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Let the flat face make angle $\theta$ with horizontal while the rounded face contacts a horizontal plane. The centroid has height $a-(3a/8)\cos \theta$ and contact distance squared $b^2=a^2+(3a/8{)}^2-2a(3a/8)\cos \theta$. For rolling without slipping, its speed is $b|\dot{\theta }|$. Combining translation and rotation gives kinetic energy $\frac{1}{2}(I_G+m{b}^2){\dot{\theta }}^2$. Conservation from rest at $\theta =\pi /2$ gives the displayed result until the flat-face/contact geometry changes.

For a rigid body rolling from rest down a straight incline with fixed rolling radius $R$ and axial moment of inertia $I$, translation and rotation give $(M+I/R^2)\ddot{s}=Mg\sin \theta$. A smooth slide has $\ddot{s}=g\sin \theta$. Travel over the same distance from rest therefore gives the displayed time ratio. Static friction must suffice for rolling without slipping; its contact point is at rest, so this ideal friction does no work. A uniform solid cylinder has $I=M{R}^2/2$ and ratio $\sqrt{3/2}$.

A body ejecting mass at constant rate $Q>0$ and relative exhaust speed $V>0$, against quadratic drag $\alpha u^2$, has remaining mass $m=m_0-Qt$. A short-interval momentum balance gives $m\dot{u}=QV-\alpha u^2$. From rest, separation of variables givesFor final mass $m_f>0$, this formula applies until $t=(m_0-m_f)/Q$. The zero-drag limit is the rocket equation $u_f=V\log (m_0/m_f)$. Two dimensionless variables are $\alpha V/Q$ and $m_0/m_f$.

For fuel mass rate $s>0$ and backward relative exhaust speed $u$, a rocket sliding on a horizontal plane has remaining mass $q(t)$ and satisfies the displayed rocket equation with kinetic friction. Integration over a sliding interval gives $v(t)-v(t_0)=u\log [q(t_0)/q(t)]-\mu g(t-t_0)$. From initial rest, immediate forward sliding additionally requires thrust to overcome static friction. If both friction coefficients equal $\mu$, motion starts only when $su>\mu gq(t)$. An algebraically negative sliding speed is not the physical solution at rest.

Rocket thrust is the propulsive force associated with ejecting rocket propellant. In the ideal rocket equation with no pressure correction, $T=u(-\dot{M})$, where $\dot{M}=dM/dt<0$ is the signed rate of change of rocket mass and $-\dot{M}$ is the positive mass loss rate. For $u=2000m/s$ and $\dot{M}=-5kg/s$, $T=10000\mathrm{N}$. With no other force, the acceleration is $T/M$: initially $10m/{\mathrm{s}}^2$ for $M=1000\mathrm{kg}$, rising as mass falls if this rocket thrust stays constant.

A real exhaust outlet can also contribute a pressure term: $T=u(-\dot{M})+A_e(p_e-p_a)$, with exit area $A_e$, exit pressure $p_e$ and ambient pressure $p_a$. The ideal formula here assumes that term vanishes; it must not silently absorb this correction while still interpreting $u$ as the actual relative rocket exhaust speed. ^[ref]

The rocket mass ratio compares the initial and final total rocket masses for a burn. It is dimensionless, and the ideal rocket equation becomes $\Delta v=u\ln R$. For the same fixed rocket exhaust speed, two successive burns with rocket mass ratios $R_1$ and $R_2$ have combined rocket mass ratio $R_1{R}_2$ and total velocity gain $u\ln R_1+u\ln R_2$.

Rocket propellant is the onboard material expelled to produce rocket thrust. A chemical rocket normally carries both fuel and oxidizer; both contribute to its mass and to the expelled material. In the ideal rocket equation, $M(t)=M_{\mathrm{dry}}+M_{\mathrm{propellant}}(t)$, with fixed rocket dry mass and decreasing remaining rocket propellant. ^[ref]

The rocket exhaust speed is the magnitude of the emitted material's velocity relative to the instantaneous rocket. In the ideal rearward-ejection convention of the rocket equation, a rocket with inertial velocity $v$ emits material with inertial velocity $v-u$ to first order. A constant $u$ does not make $v-u$ constant.

A valuative criterion tests a morphism of schemes $f:X\to Y$ by lifting maps from a valuation ring. Given a valuation ring $R$ with fraction field $K$, a map $\mathrm{Spec}R\to Y$, and a compatible map $\mathrm{Spec}K\to X$, ask whether the latter extends to $\mathrm{Spec}R\to X$. For finite-type morphisms between Noetherian schemes, uniqueness of every such lift characterizes separatedness; existence together with uniqueness characterizes proper morphisms. The test treats the generic point as a moving point and asks whether it has a unique limiting position. A projective morphism satisfies the existence-and-uniqueness property; an open immersion can fail existence when the limit leaves the open set.

For a continuous map $f:X\to Y$, the inverse-image functor takes a sheaf on $Y$ to the sheafification of its local inverse-image presheaf on $X$. It is left adjoint to the direct image functor. For sheaves of modules on ringed spaces, extension of scalars gives the separate pullback of a sheaf of modules.

For a continuous map $f:X\to Y$, the direct-image functor sends a sheaf $\mathcal{F}$ on $X$ to the sheaf $V\mapsto \mathcal{F}(f^{-1}V)$ on $Y$, and sends sheaf morphisms to their induced maps on sections. Its output is a direct image sheaf.

An element $x$ of a ring is nilpotent when $x^n=0$ for some positive integer $n$. A field has no nonzero nilpotent elements.

The opposite ring has the same additive group and identity as a ring $R$, with multiplication $a{\cdot }_{\mathrm{op}}b=ba$. Left modules over it are right modules over $R$. Forming the opposite exchanges left and right chain conditions and reverses the order of composition when interpreting endomorphisms as right multiplications. For an associative algebra with its scalar action retained, this construction gives the opposite algebra.

A commutative ring of prime characteristic $p$ is perfect in the Frobenius sense when its Frobenius endomorphism is an automorphism. It is necessarily reduced: if $a$ is nilpotent, some $p$-power iterate is zero, and injectivity gives $a=0$. This use of perfect is distinct from the homological notion of a perfect ring. The Absolute Frobenius morphism of a scheme is an isomorphism exactly when all its local rings are perfect in this sense, because it is the identity on points and is an isomorphism on every stalk.

The ring with a single element, which is both its additive and multiplicative identity. If a unital ring has $0=1$, then every element $r$ equals $r1=r0=0$, so it is this ring. Under the usual definition it is excluded from integral domains and fields. The quotient ring $\mathbb{Z}/(1)$ is an example.

The matrix ring consists of all $n\times n$ matrices over a ring $R$, using entrywise addition and matrix multiplication. Its identity element is the identity matrix. Even when $R$ is a commutative ring, this ring need not have commutative multiplication.

Suppose $A\subseteq B$ are unital rings and $B$ is a finitely generated module over $A$ on the right. If $A$ is a right Noetherian ring, then $B_A$ is a Noetherian module. Every right ideal of $B$ is an $A$-submodule, hence has finitely many generators over $A$; these also generate it over $B$. Thus $B$ is a right Noetherian ring. The same argument works on the left under the corresponding hypotheses. Finite-index group ring extensions give an important example, using coset representatives as a finite module basis.

For an additive contravariant left-exact functor, apply it to a projective resolution of $M$. Reversal of the arrows gives a cochain complex. Comparison maps lift using projectivity, and any two lifts are chain homotopic, so its cohomology is independent of the resolution. The functors are contravariant, vanish on projective modules in positive degrees, and send $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$ to $0\to F(M^{\prime \prime })\to F(M)\to F(M^{\prime })\to R^{1}F(M^{\prime \prime })\to \cdots$. Degreewise split resolutions from the horseshoe lemma yield this sequence.

The finite intersections of maximal right ideals of a right Artinian ring have a minimal member by the descending chain condition. Intersecting it with any other maximal right ideal changes nothing, so it equals the Jacobson radical. If that radical is zero, the diagonal map from the regular module into a finite direct sum of its simple quotients is injective. A submodule of a finite direct sum of simple modules is semisimple and has finite length. This proves semisimplicity of the regular module without assuming Noetherianity first.

Let $K/k$ be an infinite-dimensional field extension. For the triangular ring in the display, the right regular module has length three: $e_{1}T$ has its off-diagonal simple submodule and simple quotient, and $e_{2}T$ is simple. Thus $T$ is right Artinian. Its off-diagonal ideal is, on the left, the infinite-dimensional $k$-vector space $K$. Every $k$-subspace determines a left ideal, so an ascending sequence of finite-dimensional subspaces violates left Noetherianity. For example take $K=k(z)$ and the spans of $1,z,\dots ,z^m$.