Past exam of the algebra course of the University of Oxford / 2007 (B2, DMAB 2641, DMCN 2641, DMPB 2641)

A left module over the unital $K$-algebra $A$ is a $K$-vector space $V$ with a $K$-bilinear action $A\times V\to V$ satisfying $(ab)v=a(bv)$ and $1v=v$. For a vector $v$, the map $A\to V$, $a\mapsto av$, is an $A$-module homomorphism. Its kernel is $\mathrm{Ann}(v)$; explicitly this is a linear subspace and $b(av)=(ba)v=0$ for $a\in \mathrm{Ann}(v)$, so it is a left ideal.

If $V=Av$, this homomorphism is surjective, and the module isomorphism theorem gives $V\cong A/\mathrm{Ann}(v)$. Conversely the quotient by a left ideal is generated by $1+I$. Thus cyclic modules are exactly the quotients $A/I$ by left ideals.

For the first example set $e=\mathrm{diag}(1,1,0)$. Right multiplication by $e$ keeps the first two columns and zeroes the third. Every matrix with third column zero equals its product with $e$. Hence $V=Ae$ is cyclic, generated by $e$.

For a matrix $v$ supported in its first row and any upper triangular matrix $a$, multiplication gives $av=a_{11}v$. Thus $Av=Kv$ has dimension at most one for every $v$, whereas the entire module has dimension three. Hence this module is not cyclic, over every field $K$.

Coordinatewise multiplication gives $V=A(1,1,0)$, so the third module is cyclic.

A submodule of a cyclic module need not be cyclic. In the second example the first-row space is a left ideal and hence a submodule of the regular module $A$. The regular module is cyclic, generated by $1_A$, whereas its first-row submodule has just been shown not to be cyclic. Thus the proposed inheritance property fails even for a finite-dimensional algebra.

A simple module is nonzero and has no submodules other than zero and itself. A semisimple module is a direct sum of simple submodules; the zero module is allowed as the empty sum.

For (1)$\Rightarrow$(2), argue by dimension. Choose a minimal nonzero submodule $S$, which is simple, and split $V=S\oplus C$ using (1). The property passes to $C$: for $U\subseteq C$, write $V=U\oplus D$; then $C=U\oplus (C\cap D)$. Apply induction to $C$ and append $S$. The zero-dimensional case starts the induction. The implication (2)$\Rightarrow$(3) is immediate. For (3)$\Rightarrow$(2), repeatedly append a simple submodule not contained in the sum already chosen. Its intersection with that sum is zero by simplicity. Dimension strictly increases, so after finitely many steps their direct sum is $V$.

For (2)$\Rightarrow$(1), write $V={⨁}_i{S}_i$ and fix $U$. Choose a sum $C$ of a maximal subset of these $S_i$ whose sum is disjoint from $U$. If $U+C\ne V$, some $S_j$ is not contained in $U+C$, so simplicity gives $S_j\cap (U+C)=0$. Then $C+S_j$ is a larger direct sum still disjoint from $U$, a contradiction. Hence $V=U\oplus C$.

Condition (1) implies (1*). Conversely suppose (1*) and let $W$ be the sum of all simple submodules. If $W\ne V$, finite dimension supplies a maximal proper submodule $M$ containing $W$. By (1*), $V=M\oplus C$. Projection to the simple quotient $V/M$ identifies the nonzero complement $C$ with a simple module. Thus $C\subseteq W\subseteq M$, contradicting the direct sum. Consequently $W=V$, which is (3). This proves all four conditions are equivalent.

For the concrete complement, $x^2=x$. Orthogonal idempotents $x$ and $1-x$ give $A=Ax\oplus A(1-x)$: the sum follows from $1=x+(1-x)$, and right multiplication by $x$ shows their intersection is zero. In explicit matrix form,The decomposition of $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ uses second-column remainders $b-\lambda a,d-\lambda c$ in $C$.

Each $M_i$ is an ideal because the algebra is commutative. For $m\in M_i$, the product $m{e}_j$ belongs both to $M_i$ and $M_j$, so is zero when $j\ne i$. Multiplying $1={\sum }_j{e}_j$ by $m$ therefore gives $m=m{e}_i$. In particular $e_i^2=e_i$, different $e_i$ have zero product, andFor any idempotent $e$, the set $Ae$ is closed under multiplication and $e$ acts as its identity. With $f=1-e$ the map $a\mapsto (ae,af)$ is an algebra isomorphism $A\to Ae\times Af$: multiplication is componentwise because $ef=0$, and its inverse sends $(u,v)$ to $u+v$.

The $A$-submodules of $Ae$ are exactly its ideals as a unital algebra, since acting by $a$ on $Ae$ is the same as acting by $ae$. A nonzero commutative unital algebra is a simple module over itself exactly when every nonzero principal ideal is the whole algebra, which is equivalent to every nonzero element having an inverse. Thus a nonzero component $Ae$ is simple exactly when it is a field. If $e=0$, both properties fail under the nonzero-field convention.

A semisimple regular module is a finite direct sum of simple ideals. Applying the preceding decomposition gives a finite direct product of fields. Conversely the coordinate ideals of a finite product of fields are simple modules, so its regular module is semisimple. The fields are finite extensions of $K$, because $A$ has finite $K$-dimension.

For a generator $g$ of the cyclic group of order three, evaluation $X\mapsto g$ gives a surjection $\mathbb{Q}[X]\to \mathbb{Q}G$ with kernel $(X^3-1)$, by division and independence of $1,g,g^2$. The factors $X-1$ and $X^2+X+1$ are coprime, and the quadratic is irreducible over $\mathbb{Q}$. The Chinese remainder theorem yieldsThe explicit map sends $g$ to $(1,{\zeta }_3)$, where ${\zeta }_3=(-1+\sqrt{-3})/2$.

An $R$-module isomorphism $T:V_{\alpha }\to V_{\beta }$ is a $K$-linear isomorphism satisfying $T\alpha =\beta T$, by applying it to multiplication by $X$. Conversely that identity implies $Tf(\alpha )=f(\beta )T$ for every polynomial, so a similarity is an $R$-module isomorphism.

The cyclic decomposition theorem over the principal ideal domain $R=K[X]$ states that any finitely generated module iswith nonunit monic invariant factors, uniquely determined, and free rank $r$ uniquely determined. A finite-dimensional $K$-space has $r=0$. In each cyclic summand the basis $1,X,\dots ,X^{\mathrm{deg}{d}_i-1}$ makes multiplication by $X$ the companion matrix of $d_i$. Their block diagonal sum is the rational canonical form.

Here the two polynomials are already in divisibility order. With the companion convention having ones below the diagonal,Integer coefficients are interpreted in $K$, so this remains valid in characteristics two and three.

In the cyclic case there is one companion block $C(d)$. Its defining polynomial is both characteristic and minimal: the vector represented by $1$ is cyclic and a polynomial annihilates it exactly when divisible by $d$. Thus the minimal polynomial is read directly from that block.

The eigenvalues in $K$ are precisely the roots in $K$ of the companion polynomial $d$; over an algebraic closure take all its roots. Thus the form determines their root set, but an explicit list generally requires factoring or solving $d$ and cannot be read off by inspection. This is the relevant distinction in the qualification “with no further calculations”. Over a general field the polynomial may have no roots in $K$ at all, and its diagonal entries are not its eigenvalues. If “determinable” means only uniquely specified rather than immediately readable, all the information is of course fixed by the matrix, including this root set.

For $d(X)=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0$, the determinant of its companion matrix is $(-1{)}^n{a}_0$. ThereforeThus invertibility is readable from the constant coefficient, with no factorization or elimination.

In the cyclic quotient $R/(d)$, the kernel is formed by residue classes of polynomials $h$ such that $d\mid Xh$. If $X\nmid d$, coprimality forces $d\mid h$, so the kernel is zero. If $d=Xe$, divisibility is equivalent to $e\mid h$, and the residue classes of multiples of $e$ modulo $Xe$ form a one-dimensional $K$-space. HenceThis too is immediately readable in the cyclic case, even if zero is a repeated root. The zero-dimensional module has kernel dimension zero separately.

For $|G|=p^{r}m$ with $p\nmid m$, Sylow's theorem gives subgroups of order $p^r$, says that every $p$-subgroup lies in a conjugate of one of them, and that all such Sylow subgroups are conjugate. Their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$. A group is soluble when its derived series, obtained by repeatedly taking the commutator subgroup, reaches the identity. Equivalently it has a subnormal series with abelian factors.

For a group of order $p^2$, the class equation shows its center is nontrivial: noncentral conjugacy classes have sizes divisible by $p$. If the center had order $p$, the quotient by the center would be cyclic; a group with cyclic central quotient is abelian, since representatives are powers of one element times central elements. This would make the center the whole group. Thus every group of order $p^2$ is abelian.

Now consider $|G|=p{q}^a$ with $0\le a\le 2$. Cases with a normal Sylow subgroup are soluble: that subgroup has prime or prime-square order and is abelian, and its quotient also has such order and is abelian. This gives a subnormal series with abelian factors.

If $p<q$, the Sylow number $n_q$ divides $p$ and is $1$ modulo $q$, so is one. If $p>q$ and $a\le 1$, $n_p$ divides $q^a<p$, so again is one. For $p>q$ and $a=2$, a nontrivial $n_p$ can only be $q^2$, since $q<p$. The congruence would force $p\mid (q-1)(q+1)$, hence $p\mid q+1$. Since $p>q$, this means $p=q+1$, and primality forces $(p,q)=(3,2)$.

In the remaining order-twelve case, either a Sylow subgroup is already normal, or there are four Sylow subgroups of order three. Those have pairwise trivial intersection and contribute eight distinct elements of order three. Only four elements remain, including the identity. Every Sylow subgroup of order four lies entirely among those four elements, so it is unique and normal. It and its quotient are abelian. Therefore all groups of the required orders are soluble, by elementary Sylow arguments alone.

A composition series is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$ with simple nontrivial factors. Every finite group has one, by recursively choosing a maximal proper normal subgroup. The Jordan–Hölder theorem states that any two such series have the same length and the same multiset of simple factors up to isomorphism.

For a proof induct on $|G|$. Let $M,N$ be the penultimate terms in two series. If $M=N$, apply induction there. Otherwise their maximal normality gives $MN=G$. Set $D=M\cap N$. The second isomorphism theorem givesso these quotients are simple. Choose a composition series of $D$ and extend it once to $M$ and once to $N$. Induction in the smaller groups $M,N$ compares these with the respective original prefixes. Appending $G/M$ or $G/N$ gives in both cases the factors of $D$ together with $G/M,G/N$. This completes the proof, including the case $D=1$.

Exactly one: take $S_5$. Its only nontrivial proper normal subgroup is $A_5$. Indeed a normal $N$ has $N\cap A_5$ either $1$ or $A_5$ by the supplied simplicity. In the first case $N$ injects into $S_5/A_5\cong C_2$, so a nontrivial $N$ would be a normal subgroup of order two and hence central. But $S_5$ has trivial center: conjugating every transposition trivially forces preservation of each unordered pair, and therefore of every point. Thus the unique series is $1◃A_5◃S_5$.

Exactly two: take $A_5\times C_2$. A normal subgroup containing $A_5\times 1$ is that subgroup or the whole product. For a normal subgroup $N$ meeting $A_5\times 1$ trivially, commutators of its elements with $A_5\times 1$ vanish; its first coordinates therefore lie in $Z(A_5)=1$. Hence $N\subseteq 1\times C_2$. The two nontrivial proper normal subgroups are exactly the two factors, each simple, so the two series pass respectively through $A_5\times 1$ and $1\times C_2$.

More than two: take $C_{120}$. A cyclic group has exactly one subgroup of each divisor order, and composition steps have prime indices. The orders of those indices are the distinct permutations of $2,2,2,3,5$, givingAll three example groups have order $120$.

An extension of $A$ by $B$ is a short exact sequence $1\to A\to G\to B\to 1$. It splits if the quotient map has a homomorphic section. Equivalently the embedded $A$ has a subgroup complement isomorphic to $B$, and $G$ is the corresponding semidirect product.

For $t\notin K$ and $a\in K$, both $t$ and $ta$ lie outside $K$, so $t^2=(ta{)}^2=1$. It follows that $tat=a^{-1}$, and $t=t^{-1}$ gives the claimed conjugation identity. Conjugation is an automorphism of $K$, so inversion is an automorphism there. Comparing the image of $ab$ with the product of the images gives $b^{-1}{a}^{-1}=a^{-1}{b}^{-1}$. Therefore $K$ is abelian.

Conversely inversion is an automorphism of any abelian group $A$, of order dividing two. Form the semidirect product $G=A⋊C_2$ with multiplicationFor completeness associativity follows because $b\mapsto b^{(-1{)}^{\epsilon }}$ is an automorphism and these automorphisms compose according to addition modulo two; $(1,0)$ is the identity and inverses follow from this formula. The subgroup $H=A\times \{0\}$ is normal with quotient $C_2$, and every outside element satisfies $(a,1{)}^2=(a{a}^{-1},0)=1$. Thus the generalized dihedral group supplies the required extension, including when $A$ is infinite or has elements of order two.

The degree-sum formula for irreducible characters givesThe number of linear characters is the order of the abelianization, so $|G/G^{\prime }|=3$ and $|G^{\prime }|=7$. The intersection of the kernels of all linear characters is $G^{\prime }$. Consequently $g_4,g_5$ are nonidentity elements of this order-seven subgroup and have order seven. The images of $g_2,g_3$ under a displayed linear character have order three, so their orders are divisible by three. Lagrange's theorem leaves three or twenty-one; an element of order twenty-one would make $G$ cyclic, contradicting its degree-three irreducible characters. ThereforeThe center is trivial. A nontrivial proper center would have order three or seven, giving a cyclic quotient and hence an abelian group, again a contradiction. Thus a centralizer of a nonidentity element is proper. It has order three for an order-three element, and seven for an order-seven element. The class sizes are therefore $1,7,7,3,3$.

Column orthogonality says ${\sum }_i|{\chi }_i(g){|}^2=|C_G(g)|$. For $g_2,g_3$, the three linear characters already contribute three, so the remaining two entries must both vanish. Orthogonality with the identity column says ${\sum }_i{\chi }_i(1)\overline{{\chi }_i(g)}=0$ for $g\ne 1$. At $g_4$ this gives $1+\alpha +{\chi }_5(g_4)=0$, so ${\chi }_5(g_4)=\beta$, using $\alpha +\beta =-1$. Similarly ${\chi }_5(g_5)=\alpha$. Thus the completed table isHere $\beta =\overline{\alpha }$, $|\alpha {|}^2=|\beta {|}^2=2$ and $\alpha \beta =2$. These relations, together with $1+\omega +{\omega }^2=0$, verify all weighted row orthogonality relations and the centralizer column norms.