Past exam of the algebra course of the University of Oxford / 2007 / A

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.