port notes on propositions 5.3, 5.7, 5.8

commutative-algebra-notes.tex

@@ -1434,6 +1434,8 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
 
 
 \vspace{0.5cm}
+\subsection{Variety}
+
 \begin{defn}{5.3}[Variety]
   A \emph{variety} $V \subset k^n$:
   $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$
@@ -1445,9 +1447,45 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
 \end{defn}
 
 \begin{prop}{5.3} \label{5.3}
-  TODO
+  $k$ 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}, $V(J)$.
 \end{proof}
 
 \begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5}
@@ -1468,6 +1506,10 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$
 \end{prop}
 
+
+\vspace{0.4cm}
+\subsection{Nullstellensatz}
+
 \vspace{0.5cm}
 \begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
   Let $k$ algebraically closed field.
@@ -1545,6 +1587,100 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
 \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}
 
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