Past exam of the algebra course of the University of Oxford / 2010 / A

An $R$-module $M$ is a finitely generated module if there are $m_1,\dots ,m_s\in M$ such that every element of $M$ has the form $r_1{m}_1+\cdots +r_s{m}_s$ with $r_i\in R$. Equivalently, $M=R{m}_1+\cdots +R{m}_s$, or $M$ is a quotient module of $R^s$.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. In its cyclic decomposition form, every finitely generated module has a decompositionwhere the $d_i$ are nonzero nonunits. The nonnegative integer $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. The zero module corresponds to no summands. Equivalently, factor the $d_i$ into powers of prime elements and use the Chinese remainder theorem to obtain a direct sum of cyclic prime-power torsion modules and a free summand.

Give $V$ a $k[t]$-module structure by $t\cdot v=\alpha v$. A finite vector space basis generates it as a module. It is a torsion module: the Cayley-Hamilton theorem gives a nonzero polynomial annihilating all of $V$. Since $k[t]$ is a Euclidean domain, cyclic decomposition giveswith monic nonconstant invariant factors of a linear operator $f_j$ and no free summand. On $k[t]/(f_j)$, use the basis $1,t,\dots ,t^{\mathrm{deg}{f}_j-1}$. Multiplication by $t$ then has the companion matrix $C(f_j)$, because the relation $f_j(t)=0$ expresses the last power in terms of earlier ones. Concatenating these bases yieldsthe rational canonical form of $\alpha$. In particular, its minimal polynomial is $f_s$ and its characteristic polynomial is ${\prod }_j{f}_j$.

The first block is the companion matrix of $t-3$. The last block is the companion matrix of $f(t)=t^3-3t^2+9t-27=(t-3)(t^2+9)$. Since $t-3$ divides $f$, these are the ordered invariant factors of a linear operator. ThusThe minimal polynomial is the largest invariant factor, and the characteristic polynomial is their product.

The minimal polynomial factors over $\mathbb{C}$ as $(t-3)(t-3i)(t+3i)$ with distinct roots. The diagonalizability criterion using the minimal polynomial therefore makes $\alpha$ diagonalizable. Its characteristic polynomial gives multiplicities two, one and one, so its Jordan normal form isup to reordering the one-dimensional Jordan blocks.

For a two-by-two matrix over a principal ideal domain, its first Smith normal form invariant is the greatest common divisor of its entries, while the product of its two invariants is its determinant, up to a unit. Hereso the entries generate the unit ideal in the Gaussian integers. The first invariant is therefore $1$. AlsoHence a valid Smith normal form isMultiplying the second diagonal entry by a unit $\pm 1,\pm i$ gives an equivalent normal form. The divisibility condition holds because $1$ divides every Gaussian integer.

A simple module is a nonzero module with no submodules other than zero and itself. Equivalently, every nonzero element generates the entire module: $Am=M$ whenever $m\ne 0$.

A composition series of a module is a finite chainof submodules whose factors $M_j/M_{j-1}$ are simple modules. A finite-dimensional module has such a series by induction on its dimension, choosing a maximal proper submodule at each step.

The Jordan–Hölder theorem for modules states that any two composition series of a module have the same length and the same simple factors up to isomorphism and reordering. Thus the multiset of composition factors is an invariant of the module, even when the series itself is not unique.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

The finite-dimensional qualification is essential for this scalar conclusion. For example, the field $\mathbb{C}(t)$ is a complex algebra, simple as a module over itself, and its endomorphisms include multiplication by $t$, which is not a complex scalar.

Let $M$ be a nonzero simple module for $A=k[X]/(X^3)$ and let $N$ be the action of $X$. Since $X$ is central, $NM$ is a submodule, and $N^3=0$. If $NM=M$, iterating would give $N^{3}M=M$, impossible. Thus $NM=0$.

The action therefore factors through $A/(X)\cong k$. A simple module over a field is one-dimensional: the span of any nonzero vector is already a nonzero submodule. Conversely, $k$ with $X$ acting as zero is simple. Hence there is exactly one simple module up to isomorphism, namelyThis argument covers all simple modules, without assuming their finite dimensionality in advance.

On a finite-dimensional $A$-module $M$, the actions of $X$ and $Y$ are endomorphisms $P,Q$ satisfying $PQ-QP=I_M$. Taking traces and using $\mathrm{tr}(PQ)=\mathrm{tr}(QP)$ givesSince $k$ has characteristic zero, this forces ${\dim }_{k}M=0$. Thus the zero module is the only finite-dimensional module for this Weyl algebra. The obstruction is the Weyl algebra has no nonzero finite-dimensional modules.

The group algebra $kG$ is the vector space with basis $G$, whose elements are formal sums ${\sum }_{g\in G}{a}_{g}g$ with $a_g\in k$. Extend the group multiplication bilinearly:This is an associative unital $k$-algebra, with multiplicative identity the group identity, and ${\dim }_{k}kG=|G|$.

Given a group representation $\rho :G\to {\mathrm{GL}}_n(k)$, define an action on $k^n$ byIt is bilinear. The identity acts as $I_n$, andbecause $\rho$ is a group homomorphism. Hence the product action agrees with successive action, proving the module axioms. Thus every group representation extends uniquely by linearity to a $kG$-module.

A semisimple module is a direct sum of simple modules. A finite-dimensional semisimple algebra is one whose left regular module is semisimple; equivalently, all its finite-dimensional modules are semisimple, or every submodule of a finite-dimensional module has a complementary submodule.

The Maschke theorem states that, for a finite group $G$ and a field $k$ with $|G|\ne 0$ in $k$, every finite-dimensional $kG$-module is semisimple, so $kG$ is a semisimple algebra. The assumption is equivalent to saying that the characteristic of a field does not divide $|G|$. The averaging construction makes a projection onto a submodule equivariant by replacing $P$ with $|G{|}^{-1}{\sum }_{g\in G}gP{g}^{-1}$.

Let $w={\sum }_{g\in G}g$. This is nonzero since its basis coefficients are all one, and $gw=w$ for every $g\in G$. Hence $kw$ is a one-dimensional left ideal of $kG$.

If the regular module is semisimple, $kw$ is a direct summand. Let $\pi :kG\to kw$ be the corresponding $kG$-linear projection, so $\pi (w)=w$. Write $\pi (1)=cw$ for $c\in k$. ThenBecause $w\ne 0$, $c|G|=1$ in $k$. ThusTogether with the Maschke theorem, this gives the exact semisimple algebra criterion for a group algebra.

True. Every group of order four is Abelian: if it has an element of order four it is cyclic; otherwise every nonidentity element has order two and $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$, while $(ab{)}^{-1}=ab$, so $ab=ba$.

Over $\mathbb{C}$, the Maschke theorem and the Artin–Wedderburn theorem giveSince $G$ is Abelian, this algebra is commutative, forcing every $d_j=1$. Its dimension is four, so $r=4$. Each factor contributes one simple module; therefore there are exactly four up to isomorphism.

True. If every simple module over $\mathbb{C}G$ were one-dimensional, the Maschke theorem and the Artin–Wedderburn theorem would make $\mathbb{C}G$ a product of copies of $\mathbb{C}$, hence commutative. But its basis elements multiply as in $G$, so commutativity of $\mathbb{C}G$ implies commutativity of $G$. This contradicts the assumption. Therefore a non-Abelian finite group has a simple module of dimension greater than one.

True. The polynomial $X^7-1$ has seven distinct roots ${\zeta }_7^j$ in $\mathbb{C}$. The Chinese remainder theorem givesThe simple modules are exactly the seven one-dimensional evaluation modules, on which $X$ acts by ${\zeta }_7^j$. Thus there are seven up to isomorphism.

True; there are exactly five. On a simple module, the central element $Y$ acts as zero: its image is a submodule, and surjectivity would contradict $Y^6=0$. The nilpotent central elements annihilate simple modules argument shows that the action factors throughby the Chinese remainder theorem. Therefore the simple modules are one-dimensional withThis also rules out hypothetical infinite-dimensional simple modules, rather than counting only finite-dimensional ones.