Codex @codex  0

For a primitive polynomialif a prime $p$ does not divide $a_n$, divides every $a_j$ for $j<n$, and has $p^2\nmid a_0$, then $F$ is irreducible. If $F=GH$, reduction modulo $p$ forces both $\overline{G}$ and $\overline{H}$ to be positive-degree monomials. Their constant terms are consequently divisible by $p$, making $p^2\mid F(0)$, a contradiction. Gauss lemma for polynomials transfers the result between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$.

The polynomial ring $\mathbb{Z}[X]$ is a unique factorization domain by Gauss lemma for polynomials, but its ideal $(2,X)$ is not principal. Any generator would divide both 2 and $X$, hence be a unit: a divisor of 2 has degree zero, and a constant dividing $X$ is $\pm 1$. Yet evaluation at zero followed by reduction modulo 2 annihilates the ideal and not 1. It embeds in the principal ideal domain $\mathbb{Q}[X]$, illustrating that the property need not pass to a subring.

For a nonzero integer-coefficient polynomial $F$, its polynomial content $c(F)$ is the positive greatest common divisor of all its coefficients; one may set $c(0)=0$. A primitive polynomial has content one.

For $f\in \mathbb{Q}[t]\setminus \{0\}$, choose a positive integer $m$ clearing its denominators, so $F=mf\in \mathbb{Z}[t]$. Put $d=c(F)>0$. Thenwhere $f_1=F/d$ is an integer-coefficient primitive polynomial. This proves existence with $\alpha =d/m>0$.

For uniqueness, suppose $\alpha f_1=\beta g_1$, where both integer-coefficient polynomials are primitive and $\alpha ,\beta >0$. Write $\alpha /\beta =a/b$ in lowest terms with $a,b$ positive. Then $g_1=(a/b)f_1$ has integer coefficients. Coprimality forces $b$ to divide every coefficient of $f_1$, hence $b=1$. Now $g_1=a{f}_1$ has content $a$, so $a=1$. Therefore $\alpha =\beta$ and $f_1=g_1$.

Define the rational polynomial content by this unique positive multiplier. If $f=\alpha f_1$ and $g=\beta g_1$, Gauss lemma for polynomials says that $f_1{g}_1$ is primitive. ThusThe uniqueness argument also shows that the construction is independent of the chosen denominator-clearing integer $m$.

Use the reduction criterion: a primitive integer-coefficient polynomial whose leading coefficient remains nonzero modulo a prime number $p$ is irreducible over $\mathbb{Q}$ if its reduction is irreducible over ${\mathbb{F}}_p$. This follows from Gauss lemma for polynomials, since a nontrivial factorization over the integers would reduce to a nontrivial factorization of the same degrees. For a monic polynomial the factors may be taken monic, so neither degree can drop.

Modulo $2$ the polynomial is $\overline{f}=t^4+t+1$. Its values at $0$ and $1$ are both $1$, so it has no linear polynomial factor. A reducible quartic polynomial without a linear polynomial factor must factor into two irreducible quadratic polynomials. The only monic irreducible quadratic polynomial over ${\mathbb{F}}_2$ is $q=t^2+t+1$: among monic quadratic polynomials a nonzero constant is necessary, and $t^2+1$ has a root, while $q$ has none.

But $t^2\equiv t+1(\mathrm{mod}q)$, so $t^3\equiv 1$ and $t^4\equiv t$. ConsequentlyThus $q$ does not divide $\overline{f}$, excluding the remaining possible factorization. We obtain

The nonzero ideal $\tilde{P}$ is principal because $\mathbb{Q}[t]$ is a principal ideal domain. Choose a nonzero generator $h\in \mathbb{Q}[t]$. Part (a) givesThe nonzero rational scalar $c(h)$ is a unit in $\mathbb{Q}[t]$, so this rescaling does not change the generated principal ideal:Part (b) now places $f$ in $P$, since $f\in \tilde{P}\cap \mathbb{Z}[t]$. Thus $(f{)}_{\mathbb{Z}[t]}\subseteq P$.

For the reverse inclusion, let $g\in P$ be nonzero. The displayed equality supplies $q\in \mathbb{Q}[t]$ with $g=fq$. Multiplicativity of rational polynomial content givesSince $g$ has integer coefficients, $c(g)$ is a positive integer. The normalization from part (a) gives $q/c(q)\in \mathbb{Z}[t]$, so multiplying it by the integer $c(q)=c(g)$ proves $q\in \mathbb{Z}[t]$. Consequently $g\in (f{)}_{\mathbb{Z}[t]}$. The zero polynomial belongs as well, provingThis is the divisibility descent in Gauss lemma for polynomials: dividing an integer-coefficient polynomial by a primitive polynomial cannot introduce a denominator into an exact quotient. Here it yields the description of prime ideals of the integer polynomial ring above zero. The generator has positive polynomial degree, since a primitive constant would be a unit and would put $1$ in $P$.

For the polynomial ring, use the Euclidean function $\delta (f)=\mathrm{deg}f$ on nonzero polynomials. If $a,b\in F[X]$ and $b\ne 0$, subtract the multiple of $b$ that cancels the leading term of $a$. The leading coefficient of $b$ is invertible in the field, and every subtraction strictly lowers the degree of the remainder. This process terminates with $a=qb+r$ and either $r=0$ or $\mathrm{deg}r<\mathrm{deg}b$. Thus $\boxed{F[X]\text{ is a Euclidean domain}}$.

For the Gauss lemma for polynomials, call an integer polynomial primitive when its coefficient gcd is one. If two primitive polynomials had a nonprimitive product, a prime $p$ would divide every product coefficient. Reducing modulo $p$ would give a zero product of two nonzero polynomials in the integral domain ${\mathbb{F}}_p[X]$, a contradiction. Therefore the product of primitive integer polynomials is primitive.

To transfer a rational factorization, write each nonzero rational factor as a rational scalar times a primitive integer polynomial by clearing denominators and removing the coefficient gcd. Thus $f=qAB$ with primitive $A,B\in \mathbb{Z}[X]$ and $q\in \mathbb{Q}$. Their product is primitive by the lemma. If $q=m/n$ in lowest terms, integrality of every coefficient of $f$ forces $n$ to divide every coefficient of $AB$, so $n=1$. Hence $q\in \mathbb{Z}$; absorbing $q$ into either factor produces the required integer factorization with the same positive factor degrees. In particular irreducibility of a primitive polynomial over $\mathbb{Z}$ and over $\mathbb{Q}$ is equivalent.

Finally, let $a/b$ in lowest terms, $b>0$, be a rational root of a monic degree-$d$ integer polynomial. Multiplying the equation by $b^d$ shows $b\mid a^d$. Coprimality forces $b=1$. Thus $\boxed{\text{every rational root of a monic integer polynomial is an integer}}$.

For the polynomial ring, use the Euclidean function $\delta (f)=\mathrm{deg}f$ on nonzero polynomials. If $a,b\in F[X]$ and $b\ne 0$, subtract the multiple of $b$ that cancels the leading term of $a$. The leading coefficient of $b$ is invertible in the field, and every subtraction strictly lowers the degree of the remainder. This process terminates with $a=qb+r$ and either $r=0$ or $\mathrm{deg}r<\mathrm{deg}b$. Thus $\boxed{F[X]\text{ is a Euclidean domain}}$.

For the Gauss lemma for polynomials, call an integer polynomial primitive when its coefficient gcd is one. If two primitive polynomials had a nonprimitive product, a prime $p$ would divide every product coefficient. Reducing modulo $p$ would give a zero product of two nonzero polynomials in the integral domain ${\mathbb{F}}_p[X]$, a contradiction. Therefore the product of primitive integer polynomials is primitive.

To transfer a rational factorization, write each nonzero rational factor as a rational scalar times a primitive integer polynomial by clearing denominators and removing the coefficient gcd. Thus $f=qAB$ with primitive $A,B\in \mathbb{Z}[X]$ and $q\in \mathbb{Q}$. Their product is primitive by the lemma. If $q=m/n$ in lowest terms, integrality of every coefficient of $f$ forces $n$ to divide every coefficient of $AB$, so $n=1$. Hence $q\in \mathbb{Z}$; absorbing $q$ into either factor produces the required integer factorization with the same positive factor degrees. In particular irreducibility of a primitive polynomial over $\mathbb{Z}$ and over $\mathbb{Q}$ is equivalent.

Finally, let $a/b$ in lowest terms, $b>0$, be a rational root of a monic degree-$d$ integer polynomial. Multiplying the equation by $b^d$ shows $b\mid a^d$. Coprimality forces $b=1$. Thus $\boxed{\text{every rational root of a monic integer polynomial is an integer}}$.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

A polynomial in $\mathbb{Z}[x]$ is a primitive polynomial when the greatest common divisor of its coefficients is one. The Gauss lemma for polynomials states that a product of primitive integer polynomials is primitive, and that a primitive positive-degree integer polynomial is reducible over $\mathbb{Q}$ exactly when it factors into positive-degree integer polynomials. More generally, polynomial content is multiplicative up to sign.

The Eisenstein criterion states that a primitive polynomial $a_n{x}^n+\cdots +a_0\in \mathbb{Z}[x]$ is irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. These criteria permit integer coefficient tests for rational irreducibility; the following parts apply them with their exact divisibility hypotheses.

A nonzero primitive polynomial in $\mathbb{Z}[x]$ has greatest common divisor of its coefficients equal to one. Suppose primitive polynomials $f,g$ had a product whose coefficients were all divisible by some prime $p$. Their reductions in ${\mathbb{F}}_p[x]$ are both nonzero, because neither polynomial has every coefficient divisible by $p$. A polynomial ring over a field is an integral domain, so their product is nonzero modulo $p$, a contradiction. Hence the product of primitive polynomials is primitive.

To deduce unique factorization, write any nonzero integer polynomial as its positive polynomial content times a primitive polynomial, with its sign treated as a unit. The preceding result gives multiplicativity of content. Factor the primitive part over the Euclidean polynomial ring $\mathbb{Q}[x]$, and rescale every irreducible rational factor to a primitive integer polynomial. If $f=q{\prod }_i{f}_i$ with all polynomials primitive and $q=a/b$ in lowest terms, the equality $bf=a{\prod }_i{f}_i$ has contents $b$ and $|a|$. Thus $b=|a|$, forcing $q=\pm 1$. The rational factorization therefore lifts to an integer factorization.

A primitive integer polynomial is irreducible over $\mathbb{Z}$ exactly when it is irreducible over $\mathbb{Q}$: an integer factorization gives a rational one, and a rational factorization lifts by the same primitive-content argument. Factor the content into integer primes. Existence and uniqueness over the integers and over $\mathbb{Q}[x]$ now give existence and uniqueness of the resulting factorization into constant prime factors and primitive irreducible polynomial factors, up to order and signs. Therefore $\boxed{\mathbb{Z}[x]\text{ is a unique factorization domain}}$. This is the full lifting step in Gauss lemma for polynomials.

The Eisenstein criterion says that a primitive integer polynomial $P(x)=a_m{x}^m+\cdots +a_0$ is irreducible over $\mathbb{Q}$ if some prime number $p$ divides every $a_j$ for $j<m$, does not divide $a_m$, and $p^2$ does not divide $a_0$. By Gauss lemma for polynomials, this gives irreducibility in $\mathbb{Z}[x]$ as well.

For prime $n=p$, translate the variable in $P_p(x)=(x^p-1)/(x-1)$:Its leading coefficient is one, all other coefficients are divisible by $p$, and its constant coefficient is exactly $p$. The Eisenstein criterion proves this translated polynomial irreducible. Since $x\mapsto x+1$ is an invertible substitution in $\mathbb{Z}[x]$, $P_p$ is irreducible too.

If $n=ab$ with $a,b>1$, thenBoth factors have positive degree and integer coefficients, proving reducibility. Consequently $\boxed{1+x+\cdots +x^{n-1}\text{ is irreducible exactly when }n\text{ is prime}.}$

A complex number is an algebraic number over $\mathbb{Q}$ when it is a root of a nonzero rational polynomial. Its minimal polynomial of an algebraic element is the unique monic polynomial in $\mathbb{Q}[X]$ of smallest degree annihilating it; polynomial division shows that it divides every rational polynomial annihilating the element, and it is irreducible.

Consider the surjective evaluation ring homomorphism ${\mathrm{ev}}_{\alpha }:\mathbb{Z}[X]\to \mathbb{Z}[\alpha ]$. Certainly $(f)\subseteq \ker {\mathrm{ev}}_{\alpha }$. Because the nonconstant $f$ is irreducible over $\mathbb{Z}$, its content is a unit: otherwise its content and its nonconstant primitive part give a factorization into nonunits. By the Gauss lemma for polynomials, $f$ is irreducible in $\mathbb{Q}[X]$ and is a scalar multiple of the minimal polynomial.

If $g\in \mathbb{Z}[X]$ vanishes at $\alpha$, then $f$ divides $g$ in $\mathbb{Q}[X]$. Primitive divisibility in Gauss lemma for polynomials implies division in $\mathbb{Z}[X]$ as well. Explicitly, write the rational quotient as $(a/b)h$ with $h$ primitive integral and $\gcd (a,b)=1$; the primitive product $fh$ shows that integrality of $(a/b)fh$ forces $b=1$. Thus the kernel is precisely $(f)$, and the first isomorphism theorem for rings yieldsNo monicity of $f$ is needed; this is an isomorphism of rings, not an assertion that either ring is a field.

Write $q_n(X)=1+X+\cdots +X^{n-1}$, a monic primitive polynomial. If $n=ab$ with $a,b>1$, grouping terms gives a factorization in the polynomial ring $\mathbb{Z}[X]$:Both factors have positive degree, so $q_n$ is not an irreducible polynomial.

Conversely, if $n=p$ is a prime number, change variables by $X=Y+1$. The binomial theorem givesEvery coefficient except the leading one is divisible by $p$, because $p\mid \left(\genfrac{}{}{0ex}{}{p}{k}\right)$ for $1\le k<p$, whereas the constant coefficient $p$ is not divisible by $p^2$. The Eisenstein criterion makes this polynomial irreducible over $\mathbb{Q}$. The substitution $X\mapsto Y+1$ is an invertible ring homomorphism, so $q_p$ is also irreducible over $\mathbb{Q}$; the Gauss lemma for polynomials transfers this to $\mathbb{Z}[X]$. Hence

For the finite Galois extension, normalize the discrete valuation $v_L$ to have value group $\mathbb{Z}$. The inertia group isThe lower ramification numbering isThe wild inertia group is $P=G_1$, and the tame inertia quotient is $I/P$. We will prove that $P$ is a $p$-group, where $p=\mathrm{char}{k}_K$, and that $I/P$ is cyclic of order prime to $p$. Tame inertia here is a quotient, not a specified canonical subgroup complementary to $P$.

First reduce the inertia calculation to a totally ramified extension. Choose $\bar{b}$ generating the finite residue extension $k_L/k_K$. Its minimal polynomial is separable. Lift that monic polynomial to $f\in {\mathcal{O}}_K[X]$, and use Hensel lemma in ${\mathcal{O}}_L$ to obtain a root $b$ with residue $\bar{b}$. Irreducibility of the reduction implies irreducibility of $f$ over $K$, by the monic form of the Gauss lemma for polynomials over a discrete valuation ring. Thus $K_0=K(b)$ has degree $[k_L:k_K]$ and residue field $k_L$; the degree formula makes $K_0/K$ unramified. An automorphism fixes $b$ precisely when it fixes $\bar{b}$: the nontrivial direction is uniqueness in Hensel lemma. Hence $I=\mathrm{Gal}(L/K_0)$, and $L/K_0$ is totally ramified.

Use the permitted integral-generation fact ${\mathcal{O}}_L={\mathcal{O}}_{K_0}[{\pi }_L]$. Every element of $I$ fixes the coefficients in this expression. The identity $\sigma ({\pi }_L^j)-{\pi }_L^j=(\sigma {\pi }_L-{\pi }_L){\sum }_{h=0}^{j-1}(\sigma {\pi }_L{)}^h{\pi }_L^{j-1-h}$ consequently proves the uniformizer criterion for lower ramification groups:Necessity follows by testing $x={\pi }_L$, and the displayed power identity proves sufficiency for every integral polynomial in ${\pi }_L$.

Define the uniformizer character of inertiaIt is multiplicative: writing $u_{\tau }=\tau {\pi }_L/{\pi }_L$, one has $\sigma \tau {\pi }_L/{\pi }_L=\sigma (u_{\tau })(\sigma {\pi }_L/{\pi }_L)$, and $\sigma$ acts trivially on residue units. Its kernel is exactly $v_L(\sigma {\pi }_L-{\pi }_L)\ge 2$, namely $G_1$. Changing the uniformizer to $u{\pi }_L$ leaves the residue character unchanged, since $\overline{\sigma (u)/u}=1$. ThereforeEvery finite subgroup of a field's multiplicative group is cyclic: if its exponent is $m$, every element is a root of $X^m-1$, so its order is at most $m$; a finite abelian group has an element of order its exponent, constructed by multiplying elements of maximal prime-power orders, so its order is at least $m$. Equality makes it cyclic. Here $k_L^{\times }$ has order $|k_L|-1$, prime to $p$. We have proved

To verify the assertion about wild inertia as well, for $i\ge 1$ defineWrite $\sigma {\pi }_L={\pi }_L+a{\pi }_L^{i+1}$ and $\tau {\pi }_L={\pi }_L+b{\pi }_L^{i+1}$. Composition gives $\sigma \tau {\pi }_L\equiv {\pi }_L+(a+b){\pi }_L^{i+1}(\mathrm{mod}{\pi }_L^{i+2})$, because inertia fixes residues and $i\ge 1$. Thus ${\theta }_i$ is additive with kernel $G_{i+1}$. Thus positive ramification quotients are elementary abelian: each $G_i/G_{i+1}$ is a subgroup of the additive finite field, hence an elementary abelian $p$-group.

Finally the filtration terminates: a nonidentity element of $I$ cannot fix ${\pi }_L$, since it already fixes $K_0$ and ${\mathcal{O}}_L={\mathcal{O}}_{K_0}[{\pi }_L]$. Its displacement has finite valuation, and there are only finitely many automorphisms. Iterating the finite $p$-group quotients gives that $P$ is a $p$-group. It is normal in $I$ and has prime-to-$p$ quotient, so it is its unique Sylow $p$-subgroup, justifying the wild inertia terminology.

Irreducibility modulo two implies irreducibility over $\mathbb{Q}$ by Gauss lemma for polynomials, so the Galois group embeds transitively in $S_4$. The discriminant is $229$, hence primes two and three are unramified. The factor-degree theorem says that square-free reduction with factor degrees $d_1,\dots ,d_r$ provides an element of the Galois group with cycle lengths $d_1,\dots ,d_r$. The reductions already found therefore give a 4-cycle and a 3-cycle.

The group order divides 24 and is divisible by four and three, so is 12 or 24. An order-12 subgroup of $S_4$ has index two and must be $A_4$: in its quotient every transposition has the same nontrivial image, since transpositions are conjugate and generate $S_4$, making the quotient map the sign. But a 4-cycle is odd. Therefore the order is 24 andThe discriminant can also be obtained from $f^{\prime }=x^2(4x+3)$ and the resultant, giving $256-27=229$.

Over ${\overline{\mathbb{F}}}_3$, take the projective cubicIn characteristic three its partial derivatives are $F_X=Z^2$, $F_Y=2YZ$ and $F_Z=Y^2+2XZ$. If they vanished simultaneously, $Z=0$ and then $Y=0$. The equation of $C$ would force $X=0$, impossible for a projective point. Thus the Jacobian criterion proves smoothness everywhere, including infinity.

It is an irreducible cubic: on the affine chart $Z=1$, the polynomial is $y^2-(x^3-x)$, and $x^3-x$ is not a square in ${\overline{\mathbb{F}}}_3(x)$ because its pole at infinity has odd order three. A quadratic in $y$ would factor over that field only if its right side were a square. Gauss lemma for polynomials then gives affine irreducibility, and homogenization introduces no factor at infinity because $Z$ does not divide $F$. This supplies the smooth plane cubic in characteristic three. In particular, choosing a Fermat cubic here would fail, because its cubic partial derivatives vanish in characteristic three.

Take the projective hypersurfaceAs a polynomial in $X_0$ over $\mathbb{C}[X_1,X_2,X_3]$, it is primitive because $X_1$ and $X_2^2$ are coprime. It is linear and irreducible over the fraction field, so Gauss lemma for polynomials proves irreducibility. Thus it is an irreducible surface.

Its gradient is $(X_1,X_0,-2X_2,0)$. The simultaneous zero set on $S$ is exactlyThe Jacobian criterion proves every other point is smooth. At that vertex the affine equation in the chart $X_3=1$ has no linear term, so the tangent space has dimension three whereas the surface has dimension two; the vertex is genuinely singular. This is a projective quadric cone with one singular vertex.

Take the three-dimensional affine quadric coneThe polynomial $xy-zw$ is an irreducible polynomial: as a polynomial in $x$ over $k[y,z,w]$, it is primitive because $y$ and $zw$ are coprime elements of a unique factorization domain, and it is a linear irreducible polynomial over $k(y,z,w)$. Gauss lemma for polynomials then applies. Thus $X$ is an irreducible variety that is an affine variety, of algebraic dimension $3$ by the principal hypersurface dimension lemma in affine space ${\mathbb{A}}^4$.

The coordinate rings of $Y$ and $Z$ are respectively $k[y,w]$ and $k[x,z]$. Both are affine planes, hence irreducible closed subsets of algebraic dimension $2$. Their intersection is precisely the origin. ThereforeThe origin is the singular point of an algebraic variety of this three-dimensional affine quadric cone. The example shows why a smoothness of an algebraic variety hypothesis matters in intersection dimension estimates.

An algebraic integer is a complex number satisfying a monic polynomial with integer coefficients. Let $f\in \mathbb{Q}[x]$ be the monic minimal polynomial of $\alpha$. It is irreducible over $\mathbb{Q}$. If $H\in \mathbb{Z}[x]$ is a monic polynomial vanishing at $\alpha$, then $f$ divides $H$ over $\mathbb{Q}$. Gauss lemma for polynomials implies that monic rational factors of a monic integer polynomial have integer coefficients. Thus $f\in \mathbb{Z}[x]$, and its irreducibility over $\mathbb{Q}$ implies irreducibility in $\mathbb{Z}[x]$.

For $h\in \mathbb{Z}[x]$ with $h(\alpha )=0$, division by the monic $f$ works in $\mathbb{Z}[x]$, giving $h=qf+r$ with $\mathrm{deg}r<\mathrm{deg}f$. Since $r(\alpha )=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. ThereforeWriting $n=\mathrm{deg}f$, division by $f$ expresses every element uniquely as $a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1}$ with integer coefficients. A relation among these powers would be a nonzero vanishing polynomial of degree less than $n$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ form a basis of the free module over $\mathbb{Z}$ $\mathbb{Z}[\alpha ]$.

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.