Definition
A local ring is a ring which has a unique maximal ideal
$\mathfrak{p}$ is a prime ideal of a ring $A$. $A_{\mathfrak{p}}: =S_{\mathfrak{p}}^{-1}A$ where $S_{\mathfrak{p}}: = A-\mathfrak{p}$. Then $A_{\mathfrak{p}}$ is a local ring with unique maximal ideal $\mathfrak{m}_{\mathfrak{p}}$ consists of the quotients $\frac{x}{s}$, with $x$ in $\mathfrak{p}$ and $s$ in $A$ but not in $\mathfrak{p}$
If $B$ is a ring containing $A$, we denote by $B_{\mathfrak{p}}$ the ring $S_{\mathfrak{p}}^{-1}B$
Prop
Let $A$ be a ring and $S$ a multiplicative subset. Let $\mathfrak{a'}$ be an ideal of $S^{-1}A$. Then $\mathfrak{a'}=S^{-1}(\mathfrak{a'}\cap A)$
Proposition
$A$ is a ring and $x$ an element of some field $L$ containing $A$. Then TFAE:
Proposition 2 [S-L]
If $B$ is integral over $A$ and finitely generated as an $A-$algebra, then $B$ is a finitely generated $A-$module
Proposition 3
Let $A$ be a Noetherian ring, integrally closed, $L$ is a finite separable extension of its quotient field $K$. Then the integral closure of $A$ in $L$ is finitely generated over $A$
Proposition 4
Let $A\subset B$ be two rings, and $B$ integral over $A$. Let $\sigma$ be a homomorphism of $B$. Then $\sigma (B)$ is integral over $\sigma(A)$
Remark
Let $p$ be a prime ideal of $A$ and $\mathfrak{p}$ be a prime ideal of $B$ lies above $p$ ($\mathfrak{p}\cap A=p$) then we have a commutative diagran:
$$ \begin{equation}\begin{aligned}& B \longrightarrow & B / \mathfrak{P} \\& \uparrow &\uparrow \\& A \longrightarrow &A / p\end{aligned}\end{equation} $$
where the horizontal arrows are the canonical homomorphisms.
by Proposition 4, if $B$ is integral over $A$, then $B / \mathfrak{P}$ is integral over $A/ p$.
Proposition 6
Let $A$ be a Noetherian ring, integrally closed. Let $L$ be a finite separable extension of its quotient field $K$. Then the integral closure of $A$ in $L$ is finitely generated over $A$
Proposition 7
UFD is integrally closed.
Lemma 1.2
A finitely generated torsion-free module $M$ over an integral domain $A$ embeds into a finite free $A$-module. More precisely, if $M\neq 0$, there is an embedding $M \hookrightarrow A^d$ for some $d \geq 1$ such that the image of $M$ intersects each standard coordinate axis of $A^d$
Theorem 2.1
When A is a PID, each submodule of a free $A$-module of rank $n$ is finitely generated with at most $n$ generators.
Theorem 2.2
When $A$ is a PID, each submodule of a free $A$-module of rank $n$ is free of rank $\leq n$
Corollary 2.6
When $A$ is a PID, every finitely generated torsion-free $A$-module is a finite free $A$-module
That is. A finitely generated module over a PID that has no torsion elements admits a basis.
proof:
By Lemma 1.2 such a module embeds into a finite free $A$-module, so it is finite free too by Theorem 2.2
Theorem 1
Let $A$ be a PID, and $L$ a finite separable extension of its quotient field, of degree $n$. Let $B$ be the integral closure of $A$ in $L$. Then $B$ is a free module of rank $n$ over $A$.
proof
As a module over $A$, the integral closure is torsion-free, and by the general theory of principal ideal rings, any torsion-free finitely generated module is in fact a free module. It is obvious that the rank is equal to the degree $[L: K]$.
Proposition 8 [S-L]
Let $B$ be a ring integral over its subring $A$. $S$ is a multiplicative subset of $A$, then $S^{-1}B$ is integral over $S^{-1}A$. If $A$ is integrally closed, then $S^{-1}A$ is integrally closed
Corollary
lf $B$ is the integral closure of $A$ in some field extension $L$ of the quotient field of $A$, then $S^{-1} B$ is the integral closure of $S^{-1}A$ in $L$
Nakayama’s Lemma
Let $A$ be a ring, $\mathfrak{a}$ an ideal contained in all maximal ideals of $A$, and $M$ a finitely generated $A$-module. If $\mathfrak{a}M=M$, then $M=0$.
Proposition 9
$B$ is a ring containing $A$ and integral over $A$, $\mathfrak{p}$ is a prime ideal of $A$. Then $\mathfrak{p}B\neq B$, and there exists a prime ideal $\mathfrak{m}$ of $B$ lying above $\mathfrak{p}$