diff --git a/.github/workflows/typos.yml b/.github/workflows/typos.yml index 1c0695a..b1d0263 100644 --- a/.github/workflows/typos.yml +++ b/.github/workflows/typos.yml @@ -5,7 +5,7 @@ on: types: [ready_for_review, opened, synchronize, reopened] push: branches: - - main + - master jobs: typos: diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index 8f8cf2c..3491313 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 108b445..503301d 100644 --- a/commutative-algebra-notes.tex +++ b/commutative-algebra-notes.tex @@ -162,7 +162,7 @@ \end{defn} \begin{defn}{}[radical of an ideal] - $$rad I = \{ f \in A | f^n \in I~ \text{for some} n \}$$ + $$rad I = \{ f \in A | f^n \in I~ \text{for some}~ n \}$$ $rad I$ is an 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] +\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] let $k$ a field, $K$ a $k$-algebra which \begin{enumerate} \item is finitely generated as a $k$-algebra @@ -1348,18 +1348,28 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some extension. That is, $[K:k] < \infty$. \end{thm} \begin{proof} - by Noether normalization \ref{noether-normalization}, $\exists~ z_1, \ldots, - z_m \in K$ which are algebraically independent, and such that $K$ is finite - over $A=k[z_1, \ldots, z_m]$. + $K=k[z_1, \ldots, z_m]$ a field; want to show that $K$ is algebraic over $k$. + \vspace{0.3cm} + + Since $K$ is a fingen $k$-algebra, by Noether normalization lemma (\ref{noether-normalization}),\\ + $\exists~ z_1, \ldots, z_m \in K$ such that + \begin{itemize} + \item are algebraically independent + \item $K$ is integral over the polynomial ring $A=k[z_1, \ldots, z_m]$ (which by \ref{integral-implies-finite} is finite) + \end{itemize} \\ Now we're at the situation of \ref{R.4.9}: $A \subset K$ is integral, $K$ is a field $~~\Longrightarrow~$ therefore $A$ is a field. + \vspace{0.2cm} Since $z_1, \ldots, z_m \in K$ are algebraically independent,\\ - \hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$, and $K$ is finite over $k$. -\end{proof} + \hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$ + (since in $k[z_1]$ the element $z_1$ is not invertible, since $1/z_1$ is a rational function, not a polynomial). + + So $A=k$; which by Noether normalization we saw that $K$ is integral over $A=k$, and by \ref{integral-implies-finite} that it is finite, thus $K$ is finite over $k$, ie. $[K:k],\infty$, and $K$ is algebraic over $k$. +\end{proof} @@ -1781,6 +1791,152 @@ Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bi so that it is a exact sequence, thus, $M$ has a finite presentation. \end{proof} +\subsection{Exercises Chapter 4} + +\begin{ex}{R.4.1.a} + $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$. +\end{ex} +\begin{proof} + $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts: + $$f(X) = p(X^2) + X \cdot q(X^2)$$ + + with $p(X^2),~ q(X^2) \in k[X^2]$, and\\ + $p(X^2)$: sum of all terms with even exponents\\ + $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$. + + \vspace{0.3cm} + (Observation: this is used in FRI cryptographic protocol\\ + \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf}) + + + Rearrange it + \begin{align*} + f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\ + (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\ + f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\ + f(X)^2 \underbrace{-[2 p(X^2)]}_{a_1} f(X) &+ \underbrace{[p(X^2)^2 - X^2 \cdot q(X^2)^2]}_{a_0} = 0 + \end{align*} + + Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$. + + \vspace{0.2cm} + The integral dependence relation for any $f \in k[X]$ is given by the monic polynomial from \ref{R.4.1}.ii, in this case $T^2 + a_1 T + a_0 = 0$ with $a_i \in k[X^2]$. + + We have that + \begin{align*} + a_1 &= -2 p(X^2)\\ + a_0 &= p(X^2)^2 - X^2 q(X^2)^2 + \end{align*} + + + So for example, for $f(X)= X^3 + X^2 + X + 1$: + \begin{align*} + f(X) = (X^2 + 1) &+ X (X^2 + 1)\\ + (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\ + (f(X) - p(X))^2 &= X^2 (q(X))^2 + \end{align*} +\end{proof} + + +\begin{ex}{R.4.5} + Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$. +\end{ex} +\begin{proof} +$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$: + +Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then +\begin{align*} + (tX)^2 &= X^2 + X^3\\ + t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\ + t^2 &= 1+X, ~\text{thus}\\ + X&=t^2-1 \in k[X] +\end{align*} + +Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$. + +Hence $X, Y \in k[X]$. + +Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$. + +\vspace{0.4cm} +By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$. + +From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$. + +$(1+X) \in A$, so $t$ satisfies the monic polynomial +$$P(T) = T^2 - (1+X) \in A[T]$$ +Thus $t$ is integral over $A$. + +Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$. + +Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed). + +$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$ + +and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$. + +Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$. +\end{proof} + + + +\begin{ex}{R.4.9} + $k$ a field, $A= k[X,Y,Z]/(X^2- Y^3 -1, XZ - 1)$, find $\alpha, \beta \in k$ such that $A$ is integral over $B = k[X + \alpha Y + \beta z]$, and write a set of generators of $A$ as a $B$-module. +\end{ex} +\begin{proof} + (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$) + + The relations defining $A$ are + \begin{align*} + X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\ + XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1} + \end{align*} + +Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$. + + +Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$. + +$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$. + +Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\ +set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$. + +Let $w=X+Z$; since $XZ=1$, we have +$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$ + +which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$. + +Since $Z=w-X$, $Z$ is also integral. + +\vspace{0.4cm} +Generators of $A$ as a $B$-module: + +we hadd $B=k[w]$ with $w=X+Z$. + +From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$. + +Thus any polynomial in $X$ can be reduced to a linear form $b_1 X + b_0$ with $b_i \in k[w]$. Hence it's partial basis is $\{1, X\}$. + +Fitting $X^2$ into $(*)$, +\begin{align*} + X^2 &- Y^3 -1 =0\\ + Y^3 &= X^2-1\\ + Y^3 &= wX -2 +\end{align*} + +thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$). + +So its partial basis is $\{1, Y, Y^2\}$. + +For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$. + +\vspace{0.2cm} +Combining the previous partial basis, the generators are +$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$ + +\end{proof} + \bibliographystyle{unsrt} \bibliography{commutative-algebra-notes.bib}