For the explicit basis in part (c), $q_0+t{q}_1$ has coefficients $(1+t,1-2t,t-2)$. ThusAt these values the conics factor, up to nonzero constants, asGeometrically these are exactly the three pairings of the four base points into two pairs, with a projective line joining each pair. No double line can contain four points in general linear position.

The numerical parameter values depend on the selected basis, but the three distinct singular members do not. Another pair of independent nonsingular basis conics gives an invertible projective change of pencil parameter. Its parameter infinity is nonsingular because the second basis form is nonsingular. Therefore all three singular members again occur at three distinct finite values of $t$. This proves the claimed count for any such basis.

Use part (a) to put the four projective points atA quadratic polynomial vanishing at the first three has no square terms, and vanishing at the fourth givesThese form a two-dimensional vector space. Choose the two membersTheir coefficient triples are independent and both satisfy the relation. Their symmetric bilinear form matrices are of the formFor $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a linear combination of these two. Hence every conic through the four points has matrixUndoing the projective transformation carries the entire family back to the original four points, preserving nonsingularity. This is a pencil of conics through four points.

A projective quadric is the zero locus of a nonzero homogeneous quadratic polynomial $q(v)=v^{T}Bv$, with $B$ symmetric over a field of characteristic different from two. A projective point on it is singular when all first partial derivatives vanish. Since $\nabla q(v)=2Bv$, the quadric is nonsingular exactly when $B$ is invertible.

Over the complex numbers, a nonzero quadratic form in three variables can be diagonalized by an invertible change of basis. Its rank classifies a projective conic:

The zero polynomial is excluded from the definition of a conic. Thus these last two cases are precisely the singular conics.

The required general linear position condition is that every $n+1$ of the $n+2$ representing vectors are linearly independent. Such an ordered configuration is a projective frame.

Choose representatives $p_0,\dots ,p_n$ of the first $n+1$ projective points; they form a basis. Write a representative of the last point as $p_{n+1}={\sum }_i{a}_i{p}_i$. Every $a_i$ is nonzero: otherwise the last point together with the other $n$ basis points would fail the independence condition. Rescale the basis representatives to ${\tilde{p}}_i=a_i{p}_i$, so that $p_{n+1}={\sum }_i{\tilde{p}}_i$.

Do the same for the target configuration, obtaining representatives ${\tilde{q}}_i$ with $q_{n+1}={\sum }_i{\tilde{q}}_i$. The unique linear map sending the basis ${\tilde{p}}_i$ to ${\tilde{q}}_i$ is invertible and also sends $p_{n+1}$ to $q_{n+1}$. Its induced projective transformation sends every point to its prescribed partner. Thus any two ordered projective frames are projectively equivalent.

Factor the group order as $588=2^2\cdot 3\cdot 7^2$. The Sylow theorems say that the number of Sylow subgroups for the prime seven divides $12$ and is congruent to one modulo seven. Among $1,2,3,4,6,12$, only one qualifies. Thus there is a unique normal subgroup $N$ of order $49$.

Every group of prime-square order is Abelian. Indeed its center of a group is nontrivial, and a cyclic quotient by the center of a group forces the whole group to be abelian. Consequently $N$ is either $C_{49}$ or $C_7\times C_7$, and in either case it has two composition factors, both $C_7$.

It remains to consider $Q=H/N$, of order twelve. Its number of Sylow subgroups for the prime three is either one or four:

Lift the composition series of $Q$ to $H$ using the subgroup correspondence theorem, and join it to a composition series inside $N$. This givesas the multiset of composition factors. The Jordan–Hölder theorem makes this independent of the chosen series. In particular every group of order $588$ is a solvable group, with composition length five; that is a statement about composition length, not necessarily its derived length.

Let $1◃N◃G$ be the given length-two composition series. Thus $N$ and $G/N$ are simple groups. Let $M$ be any proper nontrivial normal subgroup of $G$, such as the penultimate term of another composition series.

If $M=N$, its two factors are already the given ones. Suppose $M\ne N$. Since $M\cap N$ is normal in the simple group $N$, it is either $1$ or $N$. The latter would mean $N\subseteq M$, but then $M/N$ would be a proper nontrivial normal subgroup of $G/N$, impossible. Thus $M\cap N=1$.

Also $M$ cannot be contained in $N$, since it would then be a nontrivial normal subgroup of $N$ distinct from $N$. The product $MN$ is a normal subgroup of $G$, strictly containing $N$. Simplicity of $G/N$ therefore forces $MN=G$. The second isomorphism theorem for groups givesIn particular $M$ is simple. Every other composition series has such an $M$ as its last proper term, and no nontrivial refinement inside a simple $M$ is possible. Hence every composition series has length two.

The general Jordan–Hölder theorem states that any two composition series of a finite group have the same length and the same composition factors up to group isomorphism and permutation. The argument above proves both assertions in this length-two case: another series either uses $N$ itself or has factors isomorphic to the original pair in the opposite order. No general uniqueness theorem was used to prove this special case.

A simple group is a nontrivial group whose only normal subgroups are itself and the trivial subgroup.

A composition series is a chain $G=G_0▹G_1▹\cdots ▹G_r=1$ in which each $G_{i+1}$ is a normal subgroup of $G_i$ and each quotient group $G_i/G_{i+1}$ is simple and nontrivial. Its length is $r$, and these quotients are its composition factors. The individual terms need not all be normal in $G$.

A solvable group is a group whose derived series reaches the trivial group: starting with $G^{(0)}=G$, define $G^{(i+1)}=[G^{(i)},G^{(i)}]$. For finite groups this is equivalent to having a composition series whose factors are all cyclic groups of prime order.

Let $J=\left(\begin{array}{cc}0& -2\\ 1& 0\end{array}\right)$, so the displayed rational canonical form of $T$ is $J\oplus J$. Since $J^2=-2I_2$, we have $T^2=-2I_4$.

If $a=0$, then $U=b{I}_4$ is a scalar matrix. Its invariant factors of a linear operator are four copies of $t-b$, so its rational canonical form isHere $m_U=t-b$ and $c_U=(t-b{)}^4$, including when $b=0$.

Now let $a\ne 0$ and putOn each two-dimensional invariant block, $A=aJ+b{I}_2$ satisfies $q(A)=0$. The vectors $e_1,A{e}_1$ form a cyclic vector basis, since their coordinate matrix is $P=\left(\begin{array}{cc}1& b\\ 0& a\end{array}\right)$ with determinant $a\ne 0$. Directly,Applying this basis change to both blocks givesThus $m_U=q$ and $c_U=q^2$. In particular there are two invariant factors $q,q$, rather than a single block for $q^2$. Also $q(t)>0$ for every real $t$, so it is an irreducible polynomial over $\mathbb{R}$. This is an instance of rational canonical form under an affine change.

Finally $T^2=-2I_4$ implies $T^{-1}=-T/2$. Substituting $a=-1/2$, $b=0$ in the preceding result gives $q(t)=t^2+1/2$, and henceIts minimal polynomial is $t^2+1/2$ and its characteristic polynomial is $(t^2+1/2{)}^2$. The normalized reciprocal of $t^2+2$ is also $t^2+1/2$, agreeing with rational canonical form of an inverse.

The rational canonical form of $T$ is a block diagonal matrixsimilar to a matrix of $T$, where the $f_j\in F[t]$ are monic nonconstant polynomials and the sum of their degrees is $\dim V$. They are the invariant factors of a linear operator. The block $C(f_j)$ is the companion matrix for $f_j$, representing multiplication by $t$ on $F[t]/(f_j)$. The invariant factors are uniquely determined by $T$.

Every companion block has a cyclic vector and therefore has both minimal polynomial and characteristic polynomial equal to $f_j$, by part (a). The determinant of a block diagonal matrix is the product of its block determinants, givingA polynomial $h$ annihilates the whole matrix exactly when it annihilates every block. On block $j$, dividing $h$ by $f_j$ leaves a remainder of smaller degree; if that remainder annihilated the block without being zero, it would contradict minimality of $f_j$. Thus the annihilation condition is exactly $f_j\mid h$ for every $j$. The monic polynomial of least degree with this property is their least common multiple. The divisibility chain makes this $f_r$, soThese are the minimal and characteristic polynomials from invariant factors formulae. On the zero vector space the product is empty and both polynomials are defined to be one.

Write $d=\dim V$. A cyclic vector for $S$ is a vector $v$ whose successive iterates span $V$; equivalently, when $d>0$,are a basis. The equivalence follows because the first linear dependence in the iterates makes their preceding span invariant under $S$, so no later iterate can enlarge it.

In this basis there are unique coefficients $c_0,\dots ,c_{d-1}$ with $S^{d}v=-{\sum }_{j<d}{c}_j{S}^{j}v$. Put $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(S)v=0$, and commutation with powers of $S$ gives $f(S)S^{j}v=S^{j}f(S)v=0$. Therefore $f(S)$ kills an entire basis and is the zero map. No nonzero polynomial of degree less than $d$ annihilates $v$, because that would be a linear dependence of the displayed basis. Thus $m_S=f$.

The matrix of $S$ in the cyclic basis is the companion matrix $C(f)$ with ones on its subdiagonal and last column $(-c_0,\dots ,-c_{d-1}{)}^{\mathsf{T}}$. Expanding $\det (tI-C(f))$ along its last column givesthe cofactors are $1,t,\dots ,t^{d-1}$. This proves the characteristic polynomial of a companion matrix formula here. HenceThis proves that a cyclic operator has equal minimal and characteristic polynomials. If $V=\{0\}$, both polynomials are one by convention, so the equality remains valid.

Since ${\omega }^2=-1-\omega$, products of integer linear combinations of $1,\omega$ remain of the same form. Thus the Eisenstein integers $E=\mathbb{Z}[\omega ]$ form a subring of $\mathbb{C}$, and hence an integral domain. Their Eisenstein norm isIt is a positive integer on nonzero elements and is multiplicative, because the complex modulus is multiplicative.

For $\alpha ,\beta \in E$ with $\beta \ne 0$, write $\alpha /\beta =u+v\omega$ with real $u,v$. Choose integers $m,n$ with $|u-m|,|v-n|\le 1/2$, and let $q=m+n\omega$. With $s=u-m$ and $t=v-n$,Consequently $r=\alpha -q\beta$ belongs to $E$ and satisfies $N(r)\le \frac{3}{4}N(\beta )<N(\beta )$ when $r\ne 0$. This proves $E$ is a Euclidean domain with Euclidean function $N$, the Euclidean norm on the Eisenstein integers.

For the representation of the prime $p$, first note that $3=N(1-\omega )$. Now suppose $p>3$ and $p\mid n^2+3$. The factorizationshows that the Eisenstein integer $p$ divides a product. It divides neither factor: an element $a+b\omega$ lies in $pE$ exactly when both integer coefficients are divisible by $p$, and the second coefficients here are $2$ and $-2$. Hence $p$ is not a prime element of $E$.

To justify the necessary factorization step, in any principal ideal domain an irreducible element $\pi$ is prime. If $\pi \nmid a$, write $(\pi ,a)=(d)$. Since $d\mid \pi$, irreducibility makes $d$ a unit or an associate of $\pi$. The latter would give $\pi \mid a$, so $d$ is a unit and $u\pi +va=1$ for some $u,v$. If $\pi \mid ab$, multiplication of this identity by $b$ gives $\pi \mid b$. This proves irreducible elements of a principal ideal domain are prime without assuming unique factorization.

By part (a), $E$ is a principal ideal domain. The integer $p$ has norm $p^2>1$, so it is a nonzero nonunit. Since it is not prime, the result just proved implies that it is reducible: $p=uv$ with $u,v$ both nonunits. A nonzero element of $E$ is a unit exactly when its norm is one, since $N(z)=1$ gives inverse $\overline{z}$ and multiplicativity proves the converse. ThusThe only possible positive integer factors exceeding one are $N(u)=N(v)=p$. Writing $u=a+b\omega$ givesThis is the prime representation by the Eisenstein norm argument.

The three answers, in the given order, are no, yes, yes.

For $\mathbb{Z}[x]$, consider the ideal $I=(2,x)$. Its members have even constant coefficient, so $1\notin I$ and the ideal is proper. If $I=(g)$, then $g$ divides the constant $2$, forcing $g$ to be constant by additivity of polynomial degree in this integral domain. Since $g$ also divides $x$, it must divide the coefficient $1$, giving $g=\pm 1$. That would make $I$ the whole ring, a contradiction. Thus $\mathbb{Z}[x]$ is not a principal ideal domain, and part (a) shows it cannot be a Euclidean domain. This is the nonprincipal ideal generated by a prime and an indeterminate obstruction.

Evaluation at $i$ gives the ring isomorphismIndeed, division by the monic polynomial $x^2+1$ leaves a unique remainder $a+bx$, and its value $a+bi$ is zero only when both integer coefficients vanish. Thus the quotient maps bijectively, with operations preserved, onto the Gaussian integers. These form a Euclidean domain for the Gaussian integer norm $N(a+bi)=a^2+b^2$, a standard Euclidean example allowed in this part. Geometrically, rounding the real and imaginary coordinates gives a quotient error with squared modulus at most $1/2<1$.

Finally,via $f(x,y)\mapsto f(0,y)\mathrm{mod}2$. Every polynomial over ${\mathbb{F}}_2$ occurs. The kernel consists exactly of polynomials whose $x$-independent part has even coefficients: write $f(x,y)=xg(x,y)+h(y)$ to see that this is precisely $(2,x)$. Hence the induced quotient map is a bijective ring homomorphism. Its target is a Euclidean domain with Euclidean function $\mathrm{deg}$: a polynomial ring over a field is Euclidean by the division algorithm in part (a).

A Euclidean domain is an integral domain $R$ with a Euclidean function $\nu :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that, for $a,b\in R$ with $b\ne 0$, there exist $q,r\in R$ satisfyingThe polynomial ring $\mathbb{Q}[x]$ has this property with $\nu (f)=\mathrm{deg}f$. To perform polynomial division, subtract an appropriate multiple of the divisor to cancel the leading term of the current remainder. Its polynomial degree decreases strictly at each step, so the process terminates below the divisor's degree. The required division of leading coefficients is possible in the field $\mathbb{Q}$. Thus a polynomial ring over a field is Euclidean, with the same argument applying to every field.

Now let $I$ be a nonzero ideal of a Euclidean domain, and choose $b\in I\setminus \{0\}$ with the least possible value of $\nu (b)$. For every $a\in I$, divide $a=qb+r$. The remainder belongs to $I$, since $r=a-qb$. If it were nonzero, its smaller Euclidean value would contradict the choice of $b$. Thus $r=0$ and every $a\in I$ lies in $(b)$. Conversely $(b)\subseteq I$ because $I$ is an ideal. Hence $I=(b)$. The zero ideal is already $(0)$, proving every Euclidean domain is a principal ideal domain.