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For a degree-$n$ polynomial $p(X)=a_n{\prod }_{i=1}^n(X-r_i)$ over a field, with $a_n\ne 0$ and roots in an algebraic closure, its discriminant isIt is a polynomial with integer coefficients in the coefficients of $p$, and lies in the original coefficient field. Its vanishing is equivalent to a repeated root of a polynomial in an algebraic closure, even when the coefficient field itself contains no roots. This detects separability without solving the polynomial equation.^[ref]

For a quadratic polynomial, the expression is the quadratic discriminant $b^2-4ac$. For the cubic polynomial $X^3+bX+c$, it is $-4b^3-27c^2$. Thus $X^3-2$ has discriminant $-108$ over $\mathbb{Q}$, but after reduction to characteristic three its discriminant is zero and the resulting polynomial has a repeated root in its splitting field. Discriminants also indicate exceptional parameter values where distinct polynomial roots collide.

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

Let $K=\mathbb{Q}(\sqrt{-31})$. Its ring of integers of a quadratic field is $\mathbb{Z}[(1+\sqrt{-31})/2]$, and its fundamental field discriminant is $-31$.

First compute the class number. The ideal-form correspondence for imaginary quadratic fields identifies the ideal classes with proper equivalence classes of primitive positive definite binary quadratic forms of discriminant $-31$. Each has a unique reduced positive definite binary quadratic form representative under the convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\ge 3a^2$ we get $1\le a\le \sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactlyFor $a=3$, neither odd residue choice $b=\pm 1,\pm 3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, soThis computes the ideal class group of Q of square root minus thirty-one.

Now take a root $\theta$ of $F(T)=T^3+T-1$, and let $H$ be its splitting field over $\mathbb{Q}$. There is no rational root: the only candidates are $\pm 1$, and neither vanishes. Thus the cubic is irreducible. Its polynomial discriminant isSince this is nonsquare, the Galois group of an irreducible cubic is $S_3$. Its quadratic subfield is $K$: the product of pairwise root differences squares to $-31$ and changes sign under odd permutations. Thus $H/K$ is cyclic of degree three, and $H=K(\theta )$. Indeed the cubic field and $K$ have coprime degrees, so their compositum already has the full splitting-field degree six. The squarefree polynomial discriminant also shows, by the discriminant-index formula for an integral lattice, that $\mathbb{Z}[\theta ]$ is the full integer ring of the cubic field.

It remains to prove that $H/K$ is unramified everywhere. At any rational prime $\ell \ne 31$, the reduction of $F$ is separable. Over a finite residue extension in which it splits, all residue roots are simple and lift by Hensel lemma to the corresponding unramified extension of ${\mathbb{Q}}_{\ell }$. Therefore the local splitting field is unramified over ${\mathbb{Q}}_{\ell }$, and its extension over a completion of $K$ is unramified as well.

At the only possible ramified prime, direct reduction givesThe simple root lifts to $r\in {\mathbb{Q}}_{31}$. Write $F(T)=(T-r)Q(T)$, with $Q$ monic quadratic over ${\mathbb{Q}}_{31}$. The discriminant product formula givesHere $Q(r)=F^{\prime }(r)$ is a unit, so the quadratic discriminant has square class $-31$. Hence the entire local splitting field isIt is quadratic over ${\mathbb{Q}}_{31}$, but its extension over the completion of $K$ is trivial. Thus the prime over $31$ splits completely in $H/K$; in particular it is not ramified there. This is the local splitting-field argument behind an unramified cyclic cubic extension from a prime-discriminant cubic. There are no real places of the imaginary quadratic field $K$, so no infinite-place ramification remains to check.

The Hilbert class field theorem identifies the maximal everywhere unramified abelian extension with an extension of degree $h_K$. Our cyclic unramified extension already has that degree, so it must be the whole Hilbert class field. ThereforeEquivalently, it is the splitting field of $T^3+T-1$, of degree six over $\mathbb{Q}$ and degree three over $K$. Replacing $\theta$ by ${\theta }^{-1}$ gives the alternative defining cubic $T^3-T^2-1$. This is the Hilbert class field of Q of square root minus thirty-one.

The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

The expectation value is the mean outcome over many identically prepared measurements of the observable. For real $\lambda$, positivity of the norm in the Hilbert space givesThis real quadratic has nonpositive quadratic discriminant, soApply this to the centered observables $Q-⟨Q⟩$ and $P-⟨P⟩$ to obtain the Robertson uncertainty principle

For the quantum harmonic oscillator, $[x,p]=i\hslash$ andThe arithmetic-geometric mean inequality and $\Delta x\Delta p\ge \hslash /2$ now give

A Lagrange top is a rigid body that is symmetric about a principal axis, has a point on that axis fixed in space, and has its center of mass on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse principal moment of inertia about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.

The Euler angles for a symmetric top use $\theta$ for inclination, $\varphi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a cyclic coordinate, its generalized momentumis conserved. The coordinate $\varphi$ is also cyclic, sois a second integral. Finally, the Lagrangian has no explicit time dependence, and conservation of energy from time-translation invariance gives the independent integral

For steady precession set $\theta$ constant and $\dot{\varphi }=\Omega$ constant. The $\theta$ Euler-Lagrange equation, with $0<\theta <\pi /2$, reduces after division by $\sin \theta$ toThis quadratic has a real precession rate precisely when its quadratic discriminant is nonnegative. Hence Steady precession of a Lagrange top is possible if and only if

For any smooth projective curve of genus $g$, a canonical divisor has degree $2g-2$. If this plane curve had genus two, part (ii) would implyso $d^2-3d-2=0$. The quadratic discriminant is $17$, which is not a perfect square, so the equation has no integral solution. Thus a smooth projective plane curve cannot have genus two, as recorded by no smooth plane curve has genus two.

Write the position of a mass element relative to the centre of mass as $r$. Expand its squared velocity and integrate:The mixed term vanishes because $\int rdm=0$. Alsousing the scalar triple-product identity. ThereforeIn principal axes the inertia tensor is diagonal, so the second term is $\frac{1}{2}(A{\omega }_1^2+B{\omega }_2^2+C{\omega }_3^2)$. Substituting the three Euler-angle components given in the source gives its displayed kinetic-energy expression directly.

For $A=B$, the first two component squares have opposite mixed terms and sum to ${\dot{\theta }}^2+{\dot{\varphi }}^2{\sin }^2\theta$. Put $n=\dot{\psi }+\dot{\varphi }\cos \theta$. Using the orientation-dependent potential printed in the source therefore gives the formal modelThe cyclic coordinates $x,y,\psi$ have conserved momenta $M\dot{x},M\dot{y},Cn$. Consequently $\dot{x},\dot{y},n$ are constant. The coordinate $\varphi$ is also cyclic, withFor imposed fixed tilt $\theta =\alpha \in (0,\pi /2)$, conservation of $n$ and $p_{\varphi }$ forces $\dot{\varphi }=\Omega$ constant. Impose the fixed-height condition $z=h$ in this formal constrained model. Its $\theta$ equation isSet the tilt constant and divide by its nonzero sine. This givesA real polynomial root requires its quadratic discriminant to be nonnegative:When this condition holds the two algebraic rates are $(Cn\pm \sqrt{C^2{n}^2-4AMgh\cos \alpha })/(2A\cos \alpha )$.

There is a physical inconsistency in the printed data that must be kept distinct from this algebra. If $(x,y,z)$ are literally Cartesian coordinates of the centre of mass in uniform gravity, the gravitational potential is $Mgz$, not $Mgz\cos \theta$. The printed Lagrangian consequently does not follow from that Cartesian-height interpretation. Moreover its unconstrained $z$ equation is $\ddot{z}=-g\cos \theta$, so constant height at the indicated tilt requires a constraint reaction. The requested rate and discriminant follow from the printed orientation-dependent potential with fixed height imposed; a standard heavy-top interpretation instead uses a height proportional to an axis distance times $\cos \theta$, with the reference point and corresponding inertia specified consistently. The kinetic-energy derivation is independent of this source inconsistency.

For an odd prime $\ell$, the unique quadratic subfield of a prime cyclotomic field has no finite ramification outside $\ell$. The classification of quadratic discriminants identifies it with $\mathbb{Q}(\sqrt{{\ell }^{\ast }})$, where ${\ell }^{\ast }=(-1{)}^{(\ell -1)/2}\ell$. A fundamental discriminant factors as $D=D_2{\prod }_{\ell \mid D,\ell \text{ odd}}{\ell }^{\ast }$ with $D_2\in \{1,-4,8,-8\}$. The remaining quadratic number field is generated by $i$, $\sqrt{2}$, or $\sqrt{-2}$ and lies in $\mathbb{Q}({\zeta }_8)$. Taking the corresponding compositum proves the indicated quadratic case of the Kronecker–Weber theorem.

Consider an axially symmetric rigid body rotating about a fixed pivot, with transverse and axial principal moments of inertia $A$ and $C$ and orientation-dependent potential energy $Mgh\cos \theta$. Let the inclination $\alpha$ be nonvertical, $0<\alpha <\pi$, and let $n$ be the axial angular velocity of a symmetric top. Conservation of the two cyclic angular momenta makes both $n$ and the steady precession rate $\Omega$ constant. The tilt Euler-Lagrange equation, after division by $\sin \alpha$, givesWhen $\cos \alpha \ne 0$, a real precession rate exists exactly when the quadratic discriminant obeys $C^2{n}^2-4AMgh\cos \alpha \ge 0$. At horizontal inclination, the equation is linear: if $Cn\ne 0$, then $\Omega =Mgh/(Cn)$. If $Cn=0$ and $Mgh\ne 0$, there is no steady horizontal solution. Vertical inclinations require the original undivided equation instead. The moments of inertia and lever arm $h$ must all refer to the same fixed pivot, rather than mixing central and pivot values.