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A nonzero polynomial of degree $d$ over a field has at most $d$ distinct roots. Indeed, each root supplies a linear factor by polynomial division, and induction on $d$ gives the bound.

The evaluation map $\phi :\mathbb{R}[x]\to \mathbb{C}$, $p\mapsto p(i)$, is a ring homomorphism and is onto because $a+bx$ maps to any $a+bi$. By polynomial division, writeThen $p(i)=0$ precisely when $a=b=0$. Thus $\ker \phi =(x^2+1)$, and the first isomorphism theorem givesSince the quotient ring is a field, $(x^2+1)$ is a maximal ideal.

Evaluation at a real point defines a surjective ring homomorphism $p\mapsto p(a,b)$ from $\mathbb{R}[x,y]$ to $\mathbb{R}$. Its kernel is $(x-a,y-b)$: subtracting the constant $p(a,b)$ expresses the remainder as a multiple of $x-a$ plus a multiple of $y-b$, by successive polynomial division. The first isomorphism theorem therefore identifies the quotient ring with the field $\mathbb{R}$, proving each $I_{a,b}$ is maximal.

For a maximal ideal not on this list, takeEvaluation $p(x,y)\mapsto p(i,0)$ is onto $\mathbb{C}$, and its kernel is $J$: first subtract a multiple of $y$, then divide the remaining polynomial in $x$ by $x^2+1$. Hence $\mathbb{R}[x,y]/J\cong \mathbb{C}$, a field. Finally, if $J=I_{a,b}$ for real $a,b$, the member $x^2+1\in J$ would vanish at $(a,b)$, requiring $a^2+1=0$, which is impossible over $\mathbb{R}$.

Here is an intrinsic definition that also proves well-definedness. For a complex line bundle $L$, define its First Chern class to be the Euler class of its canonically oriented underlying real rank-two bundle: pull its integral Thom class back along the zero section after forgetting relative support. The complex orientation fixes the sign, so this construction makes no arbitrary choice of generator.

For a rank-$r>0$ complex vector bundle $E$, let $\pi :\mathbb{P}(E)\to X$ be its projective bundle of lines, let $S\subset {\pi }^{\ast }E$ be the tautological bundle, and put $h=-c_1(S)=c_1(S^{\ast })$. On every fiber ${\mathbb{CP}}^{r-1}$, $h$ is the positive degree-two generator. The Leray-Hirsch theorem says that if globally defined cohomology classes restrict to a free basis of the cohomology of every fiber, then multiplication by those classes identifies the cohomology of the total space with a free module over the base. Applied here, it givesThe finite trivializing cover in the question is sufficient for this application: the assertion holds on each trivializing open set by the Künneth theorem, since the fiber has finite free cohomology, and the Mayer–Vietoris sequence and the Five lemma glue it over the finite cover. The same local argument constructs the oriented Thom class used for line bundles. It does not require a choice of classifying map.

There are therefore unique classes $c_i(E)\in H^{2i}(X;\mathbb{Z})$ such thatDefine $c_0(E)=1$ and $c_i(E)=0$ for $i>r$. For rank zero the Total Chern class is $1$. This is the projective bundle definition of Chern classes. Existence and uniqueness follow by expressing $h^r$ in the displayed free module basis, with degrees determining each coefficient. The projective bundle and tautological bundle are intrinsic to $E$, and the line Thom class is uniquely fixed by orientation. Hence the resulting Chern classes do not depend on trivializations or other auxiliary choices. For a line bundle the relation is $h+c_1(E)=0$, agreeing with the original normalization. Pulling back this unique relation also proves naturality. This standard construction and the sum theorem are treated in Vector Bundles and K-Theory, Section 3.1.

The requested result is the Whitney sum formula for Chern classes:Here $c(E)=1+c_1(E)+\cdots +c_r(E)$, and the formula holds for complex vector bundles over a common base. In particular, a trivial bundle has total Chern class $1$.

Now take $B={\mathbb{CP}}^{n-1}$ and let $L\subset B\times {\mathbb{C}}^n$ be its tautological bundle. The standard Hermitian inner product gives the rank-$(n-1)$ complex bundle $Q=L^{\perp }$, with $L\oplus Q\cong {\underline{\mathbb{C}}}^n$. The orthogonal complex line flag manifold in the question is precisely $\mathbb{P}(Q)$: over a first line $\ell$, the second line is any line in ${\ell }^{\perp }$. Local orthonormal frames give this identification as a fiber bundle, with fiber ${\mathbb{CP}}^{n-2}$.

Let $x=c_1(L^{\ast })$ on $B$, also writing $x$ for its pullback to $\mathbb{P}(Q)$. By part 3(a), $H^{\ast }(B;\mathbb{Z})=\mathbb{Z}[x]/(x^n)$. The Whitney sum formula for Chern classes givesLet $S_2$ be the second tautological line on $\mathbb{P}(Q)$, and set $y=c_1(S_2^{\ast })$. Thus $x,y$ are exactly the pullbacks of the positive hyperplane classes from the two projective factors. The projective bundle definition of Chern classes for $Q$ givesTogether with $x^n=0$, this gives a surjective graded ring mapThere are no additional relations. Indeed the second relation is monic of degree $n-1$ in $y$, so polynomial division makes its source free over $\mathbb{Z}[x]/(x^n)$ on $1,y,\dots ,y^{n-2}$. The projective bundle formula for complex vector bundles gives exactly the same free basis on the target. The map sends each basis element to its corresponding basis element, and is therefore an isomorphism. ConsequentlyThis is the cohomology ring of the orthogonal complex line flag manifold. Its additive basis is $x^i{y}^j$ with $0\le i<n$, $0\le j<n-1$. As a symmetry check, multiplying the second relation by $y-x$ gives $y^n-x^n=0$, so $y^n=0$ as expected from the second projection. For $n=2$, every line has a unique orthogonal line, and the relations become $x^2=0$, $x+y=0$, giving the ring of ${\mathbb{CP}}^1$. If $n=1$, the space is empty and the second relation is $1=0$, so the printed formula still gives the zero cohomology ring.

Reducing coefficients modulo $p$ gives the ring isomorphismIf $d=\mathrm{deg}f>0$, its monic leading coefficient remains one modulo $p$, so $\mathrm{deg}\bar{f}=d$. Polynomial division gives a unique representative of degree less than $d$: existence follows by division, and the difference of two such representatives cannot be a nonzero multiple of a degree-$d$ monic polynomial. There are $p$ choices for each of the $d$ coefficients, provingIrreducibility of $\bar{f}$ is not needed for finiteness; without it the quotient ring need not be a field.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

The Cayley-Hamilton theorem states that every square matrix satisfies its characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)$, then ${\chi }_A(A)=0$.

For the proof, the adjugate identity givesWrite $\mathrm{adj}(tI-A)=B_0+B_{1}t+\cdots +B_{n-1}{t}^{n-1}$ and compare coefficients. Multiplying the resulting recurrence by successive powers of $A$ and summing telescopically yields ${\chi }_A(A)=0$.

By polynomial division, $p=q{\chi }_A+r$ with $\mathrm{deg}r<n$. Therefore $p(A)=r(A)$ by Cayley-Hamilton, while $p(\lambda )=r(\lambda )$ for every eigenvalue $\lambda$, since ${\chi }_A(\lambda )=0$.

For the displayed matrix,The remainder of $t^{1000}$ modulo this polynomial has the form $a{t}^2+c$, because its values at $i$ and $-i$ are both $1$. The equations $-a+c=1$ and $4a+c=2^{1000}$ giveSince $(A^2{)}_{11}=3$, the requested entry is

For fixed interpolation nodes $c_0,\dots ,c_{s-1}$, interpolate any $p\in {\mathbb{P}}_{s-1}[x]$ asIntegrating shows that the unique weights making the quadrature rule exact through degree $s-1$ are

Now let the nodes be the zeros of the degree-$s$ orthogonal polynomial $q_s$. Given any $p\in {\mathbb{P}}_{2s-1}[x]$, polynomial division givesOrthogonality gives ${\int }_a^b{q}_{s}h=0$, while $q_s(c_k)=0$ makes the quadrature sum for $q_{s}h$ vanish. The rule is already exact on $r$, so it is exact on $p$. This is the defining precision property of Gaussian quadrature.

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over the finite field ${\mathbb{F}}_p$ has at most $d$ roots. To prove it, use mathematical induction on $d$. A constant nonzero polynomial has no roots. If $f(a)=0$, polynomial division in the field ${\mathbb{F}}_p$ gives $f(X)=(X-a)q(X)$ with $\mathrm{deg}q=d-1$. Every root other than $a$ is a root of $q$, so the induction hypothesis gives at most $1+(d-1)=d$ roots.

Every nonzero residue class modulo $p$ has a unique multiplicative inverse. In the product $(p-1)!$, pair each class with its inverse. The unpaired classes satisfy $x=x^{-1}$, hence $x^2-1=0$. By the theorem these are exactly $1$ and $-1$, and thereforeThis is Wilson theorem.

Now write $p-1=kq$. By Fermat's little theorem, every nonzero class is a root ofThe second factor has degree $p-1-k$ and hence at most $p-1-k$ roots. Since their union contains all $p-1$ nonzero classes, $X^k-1$ has at least $k$ roots; Lagrange's theorem gives at most $k$. Thus $x^k\equiv 1(\mathrm{mod}p)$ has exactly $k$ solutions.

Let ${\ell }_j$ be the Lagrange interpolation polynomial for the roots ${\omega }_j$, and defineFor $\mathrm{deg}Q\le 2n-1$, polynomial division gives $Q=S{P}_n+R$ with $\mathrm{deg}S,\mathrm{deg}R<n$. Orthogonality kills the first term, while interpolation gives $R={\sum }_{j}R({\omega }_j){\ell }_j$. ThereforeTaking $Q={\ell }_j$ proves uniqueness. Taking $Q={\ell }_j^2$ also showsThis is Gaussian quadrature.

Let $Q=Q_{n+1}$. By polynomial division, every $q\in {\mathcal{P}}_{n+1+k}$ has a unique decompositionEvery node is a zero of $Q$, so $I_n(Qs)=0$. The assumed degree-$n$ exactness also gives $I_n(r)=I(r)$. HenceTherefore $I_n(q)=I(q)$ for every $q\in {\mathcal{P}}_{n+1+k}$ if and only iffor every $s\in {\mathcal{P}}_k$. This proves the nodal-polynomial criterion for quadrature exactness.