@@ -12,7 +12,7 @@ Commutative Algebra is the study of commutative Rings and related objects such a
We will use the convention that rings, (unless specified otherwise) are all commutative with unit.
We will also require that ring morphisms preserve the unit.
In particular, $2\Z\subseteq\Z$ is not a ring and$\Z\to\Z, x \mapsto0$ is not a morphism of rings.
In particular $\Z\to\Z, x \mapsto0$ is not a morphism of rings.
\subsubsection*{Why study this?}
One reason to study commutative algebra is that it is the ``local'' side of algebraic geometry, which is the study of geometric objects such as solution sets of systems of polynomial equations.
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@@ -134,6 +134,7 @@ However that is not the case. Many of the concepts we introduce are motivated by
This fact is false for maximal ideals. Consider the inclusion map $f: \Z\hookrightarrow\Q$.
As $\Q$ is a field $\{0\}$ is a maximal ideal, but $f^{-1}(\{0\})=\{0\}$ is not.
\item Krull: Let $I \neq R$ be an ideal of $R$. Then there exists a maximal ideal containing $I$. (See [ACL p. 57]
\item Let $R$ be a UFD, then $p \in R -\{0\}$ is prime if and only if $p$ is irreducible.
The following are more properties and were not part of the lecture
\begin{rem}[]
In the proof, we didn't need any special properties of $h_Y$, so the proof generalizes for any other set-valued functors $F$, i.e.
\begin{align*}
\Hom(X,Y) = h_Y(X) \iso\Nat(h_X,h_Y) \text{ generalises to }
F(X) \iso\Nat(h_X,F)
\end{align*}
Yoneda Lemma also shows functorial properties.
If $\Phi: F \to G$ and $\Psi: G \to H$ are natural transformations, we can define their \textbf{vertical composition}$\Psi\bm{\cdot}\Phi: F \to H$.
(Draw a diagram to see why it's called this way).
In particular, if we have morphisms $X \stackrel{f}{\to}Y \stackrel{g}{\to}Z$ with induced natural transformations $\Phi_f, \Phi_g$, then we have
\begin{align*}
\Phi_g \bm{\cdot}\Phi_f = \Phi_{g \circ f}
\end{align*}
So the Yoneda Lemma says that there is a functor $C \to\text{Func}(C,\text{Set})$ which preserves hom-sets and is ``surjective''. (The categoric definition here is that the functor is \textbf{fully-faithful}).
\end{rem}
\begin{cor}[] \label{cor:iso-with-hom}
The functors $h_X$ and $h_Y$ are isomorphic if and only $X$ and $Y$ are isomorphic.
\end{cor}
What this means is that to define an object $X$, it is enough to describe its hom-functor.
\begin{proof}
The Yoneda Lemma tells us that the mapping $\textsf{C}\to\textsf{Fun}(\textsf{C},\textsf{Set})$ is a fully faithul functor (See Definition \ref{edfn:fully-faithful} from Exercise sheet 1), where the objects in $\textsf{Fun}(\textsf{C},\textsf{Set})$ are functors and morphisms are natural transformations.
We show that any fully faithful functor respects isomorphisms.
Let $F: \textsf{C}\to\textsf{D}$ be a fully faithful functor and let $\Phi: F(X)\to F(Y)$ be an isomorphism.
Chose a morphism $f: X \to Y$ that induces $\Phi$. (In the case of the Yoneda Lemma, this was $f =\Phi_X(\id_X)$).
By fullness there exists a morphism $g: Y \to X$ that induces $\Phi^{-1}$.
By functoriality $f \circ g$ and $g \circ f$ induce the identity of $F(Y)$ and $F(X)$, and by faithfullness, they must be the identity on $Y$ and $X$.
\end{proof}
\begin{exr}[]
The Yoneda Lemma is also true for the covariant functor $h^{X}$.
Prove the Yoneda Lemma for the covariant functors$h^{X},h^{Y}$.
We want to define an invariant of a ring that measures its ``complexity'' and geometrically corresponds to the ``dimension'' of the zero set of polynomials $(f_i)$ over a field $K$, when $R =\faktor{K[X_{1}, \ldots, X_{n}]}{(f_{1}, \ldots, f_{m})}$.
$R$ is a field if and only if $R$ is an integral domain and $\dim R =0$.
\end{prop}
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@@ -26,7 +26,7 @@ $R$ is a field if and only if $R$ is an integral domain and $\dim R = 0$.
If we drop the condition that $R$ has to be an integral domain, then the statement is no longer true, as for $R = K_1\times K_2$ with $K_1,K_2$ fields is zero-dimensional, but not a field.
\begin{prop}[]
\begin{prop}[]\label{prop:pid-not-a-field}
If $R$ is a PID and not a field. Then $\dim R =1$, but the converse is not true, even for integral domains.
\end{prop}
\begin{proof}
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@@ -48,7 +48,7 @@ In particular, these propositions show that for a field $K$, we have
From Proposition \ref{prop:prime-ideals-localisation}, we know that there is an order-preserving bijection
\begin{align*}
\left\{
q \subseteq A_p \text{ prime}
\right\}
\iso\left\{
\tilde{q}\subseteq P \text{ prime}
\right\}
, \quad q \mapsto s^{-1}(q)
\end{align*}
which implies that prime chains in $R_p$ correspond to prime chains in $R$ that end in an ideal contained in $p$.
Any such chain cen be (if needed) lengtened to include either the maximal ideal $pR_p \subseteq R_p$, or the prime ideal $p \subseteq R$, so the conlusion $\dim R_p =\hht p$ follows.
\item Let $\pi: R \to R/p$ be the projection.
We can assume that a prime chain in $R_p$ ends in $pR_p$ because this is the uniqu maximal ideal, and that a prime chain in $R/p$ starts at $\{0\}$, so given prime chains
we can again use Proposition \ref{prop:prime-ideals-localisation} together with \ref{prop:ideals-and-quotients}
to get the prime chain
\begin{align*}
s^{-1}(p_0) \subsetneq\ldots\subsetneq s^{-1}(p_m) = p = \pi^{-1}(q_0) \subsetneq\pi^{-1}(q_1) \subsetneq\ldots\subsetneq\pi^{-1}(q_k)
\end{align*}
which has length $m + k$.
\end{itemize}
\end{proof}
One does not always have equality in (b), but it is true for most rings.
In particular, we will see that this is true if $R$ is an integral domain and a finitely-generated algebra over a field.
\begin{xmp}[] \label{xmp:ufd-principal-height}
Let $R$ be a UFD, $p \subseteq R$ prime. Then
\begin{align*}
\hht p = 1 \iff p \text{ is principal}
\end{align*}
\end{xmp}
\begin{proof}
\begin{itemize}
\item[$\implies$] We can find an irreducible element $a \in p$ since for any $a \in p -\{0\}$ and expressing it as a product of the fewest possible number of irreducible elements, $a$ cannot be factored non-trivially, as one factor would be in $p$ and involve fewer irreducibles, so $a$ is irreducible.
Then $\{0\}\subsetneq aR \subseteq p$, and since $\hht p =1$, it follows $p = aR$ is a principal ideal.
\item[$\impliedby$] Let $p = aR$ be prime. So $a$ is an irreducible element and we get $\{0\}\subsetneq aR$.
If there is a prime ideal sandwitched $\{0\}\subsetneq p \subseteq aR$, we must have $a \in p$, as otherwise, let $b \neq0\in p$ such that $b = a^{d}b'$, where $b$ is not divisible by $a$. Since $p$ is prime, we get $b' \in p$, but $b' \notin aR, \lightning$.
Thus, $a \in p \implies aR \subseteq p$, so $\hht p =1$.
\end{itemize}
\end{proof}
\begin{rem}[]
Later, we will prove two important related results.
\emph{Krull's Principal Ideal Theorem} states that if $R$ is noetherian and $a \in R$ not a zero-divisor, then there are minimal prime ideals $p \subseteq R$ such that $a \in P$ and all these satisfy $\hht p =1$.
The second result states that a noetherian integral domain $R$ is a UFD if and only if every prime ideal $p \subseteq R$ of height $1$ is principal.
\end{rem}
A somewhat counter-intuitive fact is that there exist infinite-dimensional noetherian rings, which are rings, where there aren't any infinitely long increasing chains of ideals, but prime chains of arbitrary length.
On the other hand one can show that if $R$ is notherian and local, then it has finite dimension.
For fields, one can talk about algebraic extensions, or algebraic elements.
We want to find the correct analogue for general rings.
For example, we cannot simply say that $x \in\C$ is \emph{algebraic over $\Z$}, if there exists a $p \neq0$ in $\Z[X]$ such that $p(x)=0$, because then there would be not difference with $x$ algebraic over $\Q$.
An element $a \in A$ is \textbf{integral over $R$}, if there exists a polynomial $P \in R[X]$ which is \emph{monic}\footnote{Or equivalently, has leading coefficient in $R^{\times}$}and satisfies $P(a)=0$.
\end{dfn}
\begin{xmp}[]
If $L/K$ is a field extension, then $x \in L$ is integral over $K$ if and only if $x$ is algebraic over $K$, since for fields, the leading coefficient is non-zero and thus a unit.
The element $a =\tfrac{1}{2}\in\Q$ is not integral over $\Z$, because $2X -1$ is not monic, but $\sqrt{2}$ is, since $X^{2}-2$ is monic.
The element $\omega=\tfrac{1+ i \sqrt{3}}{2}\in\C$ is integral over $\Z$, since $\omega^{2}=\tfrac{-1+ i \sqrt{3}}{2}=\omega-1$, and $P = X^{2}- X +1$ is monic.
\end{xmp}
Generally, any fraction $\tfrac{a}{b}\in\Q$ is only integral, if $b \in\Z^{\times}$.
\begin{prop}[]
Let $R$ be a UFD with fraction field $K$. Any $a \in K$ that is integral over $R$ is in $R$.
\end{prop}
\begin{proof}
Let $n \geq0$ and $P = X^{n}+ a_{n-1}X^{n-1}+\ldots+ a_1X + a_0\in R[X]$ such that $P(a)=0$.
Write $a =\tfrac{\alpha}{\beta}$ for $\alpha,\beta\in R$, $\beta\neq0$ and $\alpha, \beta$ coprime. (We can do this since $R$ is a UFD and thus a PID).
Modulo $\beta$, we deduce $\beta | \alpha^{n}$, which implies $\beta\in R^{\times}$, since $\beta$ is coprime to $\alpha^{n}$.
\end{proof}
\begin{dfn}[]
For an $R$-algebra $R \stackrel{s}{\to}A$ and an element $a \in A$, the universal property of polynomial rings (Proposition \ref{prop:polynomial-ring}) gives us a unique map
\begin{align*}
\Phi: R[X] \to A, \quad\text{such that}\quad X \mapsto a
\end{align*}
The \textbf{subalgebra generated by $a$} is
\begin{align*}
R[a] := \image\Phi = \left\{
\sum_{i=0}^{n} r_i \cdot a^{i}\big\vert n \in\N, r_i \in R
Let $P \in R[X]$ be a monic polynomial such that $P(a)=0$, let $d =\degree P$.
Because $P(a)=0$, this implies $a^{d}\in R + aR +\ldots+ a^{d-1}R$, so $R[a]$ is generated by $1,a,\ldots,a^{d-1}$ as an $R$-module.
\textbf{(b) $\bm{\implies}$ (c):} Just take $M = R[a]$. If $xM =0$, then $x \cdot1_R = x =0$.
\textbf{(c) $\bm{\implies}$ (a):} Let such an $M$ be given.
Define $u: M \to M$ by $u(m)= am$. This defines an $R$-linear map and since $M$ is finitelly-generated, there exists a monic $f \in R[X]$ with $f(u)=0$.
Then for all $m \in M$, we get $(f(u))(m)= f(a) m =0$, so $f(a)=0$, by the assumption $(\ast)$.
Such an $f$ exists because we can take $f =\det(x - u)$ by the Cayley-Hamilton Theorem.
In general, fixing a generating set $(x_{1}, \ldots, x_{n})$ of $M$ as $R$-modules, we get a commutative diagram
with $\alpha(a_{1}, \ldots, a_{n})=\sum_{i}a_Ix_i$ and $v$ determined by $v(e_i)=(u_{i_1}, \ldots, (u_{i_n})$, where $u(x_i)=\sum_{j}u_{ij}x_j$.
Then taking $f =\det(T - v)$, one gets $f(v)=0$ and it also folows that $f(u)=0$ because $\alpha$ is surjective and $f(u \circ\alpha)=\alpha\circ f(v)=0$.
\end{proof}
Note that the condition $(\ast)$ is necessary as for example, take $R =\Z$, $A =\Q$ and $a =\tfrac{1}{2}$ which is not integral over $\Z$. Then $M =\Z/3\Z$ is a finitely-generated $\Z[\tfrac{1}{2}]$-module because $2=-1$ has an inverse in $\Z/3\Z$.
An $R$-algebra $A$ is \textbf{integral over $R$} if all elements of $A$ are integral over $R$.
We say that $R \subseteq A$ is an \textbf{integral extension} and that the structure morphism $\phi: R \hookrightarrow A$ is an \textbf{integral morphism}.
\end{dfn}
\begin{prop}[]
Let $A$ be an $R$-algebra.
The set $B \subseteq A$ of all elements which are integral over $R$ forms a subalgebra of $A$.
It is called the \textbf{integral closure} of $R$ in $A$.
\end{prop}
\begin{proof}
We need to show that if the set $B$ is closed with respect to addition and multiplication.
Let $b,b' \in B$.
Clearly, $bb', b+b'\in R[b,b']= R[b][b']$ which is finitely-generated as an $R$-module, because if $(x_{1}, \ldots, x_{n})$ generate $R[b]$ as an $R$, module and $(y_{1}, \ldots, y_{m})$ generate $R[b][b']$ as an $R[b]$-module, then $(x_iy_j)_{ij}$ generate $R[b][b']$ as an $R$-module.
Moreover, $xR[b,b']=\{0\}\implies x1=0$, so the previous criterion shows that $bb'$ and $b+b'$ are indeed integral over $R$.
\end{proof}
\begin{dfn}[]
Let $R \subseteq A$ be rings. Then $R$ is \textbf{integrally closed} in $A$, it its integral closure is $R$ iteself.
If $R$ is an integral domain, then $R$ is said to be \textbf{integrally-closed} if it is so in its fraction field.
\end{dfn}
By the previous Proposition, any UFD is integrally closed.
\begin{lem}[]
If $R$ is an integrally closed integral domain, and $S \subseteq R$ multiplicative with $0\notin S$, then $S^{-1}R$ is integrally closed.
\end{lem}
\begin{proof}
The fraction field of $S^{-1}R$ is the same as the fraction field $K$ of $R$.
Let $x =\tfrac{a}{b}\in K$ be integral over $S^{-1}R$ and suppose
\item Let $A$ be a finitely-generated $R$-algebra (as an algebra).
Then
\begin{align*}
A \text{ is integral over $R$}\iff A \text{ is finitely-generated as an $R$-module}
\end{align*}
\item Let $B$ be integral over $A$ and $A$ integral over $R$. Then $B$ is integral over $R$.
\item If $A$ is integral over $R$ and $S \subseteq R$ multiplicative, then $S^{-1}A$\footnote{For the structure morphism $R \stackrel{\phi}{\to}A$, we write $S^{-1}A =\phi(S)^{-1}A$} is integral over $S^{-1}R$
\item If $R \stackrel{s}{\to}A$ is integral and $I \subseteq R$ an ideal and $J \subseteq A$ satisfies $s(I)\subseteq J$, then $R/I \to A/J$ is integral.
\item If $R \stackrel{\phi}{\to} A$ is an integral extension and $I \subseteq R$, $J \subseteq A$ ideals with $\phi(I)\subseteq J$, then $R/I \to A/J$ is an integral morphism.
\end{enumerate}
\end{prop}
\begin{proof}
\begin{enumerate}
\item[(a)]
Let $(a_{1}, \ldots, a_{n})$ be generators of $A$.
Basically, we just apply Proposition \ref{prop:integral-elements} to the generators of $A$.
\textbf{$\bm{\implies}$:}
By that Proposition, we know that $R[a_1]$ is a finitely-generated $R$-module and then so is $R[a_1,a_2]= R[a_1][a_2]$. By induction $A = R[a_{1}, \ldots, a_{n}]$ is finitely-generated.
\textbf{$\bm{\impliedby}$:} By the same Proposition, $a_{n}$ is integral over $R[a_{1}, \ldots, a_{n-1}]$ and $\ldots$ and $a_1$ is integral over $R$.
\item Because of (a), this surmounts to saying that being finitely-generated is transitive.
If $(a_i)_i$ are generators of $B$, and $(r_j)_j$ are generators of $A$, then $(a_ir_j)_{i,j}$ are generators of $B$.
\item Let $x =\tfrac{a}{s}\in S^{-1}A$.
The there is an equation $a^{n}+ r_{n-1}a^{n-1}+\ldots+ r_1 a + r_0=0$ with $r_i \in R$.
But as we will see in the next theorem, once we chose the last ideal in the chain $q_n$, there exists a prime chain that ends with $q_n$.
\end{rem}
This theorem lets us calculate many dimensions of rings. For instance, $A =\faktor{\Z[X]}{(X^{n}-1)}$ is integral over $\Z$ (see Example \ref{xmp:integral-polynomial})
By the theorem, $\dim A =\dim\Z=1$ (see Proposition \ref{prop:pid-not-a-field}).
Similarly, $\Z[\sqrt{5}]=\faktor{\Z[X]}{(X^{2}-5)}$ has dimension $1$.
We prove the theorem in five steps.
\begin{step}[]
If $A$ is an integral domain (so also $R$), then $\dim A =0\iff\dim R =0$.
By Proposition \ref{prop:field-iff-id-dim0}, this means that $A$ is a field $\iff$$R$ is a field.
\end{step}
\begin{proof}
Suppose $A$ is a field. Let $x \neq0\in R$ and $x^{-1}\in A$ its inverse.
By integrality, let $n \in\N, r_{i}\in R$ such that
