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For affine varieties $X\subset {\mathbb{A}}^m$ and $Y\subset {\mathbb{A}}^n$, a morphism of affine varieties is a map whose $n$ coordinate functions are restrictions to $X$ of polynomials in its $m$ ambient coordinates. Intrinsically, its coordinates belong to the coordinate ring $A(X)$; they must satisfy the defining relations of $Y$.

Suppose a bounded linear isomorphism $T:{\ell }^p\to Y\subset {\ell }^2$ with bounded inverse existed. There would be constants $0<c\le C$ such that $c\Vert x{\Vert }_p\le \Vert Tx{\Vert }_2\le C\Vert x{\Vert }_p$. Put $y_j=T{e}_j$. Averaging the squared Hilbert norm over all $2^n$ independent sign choices cancels cross terms, giving the generalized parallelogram identityEvery signed sum of the original $e_j$ has $p$-norm $n^{1/p}$. The left side is therefore between $c^2{n}^{2/p}$ and $C^2{n}^{2/p}$, while the right side is between $c^{2}n$ and $C^{2}n$. Thus $n^{2/p}$ and $n$ would be bounded constant multiples of one another for all $n$, impossible unless $p=2$. Consequently no such normed-space isomorphism exists for $p\ne 2$.

Use the convention that the inner product is linear in its first argument. Set $a_n=⟨x,e_n⟩$ and $P_{N}x={\sum }_{n\le N}{a}_n{e}_n$. Orthogonality makes $P_{N}x$ the nearest point in the finite span of the first $N$ basis vectors. Since an orthonormal basis has dense linear span, for every $ϵ>0$ a finite linear combination approximates $x$ within $ϵ$. For all larger $N$, the nearest-point property gives $\Vert x-P_{N}x\Vert <ϵ$. Thus the series converges in $X$ to $x$ even without completeness. Taking inner products with each $e_n$ proves uniqueness of its coefficients.

In the Hilbert-space case, Parseval's identity gives $\sum |a_n{|}^2=\Vert x{\Vert }^2$. Completeness makesconvergent. It is linear and preserves inner products and norms, so bounded with norm one. Its inverse is the corresponding map sending $f_n$ to $e_n$; equivalently every coefficient sequence of a vector in the $f_n$ basis is also square-summable in the $e_n$ basis. Hence $U$ is surjective and unitary. Any bounded map agreeing on the $e_n$ agrees on their dense span and then on all of $X$, proving uniqueness.

A finite separable field extension of degree $d$ has exactly $d$ embeddings into an algebraic closure: adjoin finitely many generators successively, and use part (i) to count extensions of each embedding by the distinct roots of the relevant separable minimal polynomial.

For $L=K(\alpha ,\beta )$, label these embeddings ${\sigma }_1,\dots ,{\sigma }_d$. For each pair $i\ne j$, the equalityexcludes at most one $c\in K$. If the images of $\beta$ agree, the images of $\alpha$ cannot agree too, so it excludes none. Since $K$ is infinite, choose $c$ avoiding the finitely many bad values and put $\eta =\alpha +c\beta$. The $d$ images of $\eta$ are distinct, hence its minimal polynomial has degree at least $d$. But $K(\eta )\subset L$ gives degree at most $d$. Therefore $K(\eta )=L$. Combining generators two at a time proves the primitive element theorem for every finite separable extension over an infinite base field.

A splitting field of $f$ over $K$ is an extension generated over $K$ by all its roots in which $f$ splits into linear factors. A nonconstant irreducible polynomial is separable polynomial when its roots in a splitting field are distinct, equivalently when it is coprime to its formal derivative. For a general polynomial, the convention compatible with the printed assertion is that all its irreducible factors are separable; repeated occurrences of the same factor are allowed under this convention.

For a finite field of characteristic $p$, the finite-field Frobenius automorphism $x\mapsto x^p$ is injective and hence surjective. If an irreducible polynomial $g$ had zero derivative, every nonzero exponent would be divisible by $p$. Taking $p$th roots of its coefficients would give $g=h^p$, contradicting irreducibility. Thus $g^{\prime }\ne 0$, and irreducibility gives $\gcd (g,g^{\prime })=1$. Every irreducible factor is therefore separable: finite fields are perfect.

If instead one defines a separable polynomial to mean that the entire polynomial has no repeated roots, the printed claim for every $f$ is false: $(t-1{)}^2$ is a counterexample over any finite field. The proof above establishes the precise intended perfect-field assertion; a polynomial has entirely distinct roots if and only if it is also squarefree.

For $\alpha \in L$ with $f(\alpha )=0$, evaluation $K[t]\to L$, $g\mapsto g(\alpha )$, has $(f)$ in its kernel. It therefore induces a $K$-algebra homomorphism ${\sigma }_{\alpha }:K[t]/(f)\to L$. Conversely, such a homomorphism is determined by $\alpha =\sigma (\bar{t})$, since every class is a polynomial in $\bar{t}$. Applying it to $f(\bar{t})=0$ shows $f(\alpha )=0$. These constructions are inverse, giving the root-embedding correspondenceIrreducibility ensures the quotient is a field, so each unital homomorphism is injective.

The cubic has no rational integer root among the divisors of four, so is irreducible. Its polynomial discriminant is $4-27\cdot 16=-428=-4\cdot 107$. Thus the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ is one or two, since its square divides the discriminant.

Set $\beta =({\alpha }^2+\alpha )/2$. Using ${\alpha }^3=\alpha +4$ givesand hence ${\beta }^3-{\beta }^2-3\beta -2=0$. Therefore $\beta$ is integral. The lattice spanned by $1,\alpha ,\beta$ contains $\mathbb{Z}[\alpha ]$ with index two. Changing basis divides the discriminant by four, giving $-107$, which is squarefree. An integral lattice of full rank with squarefree discriminant cannot have any further index in the full ring of integers. Henceis the required integral basis. The relations also give $\alpha ={\beta }^2-\beta -2$, so this ring can equally be generated by $\beta$.

Let $c_1,\dots ,c_d$ be the coefficients of $f$. Each is an algebraic integer. The ring $R=\mathbb{Z}[c_1,\dots ,c_d]$ is a finitely generated $\mathbb{Z}$-module: reduce powers of each generator by its monic integral equation. As $f$ is monic, $R[\alpha ]$ is generated over $R$ by $1,\alpha ,\dots ,{\alpha }^{d-1}$, so is a finite $\mathbb{Z}$-module. It is torsion-free as a subgroup of a characteristic-zero field, hence free of finite rank. Multiplication by $\alpha$ on it is represented by an integer matrix. Its monic characteristic polynomial, by Cayley-Hamilton theorem, annihilates $\alpha$. Thus $\alpha$ is an algebraic integer. This proves the required transitivity of integrality in this setting.

Every root of the monic $h$ is also a root of the monic polynomial $h^n\in {\mathcal{O}}_K[x]$, and is therefore an algebraic integer. The coefficients of $h$ are elementary symmetric polynomials in these roots, hence algebraic integers because algebraic integers form a ring. They also belong to $K$, so belong to ${\mathcal{O}}_K$. Consequently $h\in {\mathcal{O}}_K[x]$.

If the two-dimensional group representation were reducible, Maschke's theorem would split it into a direct sum of two one-dimensional group representations. In a basis respecting that decomposition, every representing matrix would be diagonal. All would therefore commute, contradicting the specified pair. Hence the group representation is irreducible. Complete reducibility over $\mathbb{C}$ is essential to this argument.

Count edges from the $k$ neighbours of a fixed vertex to its $n-k-1$ nonneighbours. Each neighbour contributes $k-a-1$ such edges, and each nonneighbour receives $b$. With $a=0,b=3$,The graph is incomplete, so there are nonadjacent vertices and $k\ge 3$. Substituting into the eigenvalue multiplicity formula gives a multiplicity difference of $k^2/\sqrt{4k-3}$. This must be an integer. As $k>0$, $d=\sqrt{4k-3}$ is rational and hence an integer; it is positive and odd. Moreover $d\mid k^2$ and $4k=d^2+3$, so reducing $16k^2$ modulo $d$ gives $d\mid 9$. Thus $d\in \{1,3,9\}$. The value $d=1$ gives $k=1$, impossible here; the other two give $(k,n)=(3,6)$ and $(21,162)$. Consequently $|G|\in \{6,162\}$. This is a necessity argument, not a claim that integrality alone constructs either graph.

Let $A$ be the adjacency matrix of a graph, $J$ the all-ones matrix and $1$ the all-ones vector. Counting two-step walks givesBecause $b\ge 1$, any two nonadjacent vertices have a common neighbour, so the graph is connected. The eigenvalue $k$ is simple: for a real eigenvector at $k$, a maximal entry and the averaging equation force its neighbours, and then every vertex, to have the same value. Since $A$ is real symmetric, its other eigenvectors lie in $1^{\perp }$, where $J$ vanishes. Their eigenvalues are thereforeHere $d>0$ for the present incomplete connected graph. Indeed $k\ge b$; if $d=0$, then $k=b=a$, which conflicts with $a\le k-1$ for an edge. If $r,s$ have multiplicities $m_r,m_s$, dimension and zero trace give $m_r+m_s=n-1$ and $k+m_{r}r+m_{s}s=0$. ThusThese are dimensions of eigenspaces, hence integers. A multiplicity may be zero; that does not affect the conclusion.

A strongly regular graph with parameters $(k,a,b)$ is a simple graph in which each vertex has degree $k$, adjacent vertices have exactly $a$ common neighbours, and distinct nonadjacent vertices have exactly $b$ common neighbours. Its order $n$ is often included separately as a fourth parameter.

Let $j:B\to \mathbb{N}$ be an injection witnessing that $B$ is a countable set. The composition $j\circ f:A\to \mathbb{N}$ is an injection, so $A$ is a countable set. This proof includes finite and empty sets.

Write both nonnegative integers $n,k$ in base-p expansions, padding with zero digits to a common length $L$. By the preceding polynomial identity,Expanding the last product with the binomial theorem, each term is obtained by choosing an integer $j_i$ with $0\le j_i\le n_i<p$. Its exponent is ${\sum }_i{j}_i{p}^i$ and its coefficient is ${\prod }_i\left(\genfrac{}{}{0ex}{}{n_i}{j_i}\right)$. Uniqueness of base-p expansions says that the only way this exponent can equal $k$ is to choose $j_i=k_i$ for every $i$.

If some $k_i>n_i$, no such term exists and the coefficient is zero. Otherwise its coefficient is precisely the product below. Comparing the coefficient of $x^k$ proves Lucas theorem:with the convention $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=0$ for $b>a$. This also covers $k>n$, and padding either expansion by further zero digits multiplies the product only by $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$.

For $1\le j\le p-1$, the identity for binomial coefficientsshows that $p$ divides $j\left(\genfrac{}{}{0ex}{}{p}{j}\right)$. Since $p$ is a prime number and $j$ is not divisible by $p$, cancellation modulo $p$ givesThe binomial theorem consequently gives $(1+x{)}^p\equiv 1+x^p$ as a coefficientwise polynomial modular congruence. Apply the same identity with $x$ replaced by $x^{p^r}$ and use mathematical induction to obtainThis is an iteration of the Frobenius endomorphism in characteristic $p$. Comparing the intermediate coefficients gives the prime-power binomial divisibility result