7 250408

Let $1=e_1+\cdots +e_n$ be a primitive idempotent decomposition in $A/J(A)$. Let $1=f_1+\cdots +f_n$ be a primitive idempotent decomposition in $A$ with $e_i=f_i+J(A)$.

Definition

We denote by $c_{ij}$ the multiplicity of $(A/J(A))e_i$ in $A{f}_j$, which is called the Cartan variant in $A$. The $n\times n$ matrix $(c_{ij}{)}_{n\times n}$ is called the Cartan matrix of $A$.

Theorem

$c_{ij}\ne 0$ iff $f_{i}A{f}_j\ne 0$.

\begin{proof} See ^e4tl53. \end{proof}

For convenience, if $S$ and $T$ are two simple $A$-modules, we denote by $C_{ST}$, the corresponding Cartan invariant, i.e., the multiplicity of $S$ in $P_T$.

Proposition

Let $A$ be a finitely generated $k$-algebra over a field $k$, and $S$ a simple $A$-module with projective over $P_S$. Let $A$ be a finitely dimensional $A$-module.

• If $T$ is a simple $A$-module, then

  $${\dim }_k{\mathrm{Hom}}_A(P_S,T)=\{\begin{array}{ll}\dim {\mathrm{End}}_A(S)& \text{if }S\simeq T,\\ 0& \text{otherwise.}\end{array}$$

• The multiplicity of $S$ in $U$ is $\dim {\mathrm{Hom}}_A(P_S,U)/\dim {\mathrm{End}}_A(S)$.

• If $e\in A$ is an idempotent, then $\dim \mathrm{Hom}(Ae,U)=\dim eU$.

• Let $e_S$ and $e_T$ be idempotents so that $P_S=A{e}_S$ and $P_T=A{e}_T$ are projective covers of $S$ and $T$. Then

  $$C_{ST}=\dim {\mathrm{Hom}}_A(P_S,P_T)/{\mathrm{End}}_A(S)=\dim e_{S}A{e}_T/\dim {\mathrm{End}}_A(S).$$

  If moreover $k=\overline{k}$, then $C_{ST}=\dim {\mathrm{Hom}}_A(P_S,P_T)=\dim e_{S}A{e}_T$.

\begin{proof} See ^qi7oer. \end{proof}

Let $R$ be a commutative ring with $1$. For $R$-modules $X,Y$ and $M$, we have an $R$-morphism

$$\begin{array}{rl}{\mathrm{Hom}}_R(X,M){\otimes }_R{\mathrm{Hom}}_R(M,Y)& \to {\mathrm{Hom}}_R(X,Y),\\ \phi \otimes \psi & \mapsto \psi \phi .\end{array}$$

Proposition

The image of this morphism is comprised of those morphism from $X$ to $Y$ which factor through some finite power of $M$.

\begin{proof} See ^v26ddz. \end{proof}

projective morphisms

Recall that the dual of an $R$-module $X$ is defined as $X^{\ast }={\mathrm{Hom}}_R(X,R)$. Taking $M=R$, we obtain a natural $R$-linear map

$${\tau }_{X,Y}:X^{\ast }{\otimes }_{R}Y⟶{\mathrm{Hom}}_R(X,Y),\phi \otimes y\mapsto (x\mapsto \phi (x)\cdot y).$$

The image of ${\tau }_{X,Y}$ is denoted by ${\mathrm{Hom}}_R^{\text{pr }}(X,Y)$, and its elements are called projective morphisms from $X$ to $Y$.

Lemma

For any $R$-module $M$, ${\mathrm{Hom}}_R^{\mathrm{pr}}(M,M)$ is a $2$-sided ideal of the ring ${\mathrm{End}}_R(M)={\mathrm{Hom}}_R(M,M)$.

\begin{proof} See ^smhffy. \end{proof}

Theorem

Let $P$ be an $R$-module. TFAE:

• $P$ is a finitely generated projective module.
• ${\mathrm{Hom}}_R^{\mathrm{pr}}(P,P)={\mathrm{End}}_R(P)$.
• For any $R$-module $X$, the morphism ${\tau }_{X,P}:X^{\ast }{\otimes }_{R}P\to {\mathrm{Hom}}_R(X,P)$ is an isomorphism.
• For any $R$-module $X$, the morphism ${\tau }_{P,X}:P^{\ast }\otimes X\to {\mathrm{Hom}}_R(P,X)$ is an isomorphism.
• The morphism ${\tau }_{P,P}:P^{\ast }{\otimes }_{R}P\to {\mathrm{End}}_R(P)$ is an isomorphism.

\begin{proof} See ^joizl5. \end{proof}

Integrality

Let $S$ be an commutative integral domain with $1$.

Definition

Let $R$ be a subring of $S$ containing $1_S$. We say $\alpha$ is an integral element over $R$, if there exists a monic polynomial $f(x)\in R[x]$ such that $f(\alpha )=0$.

Proposition

Let $S,R$ be rings and $\alpha \in S$. TFAE:

• $\alpha$ is an integral domain over $R$
• $R[\alpha ]$ is a finitely generated $R$-module.
• There exists a subring $T$ of $S$ with $R\subseteq T$ such that $T$ is a finitely generated $R$-module and $\alpha \in T$.

\begin{proof} See ^274bb5. \end{proof}

Lemma

$S,R$ are as above. $R\subseteq M_i\subseteq S$ where $M_i$ are subrings of $S$. If $M_1,M_2$ are finitely generated $R$-modules, then the ring $R[M_1,M_2]$ is finitely generated as an $R$-module.

\begin{proof} See ^0f6d0e. \end{proof}

Theorem

$S,R$ are as above. The integral elements of $S$ over $R$ form a subring of $S$ containing $R$.

\begin{proof} See ^b8xi28. \end{proof}

Definition

• $R,S$ are as above. The subring of $S$ containing of integral of $S$ over $R$ is called the integral closure of $R$ in $S$.
• $R$ is called integral closed in $S$ if every integral elements of $S$ over $R$ is containing in $R$.
• We say an integral domain $R$ is integral closed if it is integral closed in tis fractional field.
• $\mathbb{Z}\subseteq \mathbb{C}$, if $z\in \mathbb{C}$ is integral over $\mathbb{Z}$, we call $z$ an algebraic integer. All algebraic integer form a ring.

Theorem

A UFD is integral closed.

\begin{proof} See ^437xgo. \end{proof}

Torsion and torsion free elements

Definition

Let $R$ be a ring, and let $M$ be a $R$-module. $m\in M$ is said to be

• a torsion element if ${\mathrm{Ann}}_R(m)\ne 0$
• a torsion free element if ${\mathrm{Ann}}_R(m)=0$.

Example.

• $\mathbb{Z}$-module $\mathbb{Q}$ has no torsion element but $0$.
• $R$ is commutative,
  • ${}_{R}R$ has no torsion element but $0$ iff $R$ is an integral domain;
  • if $I$ is a proper ideal, all element if $R$-module $R/I$ are torsion;
  • $R/I$ has no torsion element but $0$, iff $I$ is prime.

Lemma

If $R$ is an integral domain, the subset of $M$

$$\mathrm{tor}(M)=\{m\subseteq M:{\mathrm{Ann}}_R(m)\ne 0\}$$

is a submodule of $M$, which is called the torsion submodule of $M$.

Definition

We say $M$ is a torsion module if $M=\mathrm{tor}(M)$, and $M$ is torsion free if $\mathrm{tor}(M)=0$.

Examples.

• $\mathbb{Z}$-modules: $\mathbb{Z},\mathbb{Q},\mathbb{Z}/2\mathbb{Z}$, $\mathrm{tor}(\mathbb{Q})=0$ and $\mathrm{tor}(\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$
• For an integral domain $R$ and a $R$-module $M$, $\mathrm{tor}(M^{\ast })=0$ where $M^{\ast }={\mathrm{Hom}}_R(M,R)$.

Proposition

Let $R$ be an integral domain. Free modules are torsion free.

Proposition

Let $R$ be an integral domain. Any finitely generated torsion free $R$-module is isomorphic to a submodule of a free $R$-module of finite rank.

\begin{proof} See ^czedmd. \end{proof}

Localization

f:S^{-1}A\otimes_A M\simeq S^{-1}M,a/s\otimes m\mapsto am/s.

Link to original

Set $FM=F{\otimes }_{R}M=S^{-1}M$ with $S=R\setminus \{0\}$. The kernel of $M\to FM,m\mapsto 1\otimes m$ is $\mathrm{tor}(M)$.

Corollary

• $F{\otimes }_R\mathrm{tor}(M)=0$
• TFAE:
  • the natural $R$-morphism $M\to FM,m\mapsto 1\otimes m$ is injective
  • $M$ is torsion free.

Corollary

If $M$ is finitely generated projective $R$-module, then $M\to FM,m\mapsto 1\otimes m$ is injective.

\begin{proof} See ^952afe. \end{proof}

Definition

The rank of a finitely generated torsion free $R$-module is the dimension of the $F$-space $FM$.

Remark. A finitely generated torsion free $R$-module $M$ of rank $1$ is isomorphic to a finitely generated $R$-submodule of $F$.

• Indeed, if $FM$ is a $1$-dimensional vector space over $F$, then $FM=Fe$. Define $I=\{\lambda \in F:\lambda e\in M\}$, then $I$ is a $R$-submodule of $F$, and we have $M=Ie$.
• In other words, a finitely generated torsion-free $R$-module of rank $1$ can be viewed as a fractional ideal of $R$, that is, as a finitely generated $R$-submodule of its field of fractions $F$.

Let $M$ be a finitely generated torsion free $R$-module. By ^952afe, $M\to FM,m\mapsto 1\otimes m$ injective. So $M$ can be viewed as an $R$-submodule of $FM$. Given an $F$-basis $(e_i{)}_{1⩽i⩽r}$ of $FM$, assume $M$ is generated by $m_1,\cdots ,m_n$ as $R$-module. Then each $m_i$ can be written as

$$m_i=\frac{s_1}{t_1}{e}_1+\cdots +\frac{s_r}{t_r}{e}_r,$$

where $s_i\in R$ and $t_i\in R\setminus \{0\}$. There exists $\lambda \in R\setminus \{0\}$ such that $\lambda M\subseteq R{e}_1\oplus \cdots \oplus R{e}_r$. Additionally, $M$ generated $FM$. This motivates the following definition:

Definition

Let $V$ be an $F$-vector space of finitely generated $r$. A fractional $R$-module in $V$ is an $R$-submodule of $V$ such that

• $M$ generates $V$;
• given any basis $(e_i{)}_{1⩽i⩽r}$ of $V$, there exists $\lambda \in R$ with $\lambda \ne 0$ such that $\lambda M\subseteq {\oplus }_{i=1}^{r}R{e}_i$.

In the case $V=F$, fractional modules are called fractional ideals of $R$. It is an $R$-submodule $I$ of $F$ such that there exists nonzero $\lambda \in R$ such that $\lambda I\subseteq R$. Hence $\lambda I$ is an ideal of $R$.

• every finitely generated $R$-submodule of $F$ is a fractional ideal
• if $R$ is Noetherian, then these are all fractional ideals. That is, when $R$ is Noetherian, finitely generated $R$-module $⟺$ fractional ideal.

the proof of the second statement

Let $R$ be a Noetherian integral domain and $F=\mathrm{Frac}(R)$. Let $I$ be an $R$-submodule of $F$. We want to show that $I$ is a fractional ideal if and only if $I$ is finitely generated.

($⟹$) Assume $I$ is finitely generated. Let $I={\sum }_{k=1}^{n}R{x}_k$ for some $x_k\in I$. Since $x_k\in F$, we can write $x_k=p_k/q_k$ where $p_k,q_k\in R$ and $q_k\ne 0$. Let $\lambda =q_1{q}_2\dots q_n$. Then $\lambda \in R$ and $\lambda \ne 0$. For each $k$, $\lambda x_k=(q_1\dots q_{k-1}{q}_{k+1}\dots q_n)p_k\in R$. Then $\lambda I={\sum }_{k=1}^{n}R(\lambda x_k)\subseteq R$. By definition, $I$ is a fractional ideal.

($⟸$) Assume $I$ is a fractional ideal. By definition, $I$ is an $R$-submodule of $F$ and there exists $\lambda \in R\setminus \{0\}$ such that $\lambda I\subseteq R$. Let $J=\lambda I$. Note that $J$ is an $R$-submodule of $R$, which means $J$ is an ideal of $R$. Since $R$ is a Noetherian ring, the ideal $J$ must be finitely generated. Let $J={\sum }_{k=1}^{n}R{y}_k$ for some $y_k\in J$. Since $y_k\in J=\lambda I$, we can write $y_k=\lambda x_k$ for some $x_k\in I$. We claim $I={\sum }_{k=1}^{n}R{x}_k$.

Take any $x\in I$. Then $\lambda x\in J$. So, $\lambda x={\sum }_{i=1}^n{r}_i{y}_i$ for some $r_i\in R$. Substituting $y_i=\lambda x_i$, we get $\lambda x={\sum }_{i=1}^n{r}_i(\lambda x_i)=\lambda ({\sum }_{i=1}^n{r}_i{x}_i)$. Since $\lambda \ne 0$ and $F$ is a field, we can multiply by $1/\lambda$ to get $x={\sum }_{i=1}^n{r}_i{x}_i$. This shows $I\subseteq {\sum }_{k=1}^{n}R{x}_k$. The inclusion ${\sum }_{k=1}^{n}R{x}_k\subseteq I$ is clear since each $x_k\in I$ and $I$ is an $R$-module. Thus, $I={\sum }_{k=1}^{n}R{x}_k$, which means $I$ is finitely generated.