Past exam of the rings and modules course of the University of Oxford / 2023 / 3

Since $A^2+I=0$, its minimal polynomial $m_A$ divides $X^2+1$. The polynomial $X^2+1$ is an irreducible polynomial over the rational numbers: a reducible quadratic would have a rational root, whereas $r^2+1>0$ for every $r\in \mathbb{Q}$.

Because $n>0$, the constant polynomial one evaluates to the nonzero identity matrix $I$, so $m_A\ne 1$. Both $m_A$ and $X^2+1$ are monic polynomials; hence the only remaining divisor is

The nonconstant invariant factors of a linear operator satisfy $d_1\mid \cdots \mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an irreducible polynomial, so every $d_j$ must equal $X^2+1$. Their polynomial degrees sum to $n$, giving $n=2k$.

The companion matrix of $X^2+1$, in the basis $(1,X)$ of $\mathbb{Q}[X]/(X^2+1)$, isTherefore, for each fixed even $n$, every such $A$ has rational canonical formThere are no such matrices when $n$ is odd. This is an instance of the rational canonical form with irreducible minimal polynomial principle; equality of minimal polynomials alone does not determine rational canonical form in general.

Let $v\in {\mathbb{Q}}^2$ be nonzero. If the columns $v,Av$ were linearly dependent over the rational numbers, then $Av=\lambda v$ for some $\lambda \in \mathbb{Q}$. Applying $A$ once more would giveso $({\lambda }^2+1)v=0$. Since $v\ne 0$, this forces ${\lambda }^2=-1$, impossible in $\mathbb{Q}$. Thus the two columns are linearly independent, and the square matrixis an invertible matrix.

In the basis $(v,Av)$, the linear map given by $A$ sends $v$ to the second basis vector and sends $Av$ to $A^{2}v=-v$. Equivalently,Multiplying on the left by the matrix inverse $E^{-1}$ givesThis change of basis produces the companion matrix found in part (a)(ii).

The Gaussian integers form an integral domain, since they are a subring of the complex numbers. Use the Gaussian integer normwhich is a positive integer for nonzero $a+bi$ and satisfies $N(\alpha \beta )=N(\alpha )N(\beta )$.

For $\alpha ,\beta \in \mathbb{Z}[i]$ with $\beta \ne 0$, write $\alpha /\beta =x+iy$. Choose integers $m,n$ with $|x-m|\le 1/2$ and $|y-n|\le 1/2$, put $q=m+in$, and set $r=\alpha -\beta q$. ThenThus division with remainder has either $r=0$ or a strictly smaller Gaussian integer norm. This proves that $N$ is a euclidean function and

Over a principal ideal domain $D$, an $m$-by-$n$ matrix is in Smith normal form if all entries off the main diagonal vanish and its nonzero diagonal entries $d_1,\dots ,d_r$ occur first and satisfyAll remaining diagonal entries are zero. The zero matrix is allowed, with $r=0$. For an arbitrary matrix $M$, the Smith normal form theorem gives invertible matrices $P\in {\mathrm{GL}}_m(D)$ and $Q\in {\mathrm{GL}}_n(D)$ withwhere the right side is interpreted as a rectangular diagonal matrix when necessary. The $d_j$ are determined up to multiplication by units. Positivity can be required over $\mathbb{Z}$, but is not a condition over the Gaussian integers.

Regard $A$ as a matrix over the rational numbers. Part (a)(ii), with $n=2$, gives its characteristic polynomial $X^2+1$. Therefore, writing $i$ for the imaginary unit,The matrix $M=iI-A$ is not zero: otherwise $A=iI$ would have noninteger diagonal entries. It consequently has rank of a matrix one over the fraction field $\mathbb{Q}(i)$ of the Gaussian integers.

By part (b)(i), $\mathbb{Z}[i]$ is a euclidean domain, hence a principal ideal domain, so apply Smith normal form. There exist $P,Q\in {\mathrm{GL}}_2(\mathbb{Z}[i])$ such thatIndeed, invertible multiplication preserves rank of a matrix, so exactly one diagonal entry is nonzero. For $e_2=(0,1{)}^{\mathsf{T}}$, multiplication on the right gives $P(iI-A)Q{e}_2=0$. Since $P$ is an invertible matrix,The Smith normal form supplies a kernel basis lemma ensures that the null vector $Q{e}_2$ is a column of an invertible matrix over $\mathbb{Z}[i]$; an arbitrary nonzero solution over $\mathbb{Q}(i)$ would not itself establish that requirement.

Each entry of the matrix $Q$ from part (b)(iii) has unique real and imaginary integer coefficients. WriteIt already belongs to the general linear group ${\mathrm{GL}}_2(\mathbb{Z}[i])$. Expanding the relation from part (b)(iii) givesThe real and imaginary parts must both vanish, soThese are the required identities, with $e_2=(0,1{)}^{\mathsf{T}}$.