Simple modules over matrix rings

hbghlyj

Let $k$ be a field, and let $R=M_n(k)$, the non-commutative ring of $n \times n$ matrices over $k$.

• Give examples of a simple left $R$-module $M$ and a simple right $R$-module $N$.
• Can you find another simple left $R$-module $M'$ that is not isomorphic to $M$? (Either find one or explain why there is none such.)
• Compute $\dim_k(N \otimes_R M)$.

hbghlyj

• A module is simple if it is nonzero, and contains no proper, nonzero modules.
  
  • A simple left $R$-module: $M=k^n$ (column vectors) with the usual left action.
  • A simple right $R$-module: $N=(k^n)^\ast$ (or row vectors) with $(\phi\cdot A)(v)=\phi(Av)$.
• There’s only simple left $R$-module up to isomorphism.
  Take $v=e_1\in k^n$ and let $e=e_{11}$.
  Then $\operatorname{Ann}_R(v)=\{A:Av=0\}= \{A:Ae=0\}=R(1-e)$.
  Pick any nonzero $m_0\in M'$. Since $1=\sum_i e_{ii}$, some $e_{ii}m_0\neq 0$, rename the basis so $i$ becomes $1$, we may assume $em_0\neq 0$. Now set $m:=em_0\in eM'$ (so $(1-e)m=0$).
  Now $\operatorname{Ann}_R(v)=R(1-e)\subseteq \operatorname{Ann}_R(m)$, so defining $f(rv)=rm$ is well-defined and $R$-linear. Because $M'$ is simple, $f$ is surjective; since $k^n$ is simple too, $f$ is an isomorphism. Thus $M'\cong k^n$.
• The multiplication map
  \[
  \mu:\ eR\otimes_R Re\longrightarrow eRe,\qquad (er)\otimes(r'e)\ \mapsto\ e r r' e.
  \]is balanced over $R$ since $μ((xr)⊗y)=xry=μ(x⊗(ry))$.
  With $e=e_{11}$ in $R=M_n(k)$, what is the subring $eRe$ inside $M_n(k)$ is just the scalars on the $(1,1)$-entry.
  Last step: show $\mu$ is an isomorphism. A handy candidate for an inverse is
  $$
  \nu:\ eRe\to eR\otimes_R Re,\qquad eae\ \mapsto\ (ea)\otimes e .
  $$The identities
  $$\mu\circ\nu(eae)=\mu((ea)\otimes e)=eae$$$$\nu\circ\mu\big((er)\otimes(r'e)\big)=\nu(err'e)=(err')\otimes e=(er)\otimes(r'e)$$show $\mu$ and $\nu$ are mutual inverses, so$$
  N\otimes_R M \;\cong\; eR\otimes_R Re \;\xrightarrow{\ \mu\ }\; eRe \;\cong\; k.
  $$Therefore $\dim_k(N\otimes_R M)=1$.

hbghlyj

Schur’s lemma (module/representation version).
If $M,N$ are simple $R$-modules and $f:M\to N$ is $R$-linear, then $f=0$ or $f$ is an isomorphism.
In particular, $\mathrm{End}_R(M)$ is a division ring (every nonzero endomorphism is invertible).
Over many contexts (e.g. complex representations) this division ring is just the base field.

In our case $R=M_n(k)$:
Let $M=k^n$ with left multiplication. Any $R$-linear endomorphism $T$ must commute with all left multiplications $A\in M_n(k)$: $T(Av)=A\,T(v)$.
Since $E_{ij}e_j=e_i$, the relation $T E_{ij}=E_{ij} T$ gives
$$
T e_i = T(E_{ij}e_j)=E_{ij}(T e_j).
$$
Write $T e_j$ as the $j$-th column of $T$. Then $E_{ij}(T e_j)= (e_j^{\top} T e_j)\, e_i = T_{jj}\, e_i$.
So for every $i$ and every $j$,
$$
T e_i = T_{jj}e_i.
$$Since the left-hand side doesn’t depend on $j$, all $T_{jj}$ must be the same scalar, say $\lambda$. Then
$$T e_i=\lambda e_i$$
so the $i$-th column of $T$ is $\lambda e_i$. That kills all off-diagonal entries and makes every diagonal entry $\lambda$. Hence $T=\lambda I$, so $\operatorname{End}_R(M)\cong k$.

hbghlyj

For any ring $R$, a left $R$-module $M$ (i.e., an $R$-$k$ bimodule), a right $R$-module $N$, and a $k$-vector space $X$, there’s a natural bijection
$$\DeclareMathOperator\Hom{Hom}\Hom_k(N\otimes_R M,X)\cong\Hom_R\big(N,\Hom_k(M,X)\big),\tilde f(n\otimes m)=f(n)(m)$$
Here $\Hom_k(M,X)$ is a right $R$-module via $(\psi\cdot r)(m)=\psi(rm)$.

One useful specialization is $X=k$, which gives
$$
\Hom_k(N\otimes_R M,k)\cong\Hom_R(N,M^∨),
$$
where $M^{*}=\Hom_k(M,k)$ with right $R$-action $(\phi\cdot r)(m)=\phi(rm)$.

When $M$ is finitely generated projective as a left $R$-module (true for $M=R e$ over $R=M_n(k)$), the adjunction upgrades to a concrete isomorphism
$$
N\otimes_R M\xrightarrow{\sim}\Hom_R(M^∨,N),\qquad
n\otimes m\mapsto\big(\phi\mapsto n\cdot \phi(m)\big).
$$Tie-back to our case: $R=M_n(k)$, $M\cong Re$, and $M^∨\cong eR\cong N$ as right $R$-modules, so
$$
N\otimes_R M\cong\Hom_R(N,N)=\operatorname{End}_R(N)\cong k.
$$

hbghlyj

• There’s always a natural map
  $$
  \theta:\ N\otimes_R M \longrightarrow \Hom_R(M^∨,N),\qquad
  n\otimes m \mapsto (\phi\mapsto n\,\phi(m)),
  $$
  but it isn’t generally bijective.
• If $M$ is finitely generated projective as a left $R$-module, then $M$ has a dual basis$\{(m_i,f_i)\}_{i=1}^r$ with $m_i\in M$, $f_i\in M^{\vee}$ such that
  $$
  \sum_i f_i(m)\,m_i = m\quad\text{for all }m\in M.
  $$
  Define the inverse
  $$
  \Psi:\ \Hom_R(M^{\vee},N) \longrightarrow N\otimes_R M,
  \qquad
  \Psi(F)=\sum_i F(f_i)\otimes m_i.
  $$
  Check $ \theta\circ\Psi = \mathrm{id}$: for $F\in\Hom_R(M^{\vee},N)$ and $\varphi\in M^{\vee}$,
  $$
  \theta(\Psi(F))(\varphi)
  = \theta\Big(\sum_i F(f_i)\otimes m_i\Big)(\varphi)
  = \sum_i F(f_i)\,\varphi(m_i)
  = \sum_i F\big(f_i\cdot \varphi(m_i)\big)
  = F\Big(\sum_i f_i\cdot \varphi(m_i)\Big).
  $$
  But $\sum_i f_i\cdot \varphi(m_i)=\varphi$ (evaluate both sides at any $m$ and use $\sum_i f_i(m)m_i=m$), hence this equals $F(\varphi)$.
  
  Check $ \Psi\circ\theta = \mathrm{id}$: for a pure tensor $n\otimes m$,
  $$
  \Psi\big(\theta(n\otimes m)\big)
  = \sum_i \theta(n\otimes m)(f_i)\otimes m_i
  = \sum_i n\,f_i(m)\otimes m_i
  = n\otimes \Big(\sum_i f_i(m)\,m_i\Big)
  = n\otimes m.
  $$
  So $\Psi$ is the desired inverse, so $θ$ becomes an isomorphism for every right $R$-module $N$.
• If $M$ is not projective, $\theta$ can fail badly. Classic counterexample:
  take $R=\mathbb Z$, $M=\mathbb Z/n\mathbb Z$ (not projective), $N=\mathbb Z$.
  Then $N\otimes_R M\cong \mathbb Z/n\mathbb Z\neq 0$, but
  $\Hom_R(M^{*},N)=\Hom_{\mathbb Z}(0,\mathbb Z)=0$
  since $M^{*}=\Hom_{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z)=0$.
  So $\theta$ cannot be an isomorphism in general.
• In our setting $R=M_n(k)$ and $M\cong Re$ with $e=e_{11}$. Because $e$ is idempotent,
  $$
  R \cong Re\oplus R(1-e)
  $$
  as left $R$-modules, so $Re$ is a direct summand of a free module—hence projective (and clearly finitely generated). That’s exactly why the tidy identification
  $$
  N\otimes_R M \cong \Hom_R(M^{*},N)
  $$goes through here.

hbghlyj

Here $M=Re$ (first-column module) is projective. Take a single pair $(m,f)$ with $m:=e=E_{11}\in Re$ and $f:Re\to R$ defined by
$$f(re)=r\qquad(\text{left $R$-linear}).$$Then for any $Ae\in Re$,$$f(Ae)m=Ae$$so $(m,f)$ is a dual basis.

With this dual basis, the inverse to
$$
\theta:\ N\otimes_R M\to \operatorname{Hom}_R(M^\vee,N),\quad
\theta(n\otimes m)(\varphi)=n\,\varphi(m),
$$
is
$$
\Psi(F)=F(f)\otimes m
$$