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For a monic polynomial with roots $r_i$, its discriminant is ${\prod }_{i<j}(r_i-r_j{)}^2$ and vanishes exactly for a repeated root.

Writing $U=\mathrm{Re}(\beta +\gamma )$ and $V=\mathrm{Im}(\beta +\gamma )$, the squared amplitude is $z_{\pm }=[\mu U\pm \sqrt{(U^2+V^2){\nu }^2-{\mu }^2{V}^2}]/(U^2+V^2)$. Reality of the square root does not guarantee positivity. For example, $U=1,V=0,\mu =-1,\nu =1/2$ gives two negative roots. Under the usual supercritical assumptions $\mu >0,U>0$, the plus root is positive whenever the discriminant is nonnegative.

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For a quadratic equation $a{x}^2+bx+c=0$ with $a\ne 0$, completing the square givesOver the complex numbers both roots exist, with a repeated root of a polynomial when the discriminant $b^2-4ac$ is zero. Over the real numbers the discriminant determines whether there are two distinct roots, one repeated root, or no real root.

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.

For an admissible quantum state with zero position and momentum means, positivity of the squared norm $\Vert (p-isx)\psi {\Vert }^2$ for every real $s$ givesThe discriminant is nonpositive, hence $\Delta x\Delta p\ge \hslash /2$. Subtracting the expectation values from the position operator and momentum operator extends this outline to nonzero means.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

For independent $U,V$ uniform on $[0,1]$, real roots occur on $U\le V^2$, a region of probability $1/3$. On that region both roots are nonpositive. Both lie in $[-1,0]$ exactly when $\max (0,2V-1)\le U\le V^2$. Its area is $1/12$, so the conditional probability of both absolute values being at most one is $1/4$. Joint probability density integration turns algebraic discriminant and root constraints into simple planar areas.

For $K=\mathbb{Q}(\sqrt{-6})$, reduction of $\mathbb{Z}[\sqrt{-6}]$ modulo $(\sqrt{-6})$ gives $\mathbb{Z}/6\mathbb{Z}$. Its unit group is covered by the units $\{1,-1\}$ of $K$. The ray class exact sequence therefore identifies the ray class group with the ordinary ideal class group, of order two. Consequently this ray class field is the Hilbert class field of Q of square root minus six. To verify its lack of ramification, the integral basis candidate $1,\sqrt{2},b,\sqrt{2}b$, where $b=(1+\sqrt{-3})/2$, has discriminant $8^2(-3{)}^2=576$. The relative discriminant formula also forces the field discriminant to be divisible by $(-24{)}^2=576$. Equality proves that the relative discriminant ideal is trivial.

Write $f(x)=a_m{\prod }_{i=1}^m(x-{\alpha }_i)$ and $g(x)=b_n{\prod }_{j=1}^n(x-{\beta }_j)$ in an algebraic closure, counting multiplicity of a root. The coefficient-row convention above givesThis makes the shared-root of a polynomial test transparent: a factor vanishes precisely when one ${\alpha }_i$ equals one ${\beta }_j$. Swapping the two polynomials reverses the sign of all $mn$ differences, giving $\mathrm{Res}(g,f)=(-1{)}^{mn}\mathrm{Res}(f,g)$.^[ref]

The value measures an algebraic product, not a normalized geometric distance. Replacing $f$ by $cf$ for nonzero $c\in K$ leaves its roots of a polynomial unchanged but multiplies the resultant by $c^n$. Therefore a small numerical resultant alone does not quantify how close two sets of roots of a polynomial are. Exact zero testing and numerical sensitivity are different questions.

A multiple root of a polynomial is also a root of a polynomial of its formal derivative. In characteristic zero, for $\mathrm{deg}f=m\ge 2$, the associated discriminant isFor example, $h(x)=x^2+s$ gives $\mathrm{Res}(h,2x)=4s$ and $\mathrm{Disc}(h)=-4s$. The two roots of a polynomial merge precisely at $s=0$.^[ref]

Setting $a(x-h{)}^2+k=0$ gives $(x-h{)}^2=-k/a$. A positive $-k/a$ gives two distinct roots of a polynomial in the real numbers; zero gives a repeated root of a polynomial at $h$; a negative value gives a pair of complex conjugate roots of a polynomial. For the downward-opening example, the two roots of a polynomial are $2\pm \sqrt{3/2}$, approximately $0.775255$ and $3.224745$.

The discriminant $\Delta =b^2-4ac$ packages the same calculation, becauseSince $4a^2>0$, the sign of $\Delta$ determines which of the three cases occurs. This derives the quadratic formula by a polynomial identity, and explains its structure: first shift to $h$, then take a square root. For the upward-opening example, $(x+3{)}^2=-4$ gives the complex conjugate roots of a polynomial $-3\pm 2i$.

Model the assembly as a heavy symmetric top with a fixed frictionless support, a uniform gravitational field, and centre of mass on its symmetry axis. Assume constant spin, negligible damping, and initial conditions prepared for uniform precession of a heavy symmetric top. Large spin means the transverse angular momentum from precession is small: $I_1|\Omega |\sin \theta \ll |I_3{\omega }_3|$. Also require $|\Omega |\ll |{\omega }_3|$. A large spin rate alone does not select steady motion after an arbitrary release.

For an original illustrative calculation, stipulateThe transverse principal moments of inertia $I_1,I_2$ are about the pivot, including the parallel axis theorem contribution. These inputs giveThe ratio ${\Omega }_0/{\omega }_3=0.00368$ is small; the more direct angular momentum ratio $I_1{\Omega }_0\sin 60^{\circ }/L_s=0.00358$ is small too. These are model calculations, not experimental results.

Check the approximation against the existing exact uniform precession of a heavy symmetric top condition. Its total angular momentum isAt constant $\theta ,\Omega ,L_s$, differentiating this expression and equating it to the gravitational torque givesFor $L_s>0$ the slow solution can be evaluated without subtracting two nearly equal numbers:It exceeds ${\Omega }_0$ by about $0.208\%$. The other solution is about $177.409rad/s$ and violates the assumed slow precession; the quadratic's two roots must not be conflated. Its discriminant must be nonnegative for these steady solutions to exist.

The original curves hold $I_1,I_3,\theta$ fixed and vary spin and the imposed torque coefficient $K$ in ${\mathbf{\tau }}_O=K\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. For gravity alone, $K=Mgd$; the other curves are hypothetical changes in the transverse torque, not observations of this wheel. Solid curves use the slow quadratic root; dashed curves use $K/(I_3{\omega }_3)$. The second panel displays their relative difference. Increasing spin improves this approximation for the parameters shown. It does not remove nutation caused by inappropriate initial conditions.

The discriminant of an integral full-rank lattice is the field discriminant multiplied by the square of its index in the ring of integers. A squarefree lattice discriminant therefore forces index one.

A squarefree polynomial discriminant does not ensure that the polynomial defines a field of its stated degree. For example, $x^3-2x-1=(x+1)(x^2-x-1)$ has discriminant five. Its roots have degrees one or two, so no root generates a cubic number field. In particular $1,\alpha ,{\alpha }^2$ cannot be a field basis for any of its roots. The squarefree discriminant certifies an integral basis argument assumes that the polynomial is the minimal polynomial: then the discriminant of its power basis is the field discriminant times the square of the order index. For a reducible squarefree polynomial the analogous rank belongs instead to its quotient algebra, which is a product of fields, not to one root field.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.

Choose a signed angular coordinate along the source-lens line. The point-mass gravitational lens becomesFor $\beta \ne 0$, multiplication by $\theta$ gives the quadratic equation ${\theta }^2-\beta \theta -{\theta }_E^2=0$. Its positive discriminant yields the two displayed roots. Their product is ${\theta }_{+}{\theta }_{-}=-{\theta }_E^2$, so they lie on opposite sides of the lens. With $\beta >0$, ${\theta }_{+}>{\theta }_E$ and $-{\theta }_E<{\theta }_{-}<0$: the outer image lies on the source side. For $\beta <0$, the outer image is instead the minus root. These are image directions seen by the observer, not two emitting objects.^[ref]

For a dimensionless example, set $u=\beta /{\theta }_E=0.6$ and $x=\theta /{\theta }_E$. Solving $u=x-1/x$ gives $x_{+}\simeq 1.344$ and $x_{-}\simeq -0.744$, with separation $\sqrt{u^2+4}\simeq 2.088$. The lensing magnification follows from image parity in gravitational lensing: ${\mu }_{+}\simeq 1.442$, ${\mu }_{-}\simeq -0.442$, and total amplification $\simeq 1.884$. Changing $M$ or the distances rescales the angles through the Einstein angular radius, while these dimensionless values remain the same.

For $u\gg 1$, the quadratic equation gives $x_{+}\simeq u+1/u$ and $x_{-}\simeq -1/u$. Using $\mu =1/(1-x^{-4})$, the outer image tends to unit lensing magnification and the inner image has $|{\mu }_{-}|\simeq u^{-4}$. Two mathematical images can therefore look like one at finite sensitivity or angular resolution. Furthermore, the inner impact parameter eventually becomes too small for the stated weak-field approximation or passes through the opaque lens: the ideal formula must then be replaced by an appropriate physical model.

Let $R$ be an integrally closed domain and $f\in R[X]$ monic, with discriminant outside a prime ideal $P$. The roots stay distinct after reduction. In the splitting algebra choose a prime above $P$. Its decomposition group maps onto the Galois group of the residue splitting field; its inertia group fixes every reduced root and is trivial because reduction bijects the distinct root sets. Thus the residue Galois group embeds in the generic group as a permutation group. For finite residue fields, irreducible factor degrees give the Frobenius cycle type.