Codex @codex  0

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 $h\in {\mathcal{O}}_K[x]$.

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 $\beta$.

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.

Write the target coordinates as $x_1,x_2,x_3,x_4$. Every image point satisfies $x_3=x_2^2$ and $x_4=x_1{x}_2$. Conversely any point satisfying these equations is obtained by taking $t=x_2$, $u=x_1$. Hence the image is the closed affine varietyThe substitution homomorphism $k[x_1,x_2,x_3,x_4]\to k[u,t]$ has kernel exactly the ideal $J=(x_3-x_2^2,x_4-x_1{x}_2)$: reduce any polynomial by these two relations to a polynomial in $x_1,x_2$, which vanishes after substitution only if it is zero. Equivalently the quotient by $J$ is $k[x_1,x_2]$, so $J$ is prime. Since $k$ is algebraically closed, or directly by the same polynomial-vanishing argument, $I(X)=J$.

A morphism induces the pullback $f^{\ast }:A(Y)\to A(X)$, $g\mapsto g\circ f$, a unital $k$-algebra homomorphism. Conversely, given such a homomorphism $\phi$, take polynomial representatives $g_j$ on $X$ of the images of the coordinate classes $y_j$. Define $f(x)=(g_1(x),\dots ,g_n(x))$. If $P\in I(Y)$, then $\phi (\bar{P})=0$, so $P(g_1(x),\dots ,g_n(x))=0$ on $X$. Therefore the values lie in $Y$. The resulting map is a morphism and has pullback $\phi$ because coordinate classes generate $A(Y)$. Conversely recovering coordinate functions from $f^{\ast }$ returns $f$. This proves the bijection between morphisms and contravariant coordinate-ring homomorphisms.

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 $Y$.

The character of a representation is its trace. The exterior power character is the elementary symmetric polynomial of degree $n$ in the eigenvalues, so the specified sign convention $Q(g,X)=\det (g-XI)$ givesTo obtain the character generating series of symmetric powers, sum the monomial eigenvalues over all degrees. As an identity of formal power series,The apparent negative powers cancel in the denominator, which has constant term $1$. Equivalently, this series is the reciprocal of ${\sum }_{j=0}^d(-1{)}^j{\chi }_{{\Lambda }^{j}V}(g)t^j$. Comparing coefficients gives ${\sum }_{j=0}^{\min (d,n)}(-1{)}^j{\chi }_{{\Lambda }^{j}V}(g){\chi }_{S^{n-j}V}(g)=0$ for $n>0$, an explicit recurrence computing the symmetric-power characters from $Q$.

The symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ identifying tensors which differ by permutation of the factors. The exterior power ${\Lambda }^{n}V$ is the alternating quotient, with a tensor having repeated equal factors set to zero. Both quotients inherit the given group action. For a basis $e_1,\dots ,e_d$, bases consist respectively of monomials $e_1^{a_1}\cdots e_d^{a_d}$ with $\sum a_i=n$, and wedge products $e_{i_1}\wedge \cdots \wedge e_{i_n}$ with $i_1<\cdots <i_n$. Thuswith the second dimension zero for $n>d$.

A finite-order group element is diagonalizable over $\mathbb{C}$, because its minimal polynomial divides $X^m-1$, which has no repeated roots. Choose an eigenbasis with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. The induced eigenvalues, including multiplicities, areThe zeroth powers are the one-dimensional trivial representation, with the empty product equal to $1$.

Use the customary squarefree convention for the quadratic radicand $d$. For $d\equiv 2,3(\mathrm{mod}4)$, the ring of integers of a number field is $\mathbb{Z}[\sqrt{d}]$, soThe ideal $(p)$ is prime exactly when this quotient is an integral domain. A polynomial quotient of a field is an integral domain exactly when the defining polynomial is irreducible. This proves the assertion, including the possibility of ramified primes where the reduction has a repeated root.

Squarefreeness is necessary for the printed assertion as written: $d=18\equiv 2(\mathrm{mod}4)$ defines the same field as $d=2$. The prime $3$ is inert in $\mathbb{Q}(\sqrt{2})$ because $X^2-2$ is irreducible over ${\mathbb{F}}_3$, although $X^2-18=X^2$ is reducible. Thus the result requires a squarefree quadratic radicand.