A projective transformation of $\mathbb{P}({\mathbb{F}}^{n+1})$ is induced by an invertible linear map $A:{\mathbb{F}}^{n+1}\to {\mathbb{F}}^{n+1}$:This is well defined because replacing $v$ by a nonzero scalar multiple does not change $[Av]$. Multiplying $A$ by a nonzero scalar also leaves $\tau$ unchanged. Conversely these are precisely the transformations in the projective linear group ${\mathrm{PGL}}_{n+1}(\mathbb{F})$. No field automorphism is included in this definition.

The second supplementary law for the Legendre symbol is $(2/p)=(-1{)}^{(p^2-1)/8}$. To multiply these signs, observe that for odd $a,b$,Both factors in the numerator are divisible by $8$, so the right side is even. Applying this identity repeatedly to the prime factorization givesAgain the convention at $Q=1$ agrees.

The supplementary law for the Legendre symbol is $(-1/p)=(-1{)}^{(p-1)/2}$ for odd prime numbers $p$. For odd integers $a,b$,and the last term is even. Induction over the prime factors of $Q$ therefore gives ${\sum }_i(p_i-1)/2\equiv (Q-1)/2(\mathrm{mod}2)$. ConsequentlyFor $Q=1$ the exponent is zero.

If $P\equiv P^{\prime }(\mathrm{mod}Q)$, then $P\equiv P^{\prime }(\mathrm{mod}p)$ for each prime factor $p$ of $Q$. The Legendre symbol depends only on the numerator's residue class, so the corresponding factors agree. Also $\gcd (P,Q)=\gcd (P^{\prime },Q)$, covering the zero cases. Multiplying givesThe convention for denominator $1$ gives the same conclusion there.

For a prime number $p$, multiplicativity of the Legendre symbol says $(P{P}^{\prime }/p)=(P/p)(P^{\prime }/p)$, including the case of a zero factor. Therefore, writing $Q={\prod }_i{p}_i$ with repetitions,This product definition still agrees with the stipulated zero when the numerator and denominator are not coprime. For $Q=1$ all three symbols are $1$.

Factor $Q={\prod }_i{p}_i$ and $Q^{\prime }={\prod }_j{q}_j$, retaining repeated prime factors. If $P$ has a common prime factor with $Q{Q}^{\prime }$, it has one with at least one of $Q,Q^{\prime }$, and both sides of the desired equality are zero. Otherwise the definition of the Jacobi symbol givesAn empty product is $1$, so the argument includes either denominator being $1$.

No. Take $P=2$, $Q=9$. The Jacobi symbol isbut the squares modulo $3$ are $0,1$. A square root of $2$ modulo $9$ would reduce to one modulo $3$, which is impossible. A Jacobi symbol of $1$ records a product of local signs; it does not assert that every local sign is positive, or that a quadratic congruence is soluble.

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 $e_2=(0,1{)}^{\mathsf{T}}$.

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.

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.

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

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).

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.

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.

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

For every $i\in {\mathbb{N}}_0$, truncate $w$ before coordinate $i$:This $f_i$ lies in $F$. For $j<i$, $(w-f_i)(j)=0$. For $j\ge i$, $(w-f_i)(j)=w(j)=\pm 2^j$, which is divisible by $2^i$. In particular the integer sequencebelongs to $B$ and satisfies $w=f_i+2^i{g}_i$.

Apply the module homomorphism $\pi$. Since $f_i\in F\subseteq \ker \pi$,Fixing any coordinate $j$, the integer $\pi (w)(j)$ is therefore divisible by $2^i$ for every $i$. The only such integer is zero: if $a\ne 0$, choose $i$ with $2^i>|a|$, which rules out $a=2^{i}b$ for any nonzero integer $b$. Thus every coordinate of $\pi (w)$ vanishes, contradicting part (b)(iii).

It follows thatThe decisive property is vanishing of infinitely divisible integer sequences, namely ${\bigcap }_{i\ge 0}{2}^{i}B=\{0\}$.

ConsiderThe map sending a sequence of signs $({\epsilon }_j)$ to $({\epsilon }_j{2}^j)$ is a bijection from $\{-1,1{\}}^{{\mathbb{N}}_0}$ onto $W$. This is an uncountable set by the Cantor diagonal argument: given a proposed list $w_0,w_1,\dots$ of its elements, define $w(j)=-w_j(j)$. Then $w\in W$ differs from $w_j$ at coordinate $j$ for every $j$.

Since $\ker \pi$ is countable, $W$ cannot be contained in it. Consequently there exists

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.

For $j\in {\mathbb{N}}_0$, let $e_j\in B$ have $e_j(j)=1$ and $e_j(k)=0$ for $k\ne j$. If $f,g\in F$ and $z\in \mathbb{Z}$, then $f+g$ can be nonzero only where $f$ or $g$ is nonzero, and $zf$ can be nonzero only where $f$ is nonzero. These are finite sets. Since $0\in F$, this proves that $F$ is a submodule.

Every $f\in F$ has the finite expressionEvaluating at coordinate $k$ recovers the coefficient of $e_k$, so the expression is unique. Thus $\{e_j:j\in {\mathbb{N}}_0\}$ is a basis of a module for $F$, andis a free module. By contrast, the ambient module of integer sequences $B={\prod }_{j\in {\mathbb{N}}_0}\mathbb{Z}$ allows arbitrary integer sequences.

Choose a countable generating family $(n_k)$ for $N$. For each $k$, the expression of $n_k$ in the basis of a module $E$ has a finite set $H_k$ of nonzero coefficients, its coordinate support in a free module. LetBy the countable union of countable sets property, $E^{\prime }$ is a countable set. Every $n_k$ belongs to $⟨E^{\prime }⟩$, so all finite $R$-linear combinations of the $n_k$ do too. ThereforeHere and below, countable includes finite. The same construction applies to a finite or empty generating family. This is the countable basis support lemma.