diff --git a/.github/workflows/typos.toml b/.github/workflows/typos.toml index 0455301..0e0e6fd 100644 --- a/.github/workflows/typos.toml +++ b/.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" diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index 3a760ed..46139b6 100644 Binary files a/commutative-algebra-notes.pdf and b/commutative-algebra-notes.pdf differ diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex index 1b35444..73efa22 100644 --- a/commutative-algebra-notes.tex +++ b/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} - Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$. + Therefore, + $$L=\frac{k[X_1, \ldots, X_n]}{m} \cong k$$ - 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 $J$.\\ - $\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$. + Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$, + $$f \in m ~\Longleftrightarrow~ f(P)=0$$ + + 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}