diff --git a/.github/workflows/typos.toml b/.github/workflows/typos.toml index 8effe42..0455301 100644 --- a/.github/workflows/typos.toml +++ b/.github/workflows/typos.toml @@ -3,9 +3,11 @@ # run: typos -c .github/workflows/typos.toml [default.extend-words] +ieal = "ideal" iddeal = "ideal" iddeals = "ideals" allpha = "alpha" +fieldd = "field" # strings that are not a typo: thm = "thm" diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index 3491313..499f599 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 503301d..60eb50b 100644 --- a/commutative-algebra-notes.tex +++ b/commutative-algebra-notes.tex @@ -99,7 +99,7 @@ \end{defn} \begin{defn}{}[prime ideal] - if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$. + if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a \in P$ or $b \in P$. \end{defn} \begin{defn}{}[principal ideal] @@ -1338,7 +1338,7 @@ 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}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] +\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski} let $k$ a field, $K$ a $k$-algebra which \begin{enumerate} \item is finitely generated as a $k$-algebra @@ -1372,7 +1372,326 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some \end{proof} +\vspace{1cm} +\section{Nullstellensatz} + +Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue field $K=k[X_1, \ldots, X_n]/m$ satisfies the Zariski's lemma (\ref{zariski}), thus $K$ is a finite algebraic extension of $k$. + +\vspace{0.3cm} +\begin{cor}{5.2} \label{5.2} + $k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots, + X_n]$ is of the form + $$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$ + + The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$. + + Thus + \begin{align*} + k^n &\longleftrightarrow m-Spec A\\ + (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n) + \end{align*} +\end{cor} +\begin{proof} + let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal. + + By fundamental property of maximal ideals, $K=A/m$ is a field. + + Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $x_i' = x_i +m$. + + By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$. + + Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\ + \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$. + + So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$. + $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$ + $$\Longrightarrow x_i - a_i \in m$$ + + 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 $(a_1, \ldots, a_n)$, + then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have + $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$ + + \vspace{0.3cm} + Let + \begin{align*} + \psi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\ + \psi: x_i &\longmapsto a_i + \end{align*} + + Since $\psi$ is a $k$-algebra homomorphism, then $\forall f \in A$: + $$\psi(f(X_1, \ldots, X_n)) = f(\psi(x_1), \ldots, \psi(x_n))= f(a_1, \ldots, a_n)$$ + + Thus there is a one-to-one correspondence: + + points in $k^n ~~~ \longleftrightarrow~~~ m-Spec A$ (maximal ideals in $k[X_1, \ldots, X_n]$ + + $(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$ +\end{proof} + + +\vspace{0.5cm} +\subsection{Variety} + +\begin{defn}{5.3}[Variety] + A \emph{variety} $V \subset k^n$: + $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$ + + \begin{itemize} + \item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$ + \item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations. + \end{itemize} +\end{defn} + +\begin{prop}{5.3} \label{5.3} + $k$ an algebraically closed field, and $A=k[X_1, \ldots, X_n]$ a fingen $k$-algebra of the form $A=k[X_1, \ldots, X_n]/J$, where $J$ is an ideal of $k[X_1, \ldots, X_n]$. + (notation: $x_i = X_i \pmod J$) + + Then every maximal ideal of $A$ is of the form + $$(x_1 -a_1, \ldots, x_n - a_n)$$ + for some point $(a_1, \ldots, a_n) \in V(J)$. + + Therefore, $\exists$ a one-to-one correspondence + \begin{align*} + V(X) &\longleftrightarrow m-Spec A\\ + \text{given by}~~~(a_1, \ldots, a_n) &\longleftrightarrow (x_1 -a_1, \ldots, x_n - a_n) + \end{align*} +\end{prop} +\begin{proof} + the ideals of $A$ are given by ideals of $k[X_1, \ldots, X_n]$ containing $J$, since for $Q=R/I$, $\exists$ one-to-one correspondence between ideals of $Q$ and ideals of $R$ that contain $I$, ie. + \begin{align*} + m \subset A \longleftrightarrow &m' \subseteq k[X_1, \ldots, X_n]\\ + &\text{s.th.}~ J \subseteq m' + \end{align*} + + Thus, every maximal ideal of $A$ is of the form + $$\underbrace{(x_1 -a_1, \ldots, x_n - a_n)}_{i.} ~\text{such that}~ \underbrace{J \subset (X_1 - a_1, \ldots, X_n - a_n)}_{ii.}$$ + + \begin{enumerate}[i.] + \item Since $k$ is algebraically closed, the maximal ideals of $k[X_1, \ldots, X_n]$ look like $m=(X_1 - a_1, \ldots, X_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$. + + Which when projected to the quotient ring $A$, $X_i \longmapsto x_i$ (residue class), giving $(x_1 - a_1, \ldots, x_n - a_n)$. + + \item for $m$ to exist in $A$, the corresponding $m'$ must contain $J$; since if it didn't contain $J$ it wouldn't "survive" the quotient process. + \end{enumerate} + + \vspace{0.3cm} + However, since $(X_1 - a_1, \ldots, X_n - a_n)$ is the kernel of the evaluation map $f \longmapsto f(a_1, \ldots, a_n)$ + + $\Longrightarrow~$ means that $m'=(X_1 - a_1, \ldots, X_n - a_n)$ consists of all polynomials that vanish at point $P=(a_1, \ldots, a_n)$. + + If $J \subseteq m'$, then $\forall~ f \in J$ must vanish at $P$. + + By definition, the set of points where all polynomials in $J$ vanish is the \emph{variety}, $V(J)$. + + \vspace{0.4cm} + Thus,\\ + every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie. + $$m-Spec A \longleftrightarrow k^n$$ + + The condition that the ideal belongs to the quotient ring $A=k[X_1, \ldots, X_n]/J$ forces that point to lie in $V(J)$, so + \begin{align*} + m-Spec A &\longleftrightarrow V(J)\\ + \text{maximal spectrum} &\longleftrightarrow \text{variety} + \end{align*} +\end{proof} + +\begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5} + A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$). + + So $V$ gives a map:\\ + $$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$ + + correspondence going the other way: + $$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$ + + defined by taking a subset $X \subset k^n$ into the ideal + $$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$ + + \vspace{0.4cm} + + $V,~I$ satisfy reverse inclusions: + $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$ +\end{prop} + + +\vspace{0.4cm} +\subsection{Nullstellensatz} + +\vspace{0.5cm} +\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz} + Let $k$ algebraically closed field. + + \begin{enumerate}[a.] + \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$ + \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$, + $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$ + \end{enumerate} +\end{thm} +\begin{proof} + \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. + + Then the only algebraic extension of $k$ is $k$ itself. Thus $L \cong k$. + + \vspace{0.3cm} + Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$. + + Let $a_i = \psi(x_i)$. Then $x_i - a \in ker(\psi) = m ~\forall~ i$. + + Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$. + + Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$. + + Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$.\\ + $\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$. + +\item $I(V(J)) = rad J$:\\ + \begin{align*} + I(V(J)) &= rad J\\ + \text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety} + \end{align*} + where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$. + + Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$. + + Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$ + + Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie. + $$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$ + + Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires + $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$ + a contradiction. + + Therefore, $V(J') = \emptyset$. + + \vspace{0.4cm} + + Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\ + \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$. + + Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$ + + $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)} g_i) + \text{(polynomial)} \cdot (1 - Yf)$$ + + which, since $1 \in J'$, + $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$ + + substitute $Y=1/f$, + $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$ + thus + $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$ + multiply by $f^n$, + $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$ + + thus $f^n$ is a linear combination of $g_i$.\\ + Thus $f^n \in J$, so $f \in rad~J$. +\end{enumerate} +\end{proof} + + +\vspace{0.4cm} +\subsection{Irreducible varieties} + +\begin{defn}{5.7}[Irreducible variety] + a variety $X \subset k^n$ is \emph{irreducible} if it is nonempty and not the union of two proper subvarieties; that is, if + $$X = X_1 \cup X_2 ~~\text{for varieties}~ X_1, X_2 ~\Longrightarrow~ X= X_1 ~\text{or}~ X_2$$ +\end{defn} + +\begin{prop}{5.7} \label{5.7} + a variety $X$ is irreducible iff $I(X)$ is prime. +\end{prop} +\begin{proof} + set $I=I(X)$. + + if $I$ not prime, then $f,g \in A \setminus I$ be such that $fg \in I$. + + Define new ideals + $$J_1=(I, f) ~~\text{and}~~ J_2=(I, g)$$ + + Then, since $f \not\in I(X)$, it follows that $V(J_1) \subsetneq X$ + + \hspace*{2em}$\Longrightarrow$ so $X=V(J_1) \cup V(J_2)$ is reducible. + + The converse is similar. +\end{proof} + +\begin{cor}{5.8} \label{cor.5.8} + let $k$ algebraically closed field. Then $V$ and $I$ induce one-to-one correspondences + + \begin{align*} + \{ \text{radical ideals}~J~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{varieties}~ X \subset k^n \}\\ + \{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \} + \end{align*} + + Therefore, + $$Spec k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$ +\end{corollary} +\end{cor} + +\begin{prop}{5.8} \label{5.8} + let $A= k[x_1, \ldots, x_n]$ a fingen $k$-algebra ($k$ an algebraically closed field). + + Write $J$ for the ideal of relations holding between $x_1, \ldots, x_n$, so that $A=k[X_1, \ldots, X_n]/J$. + + Then there is a one-to-one correspondence + $$Spec A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$ +\end{prop} +\begin{proof} + By definition, $Spec A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$. + + By Corollary \ref{cor.5.8}: + $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$ + + About varieties: + \begin{itemize} + \item $I(X)$ in $R=k[X_1, \ldots, X_n]$ is the ideal of the variety $X$\\ + \hspace*{2em}ie. the set of all polynomials that vanish on every point of $X$. + \item $I(X)$ in $A=k[X_1, \ldots, X_n]/J$, we're not looking at all possible polynomials but at the residue classes. + \end{itemize} + + If $P$ is prime in $A~~\Longrightarrow~$ it must correspond to some prime ideal $\mathfrak{P}$ in $R$ (also $J \subset R$). + + Then from Nullstellensatz (\ref{nullstellensatz}), every prime ideal $\mathfrak{P}$ in $R$ is the ideal of some irreducible variety $X$.\\ + \hspace*{2em}$\Longrightarrow~~ \mathfrak{P}=I(X)$. + + + Since we're restricted to the ring $A=k[X_1, \ldots, X_n]/J$, the ideal $P$ are the elements of $I(X)$ viewed through the lens of the quotient\\ + $$\Longrightarrow~~ P = \{ f+J ~|~ f \in I(X) \}~~ = I(X) ~\text{mod}~J$$ + + Now, $J$ is the set of equations defining our "universe" $V(J)$. + + Since $A=R/J ~~ \Longrightarrow~$ we thus have $J \subseteq R$. + + Also we have a correspondence between $A=R/J$ and $R$. + + Let $\mathfrak{P}$ be the preimage of $P$ in $R=k[X_1, \ldots, X_n]$, ie. + \begin{align*} + R &\longrightarrow A=R/J\\ + \mathfrak{P} &\longmapsto P\\ + I(X) &\longmapsto I(X) ~\text{mod}~J + \end{align*} + + $\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$. + + + Now, $J \susbset \mathfrak{P} = I(X)$;\\ + by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so + $$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$ + $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$. + + Therefore, + $$\mathfrak{P} \in Spec A \longleftrightarrow X \subseteq V(J)$$ + where $X = V(\mathfrak{P})$. +\end{proof} \newpage