Past exam of the introduction to representation theory course of the University of Oxford / 2023 / 1

An $A$-module is completely reducible if it is a direct sum of modules each of which is a simple module:Here a simple module is nonzero and has no submodules other than zero and itself. This is precisely the definition of a semisimple module. The indexing set need not be finite unless $V$ has finite dimension; the zero module is included as the empty direct sum of modules.

Construct the sum successively, retaining a simple module only when it adds a new direct summand. Suppose $S=V_1+\cdots +V_{j-1}$ is already expressed as a direct sum of modules selected from these simple modules. The intersection $S\cap V_j$ is a submodule of the simple module $V_j$, so it is either zero or all of $V_j$. In the latter case $S+V_j=S$. In the former caseStarting with $S=0$ and proceeding by mathematical induction gives a direct sum of modules equal to $V_1+\cdots +V_n$. Thus the sum is a semisimple module. The argument proves the useful finite sum of simple submodules lemma without claiming that the originally listed submodules are themselves independent.

Since $A$ is a nonzero associative algebra of finite dimension, it has a nonzero left ideal $I$ of least positive dimension. Such a minimal left ideal is a simple module under left multiplication.

For $a\in A$, right multiplication defines the module homomorphismIts kernel is a submodule of $I$, so its image $Ia$ is either zero or a simple module isomorphic to $I$. Each $Ia$ is a left ideal, since $c(xa)=(cx)a$.

ConsiderIt is a left ideal. It is also stable under right multiplication, because $(Ia)b=I(ab)$, so it is a two-sided ideal. Moreover $I=I1\subseteq J$, giving $J\ne 0$. The assumption that $A$ is a simple ring therefore forces $J=A$.

Only finitely many summands are needed: express each member of a finite basis of $A$ as a finite sum of elements of the $Ia$, and collect the finitely many indices used. After discarding zero summands, $A$ is a finite sum of simple modules. Part (a)(ii) makes its left regular module a semisimple module. This is exactly the definition of a semisimple algebra, proving that a finite-dimensional simple algebra is semisimple. The finite dimension assumption is essential to this assertion about a simple ring.

Use the left $A$-module structure $c\cdot (a\otimes b)=ca\otimes b$ on the tensor product of associative algebras. Expanding each $b\in B$ in the basis $b_1,\dots ,b_m$ shows that every tensor can be writtenThus the displayed family generates the module.

For uniqueness, let ${\beta }_j:B\to k$ be the corresponding dual basis, with ${\beta }_j(b_i)={\delta }_{ij}$. Applying the linear map ${\mathrm{id}}_A\otimes {\beta }_j$ to a linear relation ${\sum }_i{a}_i\otimes b_i=0$ gives $a_j=0$. Hence the family is linearly independent in a module, andIn particular $1\otimes b_1,\dots ,1\otimes b_m$ is a basis of a module, with coefficients uniquely in $A$.

Write $x\in Z(A{\otimes }_{k}B)$ uniquely as $x={\sum }_j{a}_j\otimes b_j$ using part (i). For each $a\in A$, the equality $(a\otimes 1)x=x(a\otimes 1)$ becomesThe coefficient uniqueness from part (i) gives $a{a}_j=a_{j}a$ for every $a$, so each $a_j$ belongs to the center of an associative algebra $Z(A)$. Consequently $x\in Z(A){\otimes }_{k}B$.

Choose a basis $z_1,\dots ,z_t$ of the vector space $Z(A)$ and rewrite $x={\sum }_{i=1}^t{z}_i\otimes d_i$. For $b\in B$, commuting with $1\otimes b$ givesExtend the linearly independent family $z_i$ to a basis of $A$ and apply its dual basis to the first factor. It follows that $b{d}_i=d_{i}b$ for every $b$, hence $d_i\in Z(B)$. This proves $Z(A{\otimes }_{k}B)\subseteq Z(A){\otimes }_{k}Z(B)$.

Conversely, if $z\in Z(A)$ and $w\in Z(B)$, thenThe pure tensors $a\otimes b$ span the tensor product, so $z\otimes w$ is central, as is every sum of such pure tensors. Therefore the center of a tensor product of associative algebras isThese tensor products are identified with their natural vector subspaces of $A{\otimes }_{k}B$; extending bases justifies these embeddings over the field $k$.

Let $J\ne 0$ be a two-sided ideal of $A{\otimes }_{k}B$. Choose a nonzero $x\in J$ of least possible two-factor tensor rank $n$, and take a shortest expressionBoth families $(a_i)$ and $(b_i)$ are linearly independent: a linear dependence in either family would combine terms and shorten the expression. In particular $a_1,b_1\ne 0$.

Because $A$ is a simple ring, the two-sided ideal generated by $a_1$ is $A$. Thus there are finitely many $u_j,v_j\in A$ with ${\sum }_j{u}_j{a}_1{v}_j=1$. FormSince the $b_i$ are linearly independent, applying a linear functional selecting $b_1$ shows $y\ne 0$. Its two-factor tensor rank is at most $n$ and hence, by the choice of $x$, exactly $n$. Consequently $c_1=1,c_2,\dots ,c_n$ are linearly independent.

Similarly, simplicity of $B$ gives finitely many $p_j,q_j\in B$ with ${\sum }_j{p}_j{b}_1{q}_j=1$. NowThe linear independence of the $c_i$ shows $z\ne 0$, and minimality again makes its two-factor tensor rank exactly $n$. We have obtainedBoth coefficient families in this expression are linearly independent. Retaining this additional property is the key to the next part. This is coefficient normalization in an ideal of a tensor product.

Keep the nonzero $z$ of minimal two-factor tensor rank constructed in part (i). For every $a\in A$, its commutator with $a\otimes 1$ lies in $J$ and equalsIf this tensor were nonzero, its two-factor tensor rank would be at most $n-1$, contradicting minimality. Thus it is zero. The linear independence of $d_2,\dots ,d_n$ forces $a{c}_i=c_{i}a$ for every $a$ and $i\ge 2$. Since $Z(A)=k$, each $c_i$ is a scalar multiple of $1$.

But $1,c_2,\dots ,c_n$ are linearly independent. Hence $n=1$, so $z=1\otimes 1\in J$. A two-sided ideal containing the multiplicative identity is the entire ring; therefore $J=A{\otimes }_{k}B$.

We have proved the stronger result that $A{\otimes }_{k}B$ is a simple ring. It has finite dimension $({\dim }_{k}A)({\dim }_{k}B)$, so part (b) provesPart (c)(ii) also gives $Z(A{\otimes }_{k}B)=k$, so this proves that a tensor product of central simple algebras is again a central simple algebra.