = \{ (2x)^n ~|~ n \geq 0 \}$$
+  $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$
+
+  \item
+    Show that $T$ is a multiplicative set:
+
+    (must contain $1$ and be closed under multiplication)\\
+    Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$.
+
+    Let $t_1,~ t_2 \in T$; by definition of $T$,
+    $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$
+
+    Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\
+    rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\
+    Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$.
+
+    Thus $T$ is a multiplicative set.
+
+    Next, we show that $T$ is maximal:
+
+    Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$.
+
+    Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
+
+    Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
+    \hspace*{2em} $\exist~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
+    \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exist s' \in S$ such that $s'(ua-s)=0$\\
+    $\Longrightarrow~ s'ua=s's \in S$
+
+    since $s,s' \in S$, their product is also in $S$.
+
+    Let $b=s'a$, then $ub \in S$.
+
+    By definition of $T$, $u \in T$.
+
+    Therefore, $U \subseteq T$; thus $T$ is maximal.
+\end{enumerate}
+\end{proof}
+
+
 \bibliographystyle{unsrt}
 \bibliography{commutative-algebra-notes.bib}