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Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.

It is enough to assume that $h_k:{\pi }_k(X)\otimes \mathbb{Q}\to H_k(X;\mathbb{Q})$ is surjective for every $k\ge 2$; an integrally surjective Hurewicz homomorphism certainly has this property. First give the minimal-model argument suggested by the hint. Let $A=(\Lambda V,d)$ be a simply connected Sullivan minimal model, with $V$ in degrees at least two. Put $A^{+}={⨁}_{k>0}{A}^k$. Its indecomposable quotient of an augmented algebra isMinimality makes the induced differential on $V$ zero, so projection is a chain map $q:A^{+}\to (V,0)$. The Sullivan model identification says that the induced mapis dual to rational Hurewicz homomorphism. Under the usual finite-type duality convention, the assumed surjectivity makes $q_{\ast }$ injective.

A product of two positive cohomology classes has a decomposable cocycle representative, so its image under $q_{\ast }$ is zero. Injectivity therefore givesThis deduction uses the hypothesis; zero cup products by themselves would not yet prove formality of a topological space.

Choose a homogeneous cohomology basis $h_{\lambda }$ and cocycle representatives $z_{\lambda }\in A$. Their indecomposable parts $q(z_{\lambda })$ are linearly independent. Extend them to a graded basis of $V$ and replace the corresponding generators by the actual cocycles $z_{\lambda }$. This is a valid change of free algebra generators: each $z_{\lambda }$ differs from its indecomposable part by products of generators of strictly smaller degree, so degree induction constructs the inverse substitution. Call these closed generators $C$, and the remaining generators $N$.

Define a map of graded algebrasIts target has zero differential. Every differential in $A$ is decomposable, and all products of positive target classes vanish, so $\varphi$ is a chain map. If $z$ is any positive cocycle, its indecomposable part is in the span of the $q(z_{\lambda })$, because these form the image of $q_{\ast }$. Write $q(z)={\sum }_{\lambda }{c}_{\lambda }q(z_{\lambda })$. Then $\varphi (z)={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$, while injectivity of $q_{\ast }$ gives $[z]={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$. Thus $\varphi$ induces the identity on cohomology and is a quasi-isomorphism. This proves that $X$ is a formal space.

The rational wedge sum assertion has a direct topological proof, which also avoids finite-dimensional duality restrictions in the preceding model argument. Choose a homogeneous basis $b_{\lambda }$ of ${\tilde{H}}_{\ast }(X;\mathbb{Q})$. By surjectivity, each basis class is the image under the Hurewicz homomorphism of an element of ${\pi }_{n_{\lambda }}(X)\otimes \mathbb{Q}$. In $X_{\mathbb{Q}}$ this element is represented by an actual sphere map $S^{n_{\lambda }}\to X_{\mathbb{Q}}$. Simple connectivity and $H_1=0$ ensure $n_{\lambda }\ge 2$. The wedge sum of these maps givesand induces an isomorphism of rational homology by the choice of basis. The rational Whitehead theorem now makes it a rational homotopy equivalence, givingA wedge sum of spheres is a formal space. For each individual sphere, its augmented minimal model maps by a quasi-isomorphism to its sphere cochain algebra, and also to its cohomology algebra, by killing the extra odd generator in the even-sphere case. Take based simplicial models of the spheres. Their wedge sum is a simplicial pushout over the common basepoint, so the algebra of rational polynomial differential forms is the corresponding fibre product of augmented sphere algebras. Taking fibre products of these model maps gives a quasi-isomorphism zigzag to the square-zero cohomology algebra. This also works for infinitely many sphere summands: the positive parts are products of the individual complexes, and products of vector-space exact sequences over a field are exact. In the finite case its minimal resolution is the minimal model of a wedge of simply connected spheres. Thus the topological proof establishes both conclusions without imposing finite rank on the sphere basis. For the zero reduced-homology case the wedge sum is a point. Both the minimal-model construction and the explicit sphere map explain why the surjectivity hypothesis is decisive.

Work over $\mathbb{C}$ with finite-dimensional Lie algebras and Lie algebra representation spaces. The adjoint form of Engel theorem isHere a nilpotent Lie algebra has terminating lower central series, ${\gamma }_1\mathfrak{g}=\mathfrak{g}$, ${\gamma }_{j+1}\mathfrak{g}=[\mathfrak{g},{\gamma }_j\mathfrak{g}]$. We prove the stronger linear statement: if $V\ne 0$, $L\subseteq \mathfrak{gl}(V)$ and every member of $L$ is a nilpotent endomorphism, then there is a nonzero $v\in V$ killed by all of $L$, and $L$ can be simultaneously represented by strictly upper triangular matrices.

For the proof of Engel theorem by induction and normalizers, first establish a useful nilpotence fact. If $x^m=0$, left and right multiplication by $x$ on $\mathrm{End}(V)$ commute. Expanding their difference givesFor $N=2m-1$, every summand contains a power of $x$ at least $m$, so $(\mathrm{ad}x{)}^{2m-1}=0$. Restrictions and induced quotient maps remain nilpotent.

Now prove the common-kernel statement, the Engel lemma, by induction on $\dim L$. The zero algebra is immediate. Choose a maximal proper Lie subalgebra $M$. Its action on $L/M$ by commutators consists of nilpotent maps by the preceding fact. Applying the lower-dimensional induction hypothesis to its image gives a nonzero coset $y+M$ with $[M,y]\subseteq M$. HenceThe normalizer of a Lie subalgebra is a Lie subalgebra by the Jacobi identity. Maximality implies $N_L(M)=L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has no proper nonzero Lie subalgebra, and every one-dimensional subspace is a Lie subalgebra. Thus $\dim L/M=1$.

Induction also provides a nonzero common annihilatorIt is stable under $L$: for $m\in M$, $x\in L$, $v\in W$,because $[m,x]\in M$. Choose $x$ representing a basis of $L/M$. The restriction of the nilpotent endomorphism $x$ to the nonzero space $W$ has nonzero kernel. A vector in this kernel is killed by both $M$ and $x$, hence by all of $L$. This finishes the induction. When an action is not faithful, induction is applied to its image, whose dimension is no larger than $\dim M$; no faithfulness assumption is hidden.

Starting with the common-kernel line, apply the same statement to each successive quotient Lie algebra representation space. It constructs a complete flag of invariant subspacesAn adapted basis represents every member of $L$ by a strictly upper triangular matrix. Every product of $d$ such matrices is zero, so every iterated commutator with $d$ entries is zero as well.

Finally, if every $\mathrm{ad}x$ on $\mathfrak{g}$ is nilpotent, apply the linear statement to the Adjoint representation. With $d=\dim \mathfrak{g}$, every product of $d$ adjoint operators vanishes, so ${\gamma }_{d+1}\mathfrak{g}=0$. Conversely, if ${\gamma }_{c+1}\mathfrak{g}=0$, then $(\mathrm{ad}x{)}^{c}y=0$ for all $x,y$, proving the other implication. This proves Engel's theorem and its simultaneous-triangularization form. The hypothesis concerns every element of the algebra, not merely a chosen set of generators.

Work with a finite-dimensional complex semisimple Lie algebra. The Dynkin diagram is obtained from its intrinsic root system, with several structure theorems entering the construction.

First choose a Cartan subalgebra $\mathfrak{h}$. The semisimple structure theory gives the root-space decompositionThe nonzero roots span ${\mathfrak{h}}^{\ast }$, root spaces have dimension one, and every root has its negative. The rank is $r=\dim \mathfrak{h}$.

Next use the Killing form. Its nondegeneracy on $\mathfrak{h}$ and its positive-definite restriction to the real coroot span identify the real root span $E=\mathbb{R}\Phi$ with a Euclidean space. Explicitly, if $t_{\lambda }={\mathcal{B}}^{-1}(\lambda )$, put $(\lambda ,\mu )=B(t_{\lambda },t_{\mu })$. The sl2 subalgebra associated with a root supplies its normalized coroot $H_{\alpha }$ and the integral pairingThe root-string theorem makes the reflection $s_{\alpha }(\beta )=\beta -\beta (H_{\alpha })\alpha$ permute $\Phi$. Together with the absence of multiples other than $\pm \alpha$, this shows that $\Phi$ is a finite reduced root system and a crystallographic root system. These properties come from semisimple Lie algebra theory, not from the diagram's definition.

Choose a vector in $E$ that is not perpendicular to any root. Define the positive roots by positive inner product with this vector. The simple roots are those positive roots that cannot be expressed as sums of two positive roots. Root-system theory shows that they form a basis $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, and each root is an integer combination of them with coefficients all of one sign. Distinct simple roots have nonpositive inner product.

Use the standard Cartan matrix conventionThere is one diagram vertex for each simple root. For $i\ne j$, the integral off-diagonal entries are nonpositive andConnect the two vertices with respectively zero, one, two or three edges. The corresponding angles are $90^{\circ },120^{\circ },135^{\circ },150^{\circ }$. A multiple edge carries an arrow toward the shorter root. For joined vertices, the length ratio is determined by $A_{ij}/A_{ji}=({\alpha }_i,{\alpha }_i)/({\alpha }_j,{\alpha }_j)$; hence the diagram records both angle and relative length.

For example, ${\mathfrak{sl}}_3$ has simple roots $L_1-L_2$ and $L_2-L_3$, giving two equal-length vertices joined by one edge: type $A_2$. For ${\mathfrak{sp}}_4$, take ${\alpha }_1=L_1-L_2$ and ${\alpha }_2=2L_2$. Thenso the two vertices have a double edge directed toward ${\alpha }_1$, the shorter root: type $C_2$.

Finally, orthogonal irreducible components of the root system correspond to the simple ideals of a Lie algebra in the semisimple Lie algebra, so its diagram is the disjoint union of their connected diagrams. Conjugacy of Cartan subalgebras and the Weyl group action on choices of positive roots make the resulting diagram independent of these choices up to isomorphism. Thus the vertices, edge multiplicities, arrows and connected components encode the simple-root geometry and the simple-ideal decomposition.

Put $X=1+T$ and $Y=X^p-1$, so that $\varphi (f)=f(Y)$. The essential integrality statement for the Frobenius substitution on cyclotomic power series is the decomposition into a finite free module:To prove it, reduce modulo $p$. Since $Y\equiv T^p$, grouping the coefficients of a formal power series according to their exponents modulo $p$ gives the unique expression ${\sum }_{j=0}^{p-1}{T}^j{h}_j(T^p)$. The change from $1,T,\dots ,T^{p-1}$ to $1,X,\dots ,X^{p-1}$ is a triangular matrix with diagonal entries $1$, so this is also a basis over ${\mathbb{F}}_p[[T^p]]$. Lift these $p$ coefficient series to $R$, subtract their contribution from $f$, divide the remaining error by $p$, and repeat. The sums of the successive lifts converge in the finer of the topologies on integral formal power series, because $R$ is complete in that topology. This proves existence of the decomposition over $R$. A relation among its summands reduces to the zero relation modulo $p$, forcing each coefficient series to be divisible by $p$. Repetition forces divisibility by every power of $p$, hence every coefficient series is zero. This proves uniqueness and, in particular, injectivity of $\varphi$.

For $f={\sum }_{i=0}^{p-1}{X}^i\varphi (f_i)$, defineThis is a ${\mathbb{Z}}_p$-linear map. To calculate the average over roots of unity, work over $\mathcal{O}={\mathbb{Z}}_p[{\zeta }_p]$. The substitutions $T\mapsto \zeta X-1$ are legitimate: their constant terms are topologically nilpotent, so the defining coefficient sums converge in $\mathcal{O}$. Since $(\zeta X{)}^p=X^p$, each $\varphi (f_i)$ is unchanged by the substitution. The sums of the powers of the roots of unity are $p$ for $i=0$ and zero for $1\le i<p$. ConsequentlyThus the average is integral and belongs to $\varphi (R)$, and the constructed Coleman trace operator has the required property. Injectivity of $\varphi$ forces uniqueness of any map satisfying that property. Finally, the decomposition of $\varphi (g)$ has $f_0=g$ and all other coefficients zero, giving $S\circ \varphi ={\mathrm{id}}_R$.

We will also use the trace of multiplication by $h$ on this finite free module. After extending scalars to split the conjugates, that trace is the sum just computed. Hence

Let $M_f$ be the matrix of multiplication by $f$ in the basis from Question 1. Define the Coleman norm operator by the norm for a finite free ring extension:The entries lie in $\varphi (R)$, and $\varphi :R\to \varphi (R)$ is an isomorphism, so this determines $Nf$ uniquely. If $f$ is a unit, $M_f$ is invertible and its determinant is a unit. Also $M_{fg}=M_f{M}_g$, giving $N(fg)=NfNg$. Splitting the conjugates, or diagonalizing multiplication after adjoining the roots of unity and inverting $p$, gives the equivalent formulaIn particular, this definition has the required target $R^{\times }$, rather than merely a ring with extra roots of unity in its coefficients.

For the requested ring congruence, observe that the reduction of $\varphi$ is the injective map $\overline{g}(T)\mapsto \overline{g}(T^p)$. If $\varphi (g)\in p^{k}R$, its reduction first gives $g=p{g}_1$. Then $\varphi (g_1)\in p^{k-1}R$; repeat to obtain $g\in p^{k}R$. Applying this to $g=f-1$ provesThis congruence reflection for Frobenius substitution is valid for all $f\in R$, without assuming $f$ is a unit.

A further property needed for the fixed points is Coleman norm contraction. If $g=1+p^{k}h$ with $k\ge 1$, expand the determinant:Here $e_j(M_h)\in \varphi (R)$ is the coefficient of degree $j$ in $\det (I+t{M}_h)$. The trace formula from Question 1 makes the linear term divisible by $p^{k+1}$; every later term is divisible by $p^{2k}$ and $2k\ge k+1$. Reflecting the resulting ring congruence through $\varphi$ givesIf $Nf=f$ and $f\in 1+pR$, iteration gives $f\in 1+p^{k}R$ for every $k$. Since ${\bigcap }_k{p}^{k}R=0$, $f=1$.

To identify all fixed points, we also need $Nf\equiv f(\mathrm{mod}pR)$ for every unit $f$. In characteristic $p$, the extension ${\mathbb{F}}_p[[T]]/{\mathbb{F}}_p[[T^p]]$ has degree $p$ and its norm for a finite free ring extension is $\overline{f}\mapsto {\overline{f}}^p$. Indeed, after passing to fraction fields it is a purely inseparable field extension of degree $p$, so multiplication has only the eigenvalue $\overline{f}$, with multiplicity $p$, after splitting. Its determinant is ${\overline{f}}^p$. Thereforeand congruence reflection for Frobenius substitution gives $Nf\equiv f(\mathrm{mod}pR)$.

Now choose any lift $f_0\in R^{\times }$ of a given $a\in A^{\times }$ and put $f_n=N^n{f}_0$. Multiplicativity givesThus $(f_n)$ is a Cauchy sequence for the finer of the topologies on integral formal power series and converges to a unit $w(a)$ reducing to $a$. The Coleman norm operator is continuous: multiplication matrices and their determinants are continuous, and the finite-free coordinate decomposition and its inverse preserve the $p$-adic filtration defining the finer of the topologies on integral formal power series. Hence $Nw(a)=w(a)$. If two fixed units reduce to $a$, their quotient is fixed and belongs to $1+pR$, so it is $1$ by the preceding argument. This proves uniqueness, independence of the initial lift, and multiplicativity of the Coleman norm-fixed lift. In particular,Every unit $f\in R$ consequently has a unique factorization $f=w(\overline{f})v$ with $v\in 1+pR$, giving the canonical Coleman decomposition of power-series units:The final assertion as printed is false. The original PDF identifies a ring with a product of unit groups. With multiplication, $A$ has a zero and is not a group. With addition, it has exponent $p$, whereas $W\times A^{\times }$ has a nonidentity element of order $2$: $-1\in W$ because $p$ is odd and $N(-1)=(-1{)}^p=-1$. Thus neither interpretation makes the printed assertion true. The two boxed isomorphisms above are the valid natural conclusions; they do not require guessing which symbols the author intended to replace.

For a profinite abelian group $G$, its Iwasawa algebra is the completed group algebrawhere $U$ runs over open subgroups. The transition map sums the coefficients over the fibers of $G/V\to G/U$ when $V\subset U$. Consequently an element gives compatible values $\mu (gU)\in {\mathbb{Z}}_p$ on cosets, equivalently a finitely additive ${\mathbb{Z}}_p$-valued p-adic measure on the clopen sets of $G$. Conversely these values specify every finite-quotient coefficient. Integration of a continuous function with values in ${\mathbb{Z}}_p$ is obtained by uniform approximation by locally constant functions; the integrals converge because all measure values have $p$-adic absolute value at most $1$. Multiplication in the Iwasawa algebra becomes convolution of p-adic measures. This is its measure realization of an Iwasawa algebra.

The Mahler theorem states that each continuous function $g:{\mathbb{Z}}_p\to {\mathbb{Z}}_p$ has the unique expansion converging in the sense of uniform convergenceIts Mahler coefficients areand $\Vert g{\Vert }_{\infty }={\sup }_n|b_n{|}_p$. The same assertion holds for continuous functions with values in ${\mathbb{Q}}_p$ with coefficients in ${\mathbb{Q}}_p$.

Given $f(T)={\sum }_{n\ge 0}{a}_n{T}^n\in R$, define the bounded linear functionalThe sum converges because $a_n\in {\mathbb{Z}}_p$ and $b_n\to 0$. Its values on indicator functions of clopen sets give a p-adic measure, denoted $\lambda (f)$, withConversely, a p-adic measure determines these coefficients $a_n\in {\mathbb{Z}}_p$, and continuity allows termwise integration of every Mahler expansion. Thus the inverse iswhich is the Amice transform. These two constructions are inverse, proving the canonical bijection $\lambda :R\simeq \Lambda ({\mathbb{Z}}_p)$. In fact it is an isomorphism of rings: the identity $\left(\genfrac{}{}{0ex}{}{x+y}{n}\right)={\sum }_{i+j=n}\left(\genfrac{}{}{0ex}{}{x}{i}\right)\left(\genfrac{}{}{0ex}{}{y}{j}\right)$ identifies convolution of p-adic measures with multiplication of formal power series. For example, $\lambda (1)={\delta }_0$ and $\lambda (1+T)={\delta }_1$, where ${\delta }_a$ is the Dirac measure at $a$.

For the moment identity, let $\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$ denote a Stirling number of the second kind. The finite polynomial identitygives $\int x^{k}d\lambda (f)={\sum }_{n=0}^k{a}_{n}n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. On the other hand, $D=(1+T)d/dT$ satisfies $D^k(1+T{)}^x=x^k(1+T{)}^x$ for every nonnegative integer $x$. Expanding $(1+T{)}^x$ and evaluating at $T=0$ yieldsComparison in the basis of binomial polynomials gives $(D^k{T}^n)(0)=n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. Terms with $n>k$ have zero constant term after at most $k$ differentiations, so no infinite-series interchange is needed. ThereforeThe case $k=0$ says that the total mass is $f(0)$.

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 $s\ge 2$.