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Let $A$ be an algebra over a commutative ring $R$, and let $U$ be an $A$-module. If $R\to R^{\prime }$ is a homomorphism to another commutative ring $R^{\prime }$, we form the $R^{\prime }$-algebra $R^{\prime }{\otimes }_{R}A$, and $R^{\prime }{\otimes }_{R}U$ becomes an $R^{\prime }{\otimes }_{R}A$-module.

• If $R$ is a subring of $R^{\prime }$, we say that the module $R^{\prime }{\otimes }_{R}U$ is obtained from $U$ by extending the scalar form $R$ to $R^{\prime }$.
  • If an $R^{\prime }{\otimes }_{R}A$-module $V$ has the form $R^{\prime }{\otimes }_{R}U$, we say that $V$ is written in $R$.
  • When $U$ is free as an $R$-module, we can identify $U$ with $1_{R^{\prime }}{\otimes }_{R}U$, and an $R$-basis of $U$ becomes an $R^{\prime }$-basis of $R^{\prime }{\otimes }_{R}U$.
• If $R^{\prime }=R/I$ with an ideal $I\subseteq R$, applying $R^{\prime }{\otimes }_R\cdot$ to an $R$-module $U$ is the same as reducing $U$ modulo $I$.
  • If $V$ is an $R^{\prime }{\otimes }_{R}A$-module of the form $R^{\prime }{\otimes }_{R}U$ for some $A$-module $U$, then we say $V$ can be lifted to $U$ and $U$ is a lift of $V$.

Proposition

Let $F\subseteq E$ be fields where $E$ is algebraic over $F$, and let $A$ be a finite-dimensional $F$-algebra. Let $V$ be a finitely-dimensional $E{\otimes }_{F}A$-module. Then there exists a field $K$ with $F\subseteq K\subseteq E$ of finite degree over $F$ so that $V$ is written in $K$.

\begin{proof} Let $a^1,\cdots ,a^n$ be an $F$-basis of $A$, and let $a^t$ acts $V$ with matrix $(a_{ij}^t{)}_d$ w.r.t. some $E$-basis $\{v_1,\cdots ,v_d\}$ of $V$. Then $K=F[a_{ij}^t:1⩽t⩽n,1⩽i,j⩽d]$ and $[K:F]<\infty$. Note that $A$ acts on $V$ by matrices over the field $K$.

To show $V$ is written in $K$, it suffices to find a $K{\otimes }_{F}A$-module $U$ such that $V\simeq E{\otimes }_{K}U$.

Define $U={\sum }_{j=1}^{d}K{v}_j={\oplus }_{j=1}^{d}K{v}_j\subseteq V$. Since $A$ acts on $V$ by matrices over the field $K$, we know $(1\otimes a^t)(\sum k_j{v}_j)\in U$ and so $U$ is a $K{\otimes }_{F}A$-module.

Since $\{v_1,\cdots ,v_d\}$ is a set of $K$-basis of $U$, $E{\otimes }_{K}U$ has $E$-basis $\{1\otimes v_j:1⩽j⩽d\}$. Define

$$\psi :E{\otimes }_{K}U\to V,e\otimes u\mapsto eu,$$

then $\psi$ is an isomorphism between $E$-vector spaces. Furthermore, notice that

$$\begin{array}{rl}\psi ((1\otimes a^t)\cdot \left(e{\otimes }_K{v}_j\right))& =\psi \left(e{\otimes }_K\left(\left(1\otimes a^t\right)\cdot v_j\right)\right)=e\cdot \left(\left(1\otimes a^t\right)\cdot v_j\right)\\ & =\left(1\otimes a^t\right)\cdot \left(e{v}_j\right)=\left(1\otimes a^t\right)\psi \left(e{\otimes }_K{v}_j\right)\end{array}$$

and so $\psi$ is an morphism between $E{\otimes }_{F}A$-modules. \end{proof}

Definition

Let $A$ be a finite-dimensional $F$-algebra where $F$ is a field. A simple $A$-module $U$ is said to be absolutely simple if $E{\otimes }_{F}U$ is simple as an $E{\otimes }_{F}A$-module for all extension field $E$ of $F$.

Definition

We say an extension field $E$ of $F$ is a splitting field of $A$ if every simple $E{\otimes }_{F}A$-module is absolutely irreducible.

Proposition

Let $A$ be a finite-dimensional $F$-algebra over field $F$, and let $U$ be a simple $A$-module. TFAE:

• $U$ is absolutely irreducible;
• ${\mathrm{End}}_A(U)=F$;
• The matrix algebra summand of $A/J(A)$ corresponding to $U$ has the form $M_n(F)$ where $n={\dim }_{F}U$.

\begin{proof} ii)→iii) is trivial.

iii)→i) Note that $A/J(A)\simeq {\mathrm{M}}_n(F)\oplus \cdots$, and we know $M_n(F)$ is an $A$-module. Since $U$ is simple, $U\simeq F^n$. Let $E\supseteq F$ be a field extension. We aim to show $E{\otimes }_{F}U$ is a simple $E{\otimes }_{F}A$-module. Since $U\simeq F^n$, there is

$$E{\otimes }_{F}U\simeq E{\otimes }_F{F}^n\simeq (E\otimes F{)}^n\simeq E^n.$$

Note that $E{\otimes }_{F}A$ acts on $E{\otimes }_F{M}_n(F)$ by $E{\otimes }_F{M}_n(F)$, where $E{\otimes }_F{M}_n(F)\simeq M_n(E)$. As $E^n$ is simple as an $M_n(E)$-module, we know $E{\otimes }_{F}U$ is simple as an $E{\otimes }_{F}A$-module. By the arbitrary of $E$, $U$ is absolutely irreducible.

i)→ii) is omitted. \end{proof}

Group Representations

Lemma

A morphism $f:(X,\rho )\to \left(X^{\prime },{\rho }^{\prime }\right)$ in $\mathrm{Rep}(G,\mathcal{C})$ is an isomorphism iff the morphism $f:X\to X^{\prime }$ is an isomorphism in $\mathcal{C}$.

Definition

Two representations $(X,\rho )$ and $\left(X,{\rho }^{\prime }\right)$ on the same object $X$ are isomorphic if there exists automorphism $f$ of $X$ such that for all $g\in G,{\rho }^{\prime }(g)=f\rho (g)f^{-1}$.

Remark. We say $\rho$ and ${\rho }^{\prime }$ are uniformly conjugate.

Let $(X,\rho )$ and $(Y,\sigma )$ be $\mathcal{C}$-representations of $G$, where $\mathcal{C}$ is a category. Then group $\mathrm{Aut}(X)\times \mathrm{Aut}(Y)$ acts on the set ${\mathrm{Mor}}_{\mathcal{C}}(X,Y)$ by $f^{(\alpha ,\beta )}=\beta f{\alpha }^{-1}$. Compose it with diagonal morphism

$$G\to \mathrm{Aut}(X)\times \mathrm{Aut}(Y),g\mapsto (\rho (g),\sigma (g)).$$

Define an action of $G$ on ${\mathrm{Mor}}_{\mathcal{C}}(X,Y)$, $f^g=\sigma (g)f\rho (g{)}^{-1}$ for all $g\in G$, and it induces the following commutative diagram.

Lemma

The set of morphisms $(X,\rho )\to (Y,\sigma )$ in $\mathrm{Rep}(G,\mathcal{C})$ is the set of fixed points of $G$ in its action on ${\mathrm{Mor}}_{\mathcal{C}}(X,Y)$.

Definition

The fixator (stabilizer) of $x$ in $G$ is subgroup $G_x=\{g\in G:gx=x\}$.

Lemma

Let $(X,\rho )$ be a $\mathcal{C}$-representation of $G$.

• $G_{gx}=g{G}_x{g}^{-1}$
• if $f:(X,\rho )\to (Y,\sigma )$ is an isomorphism of representations, then $G_{f(x)}=G_x$ for all $x\in X$.

\begin{proof}

• Exercise \end{proof}