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The Fundamental theorem of Galois theory gives an inclusion-reversing bijection between subgroups $H\le \mathrm{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto \mathrm{Gal}(L/M)$, for a finite Galois extension. One has $[L:L^H]=|H|$ and $[L^H:K]=[\mathrm{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.

Let $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The splitting field is $L=\mathbb{Q}(\alpha ,\zeta )$. A splitting field in characteristic zero is normal and separable, hence Galois. Eisenstein criterion at $2$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This field is real, whereas $\zeta$ is not, and $\zeta$ satisfies a quadratic polynomial. Thus $[L:\mathbb{Q}]=6$. The faithful permutation action on the three roots identifies the Galois group with $S_3$. The subgroup fixing $\alpha$ is generated by the transposition interchanging the other two roots. Its conjugates are different transposition subgroups, so it is not normal.

If $f\in \sqrt{I}$, a power $f^m$ vanishes at every point of $Z(I)$, so $f$ itself vanishes there. This gives $\sqrt{I}\subseteq I(Z(I))$.

For the reverse inclusion, take $f$ vanishing on $Z(I)$ and use the Rabinowitsch trick. In $k[x_1,\dots ,x_n,t]$ form $J=(I,1-tf)$. A common zero would have $x\in Z(I)$, whence $f(x)=0$ and $1-tf(x)=1$, a contradiction. The assumed Weak Hilbert Nullstellensatz gives $J=(1)$. Thus there is a finite expressionIf $f=0$ the conclusion is immediate. Otherwise substitute $t=f^{-1}$ in the localized polynomial ring. The last term vanishes, and clearing a sufficiently high power of $f$ gives $f^N\in I$. Hence $f\in \sqrt{I}$, proving the Strong Hilbert Nullstellensatz:

Apply the Gaussian quadrature rule to the constant polynomial $1$. Exactness gives ${\sum }_{j=1}^n{A}_j=2$.

Use the Lagrange interpolation polynomialIts square has degree $2n-2$, so the exactness already proved givesThe positive weights of Gaussian quadrature follow because ${\ell }_j$ is a nonzero continuous polynomial. Thus these Gaussian quadrature weights are in fact positive, rather than merely nonnegative.

A degree-$n$ Legendre polynomial $P_n$ is orthogonal, for integration on $[-1,1]$, to every polynomial of degree at most $n-1$, and has $n$ distinct roots in $(-1,1)$. Divide $p$ by $P_n$:The Orthogonality of Legendre polynomials gives $\int p=\int r$. At every quadrature node $x_j$, $p(x_j)=r(x_j)$. The assumed exactness for $r$ therefore proves exactness up to degree $2n-1$, the characteristic property of Gaussian quadrature.

To prove convergence of positive quadrature on continuous functions, for a continuous $f$, the Weierstrass approximation theorem supplies a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. For every sufficiently large $n$, the Gaussian quadrature rule is exact for this fixed polynomial. Positivity of its weights and their sum $2$ then giveSince $\epsilon$ is arbitrary, $Q_n(f)\to {\int }_{-1}^{1}f$. Notice that no differentiability of $f$ is needed; uniform polynomial approximation and positivity provide the whole argument.

Let $\mathcal{A}$ be the closure of the polynomials in $C(K)$ for the uniform norm. It is a closed algebra, since products of uniformly convergent sequences converge uniformly on the compact set $K$.

Choose a point $W$ with $|W|>{\sup }_{z\in K}|z|$. The uniformly convergent geometric seriesshows that this reciprocal belongs to $\mathcal{A}$. Join $w$ to $W$ by a continuous path in $\mathbb{C}\setminus K$. Its image is compact and disjoint from $K$, so their distance $d$ is positive. Partition the path into finitely many steps shorter than $d/2$.

Suppose $1/(b-z)\in \mathcal{A}$ and $|a-b|<d/2$, with $b$ on the path. Thenconverges uniformly on $K$, since $|b-z|\ge d$. Every partial sum belongs to $\mathcal{A}$, by its algebra property, and so does the limit. Working backwards along the partition transfers the pole from $W$ to $w$. Thus $1/(w-z)\in \mathcal{A}$: a polynomial $P$ exists withThis pole-moving polynomial approximation is the elementary mechanism behind the polynomial Runge theorem.

Work over the finite field ${\mathbb{F}}_2$. Reduce polynomials modulo $X_i^2-X_i$, so their monomials are square-free. The binary Reed-Muller code $\mathrm{RM}(r,m)$ consists of the evaluation vectors of polynomials of degree at most $r$ at all $2^m$ points of ${\mathbb{F}}_2^m$, in a fixed order. This is a linear code of length $2^m$ and dimension ${\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{m}{j}\right)$, for $0\le r\le m$. Square-free monomials give independent evaluation functions, as follows inductively by splitting according to the last variable.

The minimum Hamming distance of a linear code equals its least nonzero Hamming weight. The result isHere $r=0$ is the repetition code, while $r=m$ is the full binary vector space and has distance one. For the induction step, write $f(x,t)=u(x)+tv(x)$, where $\mathrm{deg}u\le r$ and $\mathrm{deg}v\le r-1$. Its evaluation vector is $(u,u+v)$, the Reed-Muller bar-product recursion. If $v=0$, its nonzero Hamming weight is at least $2d(r,m-1)=2^{m-r}$, with the full-space endpoint treated as above. If $v\ne 0$, every position where $v=1$ contributes exactly one to the sum of the two weights; hence the weight is at least $\mathrm{wt}(v)\ge 2^{(m-1)-(r-1)}=2^{m-r}$. The monomial $X_1\cdots X_r$ is one exactly on $2^{m-r}$ points, attaining the bound. Degrees greater than $m$ give the same full code; the zero code at negative degree has no nonzero word whose weight could define a finite minimum.

Collecting the shifted terms gives a second-order difference equation with coefficients $1-h/2$, $-(2+6h^2)$ and $1+h/2$. Substitute $y_n=r^n$ to obtain its characteristic equation of a linear recurrence:For real $h\ne 0,2$, the roots and general solution areIn the small-step regime, indeed for $0<h<2$, the polynomial is positive at $r=0$, negative at $r=1$ and positive for sufficiently large positive $r$. Thus $0<r_{-}<1<r_{+}$. The bounded solution of a second-order constant-coefficient recurrence discards the growing root, and $y_0=1$ fixes its remaining coefficient:The Taylor series establishes the displayed approximation to the root. Raising that approximation to the $n$th power is a leading approximation, rather than a uniform relative approximation for arbitrarily large $n$: its relative error is controlled when $n{h}^2$ is small. For completeness, the degenerate values give $y_n=A+Bn$ when $h=0$, and $y_n=C13^{-n}$ when $h=2$.

Start from $I_0(\lambda )=1/\lambda$. Differentiation under the integral sign givesThis interchange is justified locally around any $\lambda >0$ by the dominated convergence theorem: the differentiated integrands are bounded by an integrable polynomial times $e^{-\lambda x/2}$. Successive derivatives of ${\lambda }^{-1}$ giveThis is also the integer case of the Gamma function integral after scaling the integration variable.

Differentiate the finite binomial theorem and set $z=1$. For $n\ge 1$ this givesFor $n=0$ the sum is $0$ directly.

For the alternating binomial moments, it is useful to differentiate in a form that introduces falling factorials. Write $(m{)}_j=m(m-1)\cdots (m-j+1)$, with $(m{)}_0=1$. Differentiating the binomial theorem $j$ times and multiplying by $z^j$ yields, for $0\le j\le n$,At $z=-1$, the right side vanishes when $j<n$, and equals $(-1{)}^{n}n!$ when $j=n$.

The polynomials $(X{)}_0,\dots ,(X{)}_r$ have degrees $0,\dots ,r$ and leading coefficient $1$, so successive cancellation of leading terms expresses $X^r$ as a linear combination of them. If $r<n$, every falling-factorial term has index $j<n$ and hence its alternating sum vanishes. If $r=n$, the coefficient of $(X{)}_n$ is $1$, and all lower-index terms still vanish. Therefore the alternating moments areFor $n=0$, the latter identity uses the constant polynomial $X^0=1$, so its value at $X=0$ is $1$; thus $S_0(0)=1$ as well. This makes the edge-case convention explicit rather than treating $0^0$ as an unrelated numerical expression.

If ${\alpha }^{\ast }=g(\alpha )$, then $\alpha$ commutes with its adjoint because it commutes with every polynomial in itself. Hence $\alpha$ is a normal operator.

Conversely let ${\lambda }_1,\dots ,{\lambda }_r$ be the distinct eigenvalues of a normal $\alpha$. The orthonormal eigenbasis from part (ii) reduces the desired identity to $g({\lambda }_j)={\overline{\lambda }}_j$. Lagrange interpolation polynomial provides the polynomialRepeated eigenvalues cause no difficulty because interpolation uses only distinct values. If $V=\{0\}$, use $g=0$. This proves both directions of the third equivalence.

Eisenstein's criterion. Let $f(X)=a_n{X}^n+\cdots +a_0\in \mathbb{Z}[X]$ be a nonconstant primitive polynomial. If a prime number $p$ satisfies $p\nmid a_n$, $p\mid a_j$ for $0\le j<n$, and $p^2\nmid a_0$, then $f$ is an irreducible polynomial over $\mathbb{Q}$.

To prove the Eisenstein criterion, suppose a nontrivial factorization exists over $\mathbb{Q}$. Gauss lemma for polynomials gives $f=gh$ with $g,h\in \mathbb{Z}[X]$ of positive degree. Their leading coefficients are not divisible by $p$, because their product is $a_n$. Reduction modulo $p$ therefore preserves both degrees, andSince the only irreducible factor of the right side is $X$, both reduced factors are nonzero monomials of positive degree. Their constant coefficients are consequently divisible by $p$. This would make $p^2\mid g(0)h(0)=a_0$, a contradiction.

For the requested polynomial, shift the variable by one. The binomial theorem givesIt is monic, all coefficients below the leading one are divisible by $p$, and the constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves irreducibility of this shifted polynomial. The substitution $X\mapsto X+1$ is an automorphism of $\mathbb{Q}[X]$, with inverse $X\mapsto X-1$, so it preserves nontrivial factorizations. HenceThis is the cyclotomic polynomial for a prime index.

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

For the final request, take the following subalgebra of the two-variable Laurent polynomial ring:Let $J=X_{1}k[X_1,X_2]$, an ideal of $A$. We haveas $k$-vector spaces. In particular, the classes of $X_1{X}_2^j$, for $j\ge 0$, are linearly independent. If $J$ had finitely many generators as an $A$-module, their classes would span $J/J^2$ over $A/J=k$, contradicting its infinite dimension. Hence $J$ is not finitely generated and $A$ is not a Noetherian ring. This illustrates why the preceding result is special to one Laurent variable.

For a finitely generated module $V={⨁}_d{V}_d$ over the graded ring $S=k[X_1,\dots ,X_n]$, with each $X_i$ of degree one, its Poincare series of a graded module isThis is also its Hilbert series. One usually takes $d\ge 0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\dots ,a_h$ give a surjection ${⨁}_{j}S(-a_j)\to V$, where the graded shift is defined by $V(-a{)}_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional vector space and $V_d=0$ for all sufficiently negative $d$.

The Hilbert-Serre theorem in the standard grading saysFor a nonnegatively graded $V$, the numerator is in $\mathbb{Z}[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.

We prove the Hilbert-Serre theorem by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional graded module; thus $P_V(t)$ is a Laurent polynomial. Suppose $n>0$ and putThe Hilbert basis theorem makes $S$ a Noetherian ring, so $K$ is a finitely generated module, as is $C$. Both are graded modules killed by $X_n$, hence finitely generated over $S^{\prime }=k[X_1,\dots ,X_{n-1}]$.

The graded exact sequenceidentifies the kernel and cokernel of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yieldssoBy induction the right side has denominator $(1-t{)}^{n-1}$, which proves the required denominator $(1-t{)}^n$. Since $(1-t{)}^n{P}_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent polynomial numerator is an ordinary polynomial in that case.

For $n\ge 1$, this also implies the usual Hilbert polynomial consequence. Writing $Q(t)={\sum }_j{q}_j{t}^j$ and usinggives, for all sufficiently large $d$,a polynomial in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.

For the final request, a free resolution of $V$ is an exact sequencewith each $F_i$ a free module. In the graded setting we take finite direct sums of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The syzygy modules are the successive kernels which record the relations among generators, then relations among those relations, and so on.

Put $\mathfrak{m}=(X_1,\dots ,X_n)$, the homogeneous maximal ideal with $S/\mathfrak{m}=k$. Choose a homogeneous $k$-basis of $V/\mathfrak{m}V$ and lift it to $V$. These lifts generate $V$ by the graded Nakayama lemma: a bounded-below graded module $W$ satisfying $W=\mathfrak{m}W$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the submodule generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an isomorphism modulo $\mathfrak{m}$.

Its kernel is finitely generated because $S$ is a Noetherian ring. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a minimal graded free resolution, meaningIndeed, at each stage the free cover induces an isomorphism modulo $\mathfrak{m}$, so its kernel lies in $\mathfrak{m}$ times its source. The sequence is exact by construction, though it may at first appear infinite.

The Koszul complex on $X_1,\dots ,X_n$ iswith basis vectors $e_j$ given degree one, and differentialThe hat means omission of that factor. Every pair of terms in ${\partial }^2$ cancels with opposite signs, so this is a chain complex. The sequence $X_1,\dots ,X_n$ is a regular sequence: after quotienting by the first $j-1$ variables, the next variable is a non-zero-divisor in the remaining polynomial ring.

The Koszul complex is consequently exact in positive degrees and has $H_0=k$, so it is a Koszul resolution of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\stackrel{X_1}{\to }S\to S/(X_1)\to 0$. Appending $X_j$ constructs the mapping cone of multiplication by $X_j$ on the previous chain complex. Its homology is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\dots ,X_{j-1})$; its degree-zero homology is the further quotient. Here a mapping cone combines a chain complex with a shifted copy and adds the given multiplication map to the differential, thereby measuring its kernel and cokernel on homology.

The Tor functor ${\mathrm{Tor}}_i^S(V,k)$ is defined as the degree-$i$ homology of $F_{\bullet }{\otimes }_{S}k$ for a free resolution $F_{\bullet }$ of $V$. It can equally be computed as the homology of $V{\otimes }_S{K}_{\bullet }$, using the Koszul resolution of $k$. To see the equality, form the double complex $F_{\bullet }{\otimes }_S{K}_{\bullet }$. Taking homology first along $F_{\bullet }$ leaves $V{\otimes }_S{K}_{\bullet }$, since each $K_i$ is a free module; taking homology first along $K_{\bullet }$ leaves $F_{\bullet }{\otimes }_{S}k$, since each $F_i$ is free. The finite sums along each total degree identify both with the homology of the total complex. This also explains the symmetry used in computing the Tor functor.

Since $K_i=0$ for $i>n$, we have ${\mathrm{Tor}}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the minimal graded free resolution become zero after tensoring with $k=S/\mathfrak{m}$, soThus $F_i/\mathfrak{m}{F}_i=0$ for $i>n$, and the graded Nakayama lemma implies $F_i=0$ for those $i$. We have proved the required Hilbert syzygy theorem:where some initial $F_i$ may be zero. The argument supplies finitely generated graded free modules, which is stronger than merely giving an ungraded free resolution.