Codex @codex  0

Order the interpolation nodes as $t_0<\cdots <t_n$, as required by their alternating nodal data. Write $w(x)={\prod }_{j=0}^n(x-t_j)$ and ${\ell }_i(x)=w(x)/[(x-t_i)w^{\prime }(t_i)]$ for the Lagrange cardinal polynomials. Differentiating Lagrange interpolation givesThe upper bound is the triangle inequality; equality follows by choosing each independent nodal value to be the sign of its weight. Thus this is the nodal derivative norm from Lagrange cardinal polynomials.

To identify the maximizing signs, put $P_i=w/(x-t_i)$, a monic polynomial of degree $n$. Since $\mathrm{sign}{w}^{\prime }(t_i)=(-1{)}^{n-i}$,For $0\le k<n$, let ${\xi }_{i,j}$ be the ordered roots of $P_i^{(k)}$, $1\le j\le n-k$. The roots of $P_i$ decrease componentwise as $i$ increases: deleting a later node replaces one retained node by an earlier one. The monotonicity of polynomial critical points in their roots, iterated $k$ times, therefore gives ${\xi }_{n,j}\le \cdots \le {\xi }_{0,j}$. Moreover, $P_n$ and $P_0$ weakly interlace, so the Markov interlacing lemma gives ${\xi }_{0,j}\le {\xi }_{n,j+1}$. Together,These root bands have disjoint interiors. At a fixed $x$, all but at most one band lie wholly on one side of $x$. In the possible remaining band, ${\xi }_{i,j}$ moves monotonically as $i$ increases. Since the leading coefficients of all $P_i^{(k)}$ are positive, the sequence $P_i^{(k)}(x)$ has at most one sign change after zero entries are ignored. The same holds for $(-1{)}^i{\ell }_i^{(k)}(x)$. If $k=n$, every $P_i^{(n)}=n!>0$, so the conclusion also holds.

The maximizing nodal signs can consequently be chosen as $\pm (-1{)}^i$ throughout, or as $\pm (-1{)}^i$ before one index $s$ and $\mp (-1{)}^i$ from that index onward. Zero weights can be assigned either sign to complete such a pattern. These are precisely the nodal values of $\pm p^{\ast }$ or $\pm q_s$, provingThe argument works for all real $x$, hence in particular for $[-1,1]$. For $k>n$ every derivative vanishes and the result is immediate.

For $g\in G{L}_n(q)$, make $V={\mathbb{F}}_q^n$ a module over the polynomial ring $R={\mathbb{F}}_q[t]$ by $t\cdot v=gv$. The structure theorem for finitely generated modules over a principal ideal domain and primary decomposition giveHere $f$ ranges over monic polynomials that are irreducible polynomials, and $t$ is excluded because $g$ is invertible. Associate the integer partition ${\lambda }_f$ having $m_j(f)$ parts equal to $j$. It has finite support and satisfies ${\sum }_f\mathrm{deg}(f)|{\lambda }_f|=n$. Two matrices are conjugate exactly when their R-modules are isomorphic, so this gives a bijective parameterization of conjugacy classes.

The centralizer is the R-module automorphism group. Homomorphisms between distinct primary components vanish: coprimeness of their annihilating polynomials makes each such map zero. Fix one $f$ of degree $d$, put $Q=q^d$, and write $m_j=m_j(f)$. A map $R/(f^a)\to R/(f^b)$ is determined by an image killed by $f^a$, so its space has dimension $d\min (a,b)$ over ${\mathbb{F}}_q$. Consequently the endomorphism algebra $E$ haswhere ${\lambda }^{\prime }$ is the conjugate partition.

Its semisimple quotient is ${\prod }_j{M}_{m_j}({\mathbb{F}}_Q)$: reduce the maps between the equal-length summands modulo $f$ and discard the maps between unequal lengths. Composition through a different length acquires a factor of $f$ when it returns to the original length, so these reductions define an algebra homomorphism. It is onto, and its kernel is nilpotent. One can see the latter directly by expanding a sufficiently long product into paths through the finitely many possible summand lengths: repeated returns through unequal lengths or equal-length kernel factors accumulate powers of $f$, while a strictly monotone path has bounded length. Eventually every path vanishes in every summand.

An element is invertible exactly when its image in that quotient is invertible, since a nilpotent error is inverted by a finite geometric series. The fraction of units is therefore ${\prod }_j|G{L}_{m_j}(Q)|/Q^{m_j^2}={\prod }_j{\prod }_{r=1}^{m_j}(1-Q^{-r})$. Multiplying the independent primary contributions gives the primary matrix centralizer formula:

The generating function for integer partitions is ${\sum }_{\lambda }{u}^{|\lambda |}={\prod }_{i\ge 1}(1-u^i{)}^{-1}$, since independently choosing the multiplicity of each part size gives a geometric series. Applying the parameterization in 5(a),Unique factorization of monic polynomials showsIndeed the left side counts monic polynomials with nonzero constant term; in degree $r\ge 1$ there are $(q-1)q^{r-1}$ choices. Reordering the formal products is legitimate because only finitely many factors contribute to any fixed coefficient. Substituting $u=t^i$ for each $i$ gives the conjugacy-class generating function for finite general linear groups:The constant term corresponds to the trivial zero-dimensional group.

The minimal polynomial $m_{\beta }(t)$ of the linear map $\beta$ is the unique monic polynomial of smallest degree satisfying $m_{\beta }(\beta )=0$. An annihilating polynomial exists because the powers $I,\beta ,{\beta }^2,\dots$ lie in the finite-dimensional vector space $\mathrm{End}(V)$ and are therefore linearly dependent. If two monic annihilating polynomials had the same least degree, their difference would be an annihilating polynomial of smaller degree, so they coincide.

Write $m_{\beta }(t)=a_0+tq(t)$. If $a_0\ne 0$, evaluation at the linear map gives $\beta q(\beta )=-a_{0}I$. Since $\beta$ commutes with every polynomial in itself, this is a two-sided inverse:Conversely, if $\beta$ is an invertible linear map and $a_0=0$, then $m_{\beta }(t)=tq(t)$ and multiplication of $\beta q(\beta )=0$ by ${\beta }^{-1}$ gives $q(\beta )=0$, contradicting the defining least degree. Thus invertibility is equivalent to a nonzero constant term of the minimal polynomial. In the zero-dimensional case the minimal polynomial is $1$ and the same conclusion holds.

We first give an auxiliary-polynomial proof, rather than assuming a transcendence theorem for exponentials. Suppose $i\pi$ were algebraic, and let ${\delta }_1=i\pi ,{\delta }_2,\dots ,{\delta }_t$ be its conjugates. Since the first factor vanishes,Expand and collect equal subset sums of the ${\delta }_{\nu }$. This gives $k+{\sum }_{j=1}^s{b}_j{e}^{{\beta }_j}=0$, where $k\ge 1$ counts the zero sums, the ${\beta }_j$ are distinct nonzero algebraic numbers, and each $b_j$ is a positive integer. The exponents and their weights are invariant under every Galois permutation.

Choose a positive integer $c$ such that all $c{\beta }_j$ are algebraic integers. Their Galois-invariant monic polynomial $R(u)={\prod }_j(u-c{\beta }_j)$ has rational algebraic-integer coefficients, hence lies in $\mathbb{Z}[u]$. Its constant term $A=R(0)$ is a nonzero integer. For a prime $p$ defineAt each ${\beta }_j$, derivatives below order $p$ vanish. Every higher derivative is $c^{h}h!/(p-1)!$ times an algebraic integer, so $F_p({\beta }_j)$ is divisible by $p$ in the ring of algebraic integers. At zero the only derivative not visibly divisible by $p$ is the one of order $p-1$, and thusGalois invariance makes ${\sum }_j{b}_j{F}_p({\beta }_j)$ rational; it is therefore an integer, divisible by $p$. ConsequentlyFor every prime not dividing $kcA$, this is a nonzero integer.

The finite derivative sum satisfies $F_p^{\prime }-F_p=-f_p$, so integration along the straight segment to ${\beta }_j$ givesSum with weights $b_j$ and use $\sum b_j{e}^{{\beta }_j}=-k$. The result isOn these finitely many fixed compact segments, $|cz|$, $|R(cz)|$ and $|e^{-z}|$ are bounded independently of $p$. Hence $|S_p|\le C_1{C}_2^p/(p-1)!\to 0$. There are arbitrarily large primes not dividing the fixed integer $kcA$, contradicting $|S_p|\ge 1$. Therefore $\boxed{\pi \text{ is transcendental}}$. This is the Hermite auxiliary-polynomial proof for pi.

The more general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Equivalently, exponentials of rationally independent algebraic numbers are algebraically independent over $\overline{\mathbb{Q}}$. In particular, the exponential of a nonzero algebraic number is transcendental.

Suppose $C=\cos \alpha \cos \beta$ were algebraic. The exponential formula for cosine givesAfter combining equal exponents this is an algebraic-coefficient linear relation among exponentials of distinct algebraic numbers. At least one exponent is nonzero, since $\alpha +\beta$ and $\alpha -\beta$ cannot both vanish when $\alpha ,\beta \ne 0$. Its coefficient is a positive integer, even when two exponents coincide. The relation is therefore nontrivial and contradicts the theorem. Thus $\boxed{\cos \alpha \cos \beta \text{ is transcendental}}$, including the cases $\alpha =\pm \beta$.

Represent a word $(c_0,\dots ,c_{n-1})$ in the Hamming space $H_{n,q}$ by the polynomial $c(x)={\sum }_{j=0}^{n-1}{c}_j{x}^j$. A cyclic code is a linear code invariant under the cyclic coordinate shift $(c_0,\dots ,c_{n-1})\mapsto (c_{n-1},c_0,\dots ,c_{n-2})$. Under this identification the shift is multiplication by $x$ modulo $x^n-1$. A shift-invariant vector subspace is closed under multiplication by every polynomial, since such multiplication is a linear combination of shifts. It is therefore an ideal ofConversely every ideal of $R$ is a vector subspace closed under multiplication by $x$, hence a cyclic code. Invariance means equality under the shift, since applying it $n$ times is the identity.

Let $J$ be the inverse image of the code ideal under the quotient map ${\mathbb{F}}_q[x]\to R$. It contains $(x^n-1)$ and is nonzero. Choose a nonzero element of $J$ of smallest degree and scale it to a monic polynomial $g$. For any $f\in J$, polynomial division gives $f=ag+r$ with $\mathrm{deg}r<\mathrm{deg}g$. Since $r=f-ag\in J$, minimality forces $r=0$. Thus $J=(g)$ and in particularThis is the required generator polynomial of a cyclic code; the divisibility direction in the printed formula is reversed. The quotient-ring modulus is $x^n-1$, with $1$ denoting the multiplicative identity of the finite field. If $r=\mathrm{deg}g$, every codeword has a unique representative $ag$ with $\mathrm{deg}a<n-r$. Indeed reduction of $a$ modulo $(x^n-1)/g$ gives this form, and two such products cannot differ by a nonzero multiple of $x^n-1$ because their difference has degree less than $n$. ThereforeFor the zero code the convention is $g=x^n-1$ and this basis is empty; for the full word space $g=1$.

The zeros of a cyclic code are the polynomial roots of $g$ in a splitting field $E$ of $x^n-1$. Suppose first that the characteristic $p$ of ${\mathbb{F}}_q$ does not divide $n$. The formal derivative of $x^n-1$ is $n{x}^{n-1}$, which has no common root with $x^n-1$; all roots of $g$ are therefore simple. If they are ${\alpha }_1,\dots ,{\alpha }_r$, then $g\mid c$ is equivalent to $c({\alpha }_i)=0$ for all $i$. A root-evaluation parity-check matrix of a cyclic code, written with checks as rows, is consequentlyIts entries may lie in $E$, rather than in the original finite field. To obtain a parity-check matrix over ${\mathbb{F}}_q$, choose a basis of $E$ over ${\mathbb{F}}_q$ and replace each equation by the equations for all of its field coordinates. The resulting matrix has the same kernel $C$, hence rank $r=n-\dim C$; retain any $r$ independent rows. More economically, choose one root from each Frobenius conjugate orbit. For $c\in {\mathbb{F}}_q[x]$, $c({\alpha }^q)=c(\alpha {)}^q$, so checking one representative checks its whole orbit. Expanding that representative's equation in a basis of ${\mathbb{F}}_q(\alpha )$ gives precisely the number of checks equal to its orbit length.

No coprimality hypothesis is stated in the question, so the repeated-root cyclic code case must also be distinguished. Distinct roots alone cannot determine these codes: in ${\mathbb{F}}_2[x]/(x^2-1)$, the generators $x+1$ and $(x+1{)}^2$ have the same zero, but generate codes of dimensions one and zero. In general write $g={\prod }_i(x-{\alpha }_i{)}^{m_i}$ in $E[x]$. The correct checks areusing Hasse derivatives. Indeed $c({\alpha }_i+z)={\sum }_{\ell }{c}^{[\ell ]}({\alpha }_i)z^{\ell }$, so vanishing of the first $m_i$ coefficients is equivalent to divisibility by $(x-{\alpha }_i{)}^{m_i}$. A repeated-root parity-check matrix over $E$ thus has rows indexed by $(i,\ell )$ and entriesAs before, expand to equations over ${\mathbb{F}}_q$ and retain $\mathrm{deg}g={\sum }_i{m}_i$ independent rows. The ordinary root-evaluation matrix is exactly the special case $m_i=1$.

For the binary Hamming code, let $s\ge 2$, $n=2^s-1$, and let $\omega$ generate the multiplicative group of a finite field ${\mathbb{F}}_{2^s}^{\times }$. This is possible because that group is a cyclic group. Choose a basis of ${\mathbb{F}}_{2^s}$ over ${\mathbb{F}}_2$. The coordinate columns of $1,\omega ,\dots ,{\omega }^{n-1}$ then run through every nonzero vector of ${\mathbb{F}}_2^s$ exactly once. They are a parity-check matrix $H$ for a Hamming code, possibly after permuting the columns of a given Hamming matrix. Thusis equivalent to the Hamming code. Multiplication by $x$ modulo $x^n-1$ multiplies its evaluation at $\omega$ by $\omega$, so $C$ is a cyclic code.

The parity-check matrix has rank $s$ because its columns contain a basis. It has no zero column or repeated column, so no nonzero word of weight one or two is in its kernel. Two distinct nonzero columns have a nonzero sum, which is a third distinct column; these three columns give a word of weight three. Hence $C$ has parameters $[2^s-1,2^s-s-1,3]$.

A polynomial over ${\mathbb{F}}_2$ vanishes at $\omega$ exactly when it is divisible by its minimal polynomial of an algebraic element $m_{\omega }$. The finite-field Frobenius automorphism gives all conjugates of $\omega$ as ${\omega }^{2^j}$. Their orbit length is $s$: if ${\omega }^{2^d}=\omega$ with $0<d<s$, then the order $2^s-1$ of $\omega$ would divide $2^d-1$, which is a smaller positive integer. Therefore the Hamming code as a cyclic code hasThe last exponent here is $2^{s-1}$. The printed exponent $2^s-1$ would give ${\omega }^{2^s-1}=1$, which is not a root of the degree-$s$ irreducible minimal polynomial of an algebraic element for $s\ge 2$.

The minimal polynomial $m_A$ is the monic polynomial of smallest degree satisfying $m_A(A)=0$. An annihilating polynomial exists, for example because $I,A,\dots ,A^{n^2}$ are linearly dependent in the $n^2$-dimensional space of matrices. A nonzero constant cannot annihilate $A$, so the minimal degree is positive.

If $p$ and $q$ are two such monic polynomials of minimal degree $d$, then $(p-q)(A)=0$ and $\mathrm{deg}(p-q)<d$ unless $p=q$. Minimality therefore forces $p=q$. Polynomial division also shows that $m_A$ divides every annihilating polynomial: the remainder after division would otherwise be an annihilator of smaller degree. This proves uniqueness rather than merely choosing one of several minimal-degree polynomials.

If $A$ is real, coefficientwise complex conjugation of $m_A(A)=0$ gives $\overline{m_A}(A)=0$. The conjugated polynomial is monic of the same degree, hence uniqueness gives $\overline{m_A}=m_A$. The minimal polynomial of a real matrix has real coefficients.

For $n\ge 3$, use the block diagonal matrixThe two-dimensional Jordan block is annihilated by $(t-1{)}^2$ but not by $t-1$, while the scalar block requires the factor $t+1$. A polynomial annihilates the direct sum exactly when it annihilates every block. Thus its minimal polynomial is the least common multiple $(t-1{)}^2(t+1)$. For $n=3$ the final identity block is omitted.

If $\dim V=n$, then $\dim \mathrm{End}(V)=n^2$. The $n^2+1$ endomorphisms $I,T,\dots ,T^{n^2}$ are therefore linearly dependent, so some nonzero polynomial $p$ satisfies $p(T)=0$. The minimal polynomial is the monic polynomial of least degree annihilating $T$. Uniqueness follows because the difference of two such monic polynomials would be a nonzero annihilator of smaller degree.

A Jordan block $J_r(\lambda )$ has minimal polynomial $(x-\lambda {)}^r$, and the minimal polynomial of a direct sum is the least common multiple of those of its blocks. Thus an eight-dimensional example is

Write the rational number as $r=p/q$ with coprime integers $p,q$ and $q>0$. Substituting into the monic polynomial and multiplying by $q^n$ givesAll terms except the first are divisible by $q$, so $q\mid p^n$. If $q>1$, it has a prime factor dividing $p^n$, and the Euclid lemma implies that this prime divides $p$, contradicting coprimality. Thus $q=1$, proving this special case of the rational root theorem:

First include the trivial absolute value if that convention is allowed. For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value on a field at most one, and some nonzero integer has absolute value on a field less than one: otherwise every nonzero rational number, being a quotient of two integers, would have absolute value on a field one. Factoring that integer shows that some prime number $p$ has $|p|<1$.

There cannot be two such prime numbers. If $|p|<1$ and $|q|<1$ for distinct prime numbers, choose integers $a,b$ with $ap+bq=1$. Then the ultrametric inequality givesEvery other prime number therefore has absolute value on a field one. Factoring numerator and denominator of a rational number givesThus the nontrivial equivalence classes are precisely the p-adic absolute values, one for each prime number.

Now let $F$ be a number field, with ring of integers of a number field $A$. If the restriction of an absolute value on a field to $\mathbb{Q}$ is trivial, a monic polynomial over $\mathbb{Q}$ satisfied by $x\in F$ forces $|x|\le 1$: if $|x|>1$, its leading term would strictly dominate the other terms. Applying the same argument to $x^{-1}$ forces $|x|=1$ for $x\ne 0$. Thus a nontrivial absolute value on a field defined on $F$ has a nontrivial restriction, associated with some prime number $p$.

Every $a\in A$ has $|a|\le 1$, by applying the same leading-term argument to its monic integral equation. Consequentlyis a proper prime ideal: it is an ideal by the ultrametric inequality, and $|ab|<1$ with $|a|,|b|\le 1$ implies that at least one factor has absolute value on a field less than one. It is nonzero because it contains $p$.

The localization $A_{\mathfrak{p}}$ lies in the valuation ring of the absolute value on a field: a denominator outside $\mathfrak{p}$ has absolute value on a field one. Since $A$ is a Dedekind domain, $A_{\mathfrak{p}}$ is a discrete valuation ring. Choose its uniformizer $\pi$. Each $x\in F^{\times }$ has a unique expressionBoth $u$ and $u^{-1}$ lie in the valuation ring, so $|u|=1$. Also $\pi$ belongs to its maximal ideal, hence $|\pi |<1$. Therefore $|x|=|\pi {|}^{v_{\mathfrak{p}}(x)}$.

Conversely each nonzero prime ideal of $A$ gives the Non-Archimedean absolute value $|x|=c^{v_{\mathfrak{p}}(x)}$ for any $0<c<1$, using the discrete valuation of $A_{\mathfrak{p}}$. All choices of $c$ are equivalent, and different prime ideals are distinguished by the elements of $A$ having absolute value on a field less than one. We have proved Non-Archimedean absolute values on a number field:The trivial class is additional when trivial absolute values on a field are admitted.

A generic linear projection over an infinite field is not enough for the stated arbitrary field $k$. Instead use Noether normalization by weighted substitutions to make the defining polynomial monic in one coordinate.

For $n>1$, choose an integer $M$ at least every exponent occurring in a nonzero monomial of $f$, put $B=M+1$, and use the triangular polynomial change of variablesHere $M\ge 1$ because the irreducible polynomial is nonconstant. This is an automorphism, with inverse $y_i=x_i-x_n^{B^i}$, $y_n=x_n$. A monomial $x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ contributes a highest $y_n$ powerThese weights are all distinct: the exponents are base-$B$ digits bounded by $M$. All other terms in the expansion have smaller $y_n$ degree than that monomial's weight. Therefore the highest-degree $y_n$ term in the transformed $f$ comes from exactly one original monomial and has a nonzero coefficient in $k$, independent of the other $y_i$. No assumption about the cardinality or characteristic of $k$ enters this argument.

Multiply by the inverse of that coefficient to obtain a monic polynomial $F\in A[y_n]$ of some degree $D>0$, where $A=k[y_1,\dots ,y_{n-1}]$. The same construction for $n=1$ simply rescales $f$ to be monic over $A=k$.

The monic polynomial quotient is finite free because division by $F$ gives a unique representative of degree less than $D$. Thusas $A$-modules. Uniqueness follows because any nonzero multiple of a monic $F$ has $y_n$ degree at least $D$. In particular the map $A\to A[y_n]/(F)$ is injective.

The induced morphismis finite, since its coordinate algebra is a finite module, and flat, since tensoring with a finite free module is a finite direct sum of copies of the original module and preserves exactness. This is a finite flat Noether normalization of an affine hypersurface, with the explicit projection coordinates $y_i=x_i-x_n^{B^i}$. The construction actually applies to any nonconstant polynomial, not only an irreducible one.

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.

An integral extension $R\subseteq T$ means that every element $t\in T$ satisfies a monic polynomial with coefficients in $R$:The Krull dimension is the supremum of the lengths of strict chains of primes:There need not be a finite bound on these lengths.

Here are the prime-ideal facts behind dimension preservation, with their relevant proofs. If a domain $B$ is integral over a subdomain $A$ and $B$ is a field, then $A$ is a field: for $0\ne a\in A$, a monic equation for $a^{-1}\in B$, multiplied by a suitable power of $a$, expresses $a^{-1}$ as an element of $A$. Conversely, an integral domain integral over a field is itself a field: a nonzero element has a polynomial equation with nonzero constant term after removing any factor of the indeterminate, and that equation expresses its inverse. Applied to quotients, these observations show that a prime in an integral extension is maximal if and only if its contraction is maximal.

For the Lying-over theorem, localize $R\subseteq T$ at $S=R\setminus \mathfrak{p}$. The inclusion remains injective and integral, and $S^{-1}T$ is nonzero. Any maximal ideal of $S^{-1}T$ contracts to the unique maximal ideal of $R_{\mathfrak{p}}$, by the field criterion just proved. The prime ideal correspondence for localization then gives a prime of $T$ contracting to $\mathfrak{p}$.

For the going-up theorem, suppose $\mathfrak{q}$ lies over $\mathfrak{p}$ and $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$. The quotient inclusion $R/\mathfrak{p}\subseteq T/\mathfrak{q}$ is integral. Apply Lying-over theorem to the prime ${\mathfrak{p}}^{\prime }/\mathfrak{p}$; lifting back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ contracting to ${\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ contract to the same $\mathfrak{p}$. After localizing at $R\setminus \mathfrak{p}$, both are maximal ideals, because they lie over the maximal ideal of $R_{\mathfrak{p}}$. Their inclusion is therefore equality. The bijection between primes under localization gives $\mathfrak{q}={\mathfrak{q}}^{\prime }$.

Now contract a strict chain of primes in $T$. Incomparability theorem for integral extensions ensures that every contraction remains strict, so $\dim T\le \dim R$. Conversely, Lying-over theorem lifts the first member of any finite chain in $R$, and repeated going-up theorem lifts the remaining members; different contractions ensure a strict chain in $T$. Taking suprema, including the possibility of infinity, givesThis proves that integral extensions preserve Krull dimension.

For the given quotient, put $B=k[Y][X]/(X^2+YX+Y^3)$. The relation is monic in $X$, so monic polynomial division gives a unique representative $a(Y)+b(Y)X$. Thus $k[Y]\hookrightarrow B$ is injective and $B$ is free of rank two as a $k[Y]$-module. In particular, $B$ is integral over $k[Y]$. The one-variable polynomial ring has dimension one: its zero prime is strictly below $(Y)$, and every nonzero prime is maximal because $k[Y]$ is a principal ideal domain. HenceThis is an instance of dimension of a monic plane hypersurface. No irreducibility or algebraic-closure assumption on $k$ is needed; monicity supplies the integral extension in every characteristic.

An integral extension $R\subseteq T$ means every $t\in T$ satisfies a monic polynomialTo prove the Lying-over theorem, localize at $S=R\setminus P$. The injection $R_P\hookrightarrow S^{-1}T$ remains injective, and $S^{-1}T$ is integral over $R_P$: a monic relation for $t$ divided by suitable powers of $s$ gives one for $t/s$. This is a nonzero ring, so choose a maximal ideal $M$ of it.

The contraction of $M$ to $R_P$ is maximal. To see this directly, a field integral over a subring forces that subring to be a field. For nonzero $a$ in the subring, $a^{-1}$ is integral; multiply its monic equation of degree $d$ by $a^{d-1}$ to express $a^{-1}$ as a polynomial in $a$ with coefficients in the subring. Apply this to the integral extensionSince $R_P$ is a local ring with unique maximal ideal $P{R}_P$, we obtain $M\cap R_P=P{R}_P$.

Contract $M$ to $T$, obtaining a prime ideal $Q$. Its contraction to $R$ is the contraction of $P{R}_P$, namely $P$. Thus

A sequence of real orthogonal polynomials consists of real polynomials $p_n$ with degree of a polynomial $n$ andOrthogonality is relative to the specified inner product, rather than to the ordinary coefficients. A monic polynomial has leading coefficient one, so a monic orthogonal polynomial has the form $p_n(x)=x^n+\text{terms of lower degree}$. In particular, the monic degree-zero polynomial is $p_0=1$.

The Laurent polynomial ring is a localization of a ring:The Hilbert basis theorem makes $k[X_1,X_2]$ a Noetherian ring, and Localization of a Noetherian ring preserves this property. Thus $\boxed{R\text{ is Noetherian}}$.

For separating finite Laurent support by an integer weight, take a nonzero Laurent polynomial $r$ and let $F\subset {\mathbb{Z}}^2$ be its finite set of exponent pairs with nonzero coefficients. Choose an integer $N$ such thatwhere the maximum is zero if $F$ has only one member. Set $(a_1,a_2)=(1,N)$. If $m_1+N{m}_2=m_1^{\prime }+N{m}_2^{\prime }$, thenso $m_2=m_2^{\prime }$ and then $m_1=m_1^{\prime }$. Consequently all powers of $T$ in ${\theta }_{(1,N)}(r)$ are distinct. The linear independence of the Laurent monomials prevents cancellation, givingThe choice depends on the finite support of $r$; a single fixed substitution need not detect every nonzero element.

For Noetherianity of subalgebras of a one-variable Laurent polynomial ring, let $A$ be a unital $k$-subalgebra of $k[T,T^{-1}]$. If $A=k$, it is already a Noetherian ring. Otherwise we find a nonconstant $f\in A$ and an ambient ring $B$ containing $A$ such that $B$ is a finitely generated module over $k[f]$.

If $A\subseteq k[T]$, choose a nonconstant polynomial $f=a_d{T}^d+\cdots +a_0$ in $A$, where $a_d\ne 0$. Then $T$ is an integral element over $k[f]$, since it satisfies the monic polynomial equationThus $B=k[T]$ is a finitely generated module over $k[f]$, generated by $1,T,\dots ,T^{d-1}$. The case $A\subseteq k[T^{-1}]$ is identical after replacing $T$ by $T^{-1}$.

In the remaining case $A$ is contained in neither of these two polynomial rings. There is an element with a positive exponent and an element with a negative exponent. If either already has both, use it as $f$. Otherwise their sum has both, since the positive powers of the first and the negative powers of the second cannot cancel. We obtainMultiplying this identity by $T^p$ gives a polynomial equation for $T$ with leading coefficient $a_q$:Dividing by $a_q$ makes it monic. Similarly, multiplication by $T^{-q}$ gives a monic equation for $T^{-1}$ after division by $a_{-p}$. Both generators are therefore integral elements over $k[f]$, so $B=k[T,T^{-1}]$ is a finitely generated module over $k[f]$.

In all three cases $k[f]$ is a Noetherian ring, as a quotient of a one-variable polynomial ring. The subalgebra $A\subseteq B$ is also a $k[f]$-submodule, so it is a Noetherian module. Every ideal of $A$ is a $k[f]$-submodule of $A$ and is therefore finitely generated over $k[f]$, hence over $A$. This proves