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A field is $C_i$ if every homogeneous polynomial of degree $d\ge 1$ in more than $d^i$ variables has a nonzero zero. Algebraically closed fields are $C_0$, finite fields are $C_1$, and the Lang-Nagata theorem for Ci fields controls function-field extensions.

For a finite-field Kakeya set $A\subseteq {\mathbb{F}}_q^n$, a smaller cardinality would give a nonzero polynomial of total degree of a polynomial at most $q-1$ vanishing on $A$, by the dimension of a bounded-total-degree polynomial space. Its restriction to each complete affine line in a vector space has $q$ roots of a polynomial and degree less than $q$, hence vanishes identically. The top homogeneous polynomial part then vanishes in every direction. A polynomial of degree less than $q$ in each variable cannot vanish everywhere on ${\mathbb{F}}_q^n$ unless it is zero, giving a contradiction. For the prime field, this is the polynomial nonvanishing below the field size.

If ${\mathcal{P}}_{\ell }$ is the space of degree-$\ell$ homogeneous polynomials and ${\mathcal{H}}_{\ell }$ its harmonic polynomials, then ${\mathcal{P}}_{\ell }={\mathcal{H}}_{\ell }\oplus |x{|}^2{\mathcal{P}}_{\ell -2}$. Under the Fischer inner product, the second summand is orthogonal to the kernel of the Laplacian. Iteration expresses every polynomial restricted to the unit sphere as a sum of spherical harmonics.

For a plane curve defined by a homogeneous polynomial $F$ of degree $d\ge 2$, assume the field characteristic is different from two and $d-1$ is nonzero in the field. A smooth point of a variety is a flex exactly when the determinant of the homogeneous Hessian matrix vanishes there, provided its tangent line is not a component. Put the point at $[0:0:1]$ with tangent $y=0$ and $F_y(p)\ne 0$. The homogeneous identities give $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, so the determinant is $-(d-1{)}^2{F}_y(p{)}^2{F}_{xx}(p)$. In the completed local ring, write the curve as a formal power series $y=h(x)$. Its linear coefficient is zero and its quadratic coefficient is $-F_{xx}(p)/(2F_y(p))$. Vanishing of that coefficient is exactly intersection multiplicity at least three with the tangent, proving the criterion in these positive characteristics as well as in characteristic zero.

For a cubic homogeneous polynomial $F$, its Hessian matrix has linear entries, so its determinant is a cubic homogeneous polynomial, possibly zero. On a nonsingular plane cubic in characteristic different from two, its zeros are exactly the flexes, by the Hessian criterion for a flex. If the determinant is nonzero, Bézout's theorem supplies intersections; if it is zero, the local criterion itself supplies flexes.

For a homogeneous polynomial $F$ of degree $d\ge 2$, the determinant of its three-by-three Hessian matrix is homogeneous of degree $3(d-2)$ if nonzero. Its zero locus is the Hessian curve; it can be reducible or degenerate, and a nonzero constant for $d=2$ has empty zero locus. Over $\mathbb{C}$, its intersections with the nonsingular locus of $V(F)$ are precisely the flexes, by the Hessian criterion for a flex, as long as the tangent is not a component.

A function is homogeneous of degree $k$ when rescaling its arguments by a common scalar $\lambda$ rescales its value by ${\lambda }^k$, whenever the arguments and that power are defined. For integer $k$ over a vector space, one can require every nonzero scalar; a homogeneous polynomial gives an example. On a domain stable under positive real rescaling, the positive-scaling version allows real $k$. Degree-one positive homogeneity is the convention used for norms and many convex functions. The Euler theorem for homogeneous functions turns this scaling law into a differential identity for a differentiable function.

A degree-$n$ homogeneous polynomial satisfies Euler homogeneous function theorem $\mathbf{r}\cdot \nabla H_n=n{H}_n$. The product rule gives $\Delta (r^s{H}_n)=r^s\Delta H_n+s(2n+s+1)r^{s-2}{H}_n$ in three dimensions. Taking $s=-(2n+1)$ gives the displayed identity and the equivalence of harmonicity on $r>0$. The origin is excluded; even $1/r$ has a distributional Dirac delta function source there.

An ideal in a graded ring is homogeneous when it is generated by homogeneous elements, equivalently when every homogeneous component of each of its elements also belongs to the ideal.

The local frame $X_i^m$ on $X_i\ne 0$ identifies a section with a homogeneous rational function of degree $m$ on the punctured affine cone. Homogenizing a local rational expression and cancelling common factors writes every nonzero section as a quotient of coprime homogeneous polynomials $F/G$ with $\mathrm{deg}F-\mathrm{deg}G=m$. Conversely, such a rational function regular on the cone gives local coefficients $r/X_i^m$, which glue with the twisting sheaf's transition functions of a vector bundle.

If a degree-$2d$ homogeneous polynomial is a sum of squares polynomial, it has a representation as squares of degree-$d$ homogeneous polynomials. In any sum of squares, the highest-degree terms cannot cancel, since their homogeneous part is itself a sum of squares. Likewise the lowest nonzero degree cannot cancel. Thus every nonzero summand has only degree $d$.

For a hyperplane $H\subset {\mathbf{P}}^n$ with inclusion $i$, multiplication by its defining linear homogeneous polynomial gives the short exact sequence of sheavesThe long exact sequence in sheaf cohomology relates twists in consecutive degrees and consecutive dimensions. Surjectivity of restriction on global homogeneous polynomials, together with eventual positive-twist vanishing, proves vanishing of intermediate cohomology by descending induction. The top-degree dimensions satisfy $h_n(m-1)=h_n(m)+h_{n-1}(m)$ for $n\ge 2$, where $h_r(m)=\dim H^r({\mathbf{P}}^r,\mathcal{O}(m))$.

Write $C=V(P)$, where $P\in \mathbb{C}[x,y,z]$ is a squarefree polynomial and a homogeneous polynomial of degree $d$. A projective point $p\in C$ is a singular point precisely whenIf $p=[p_0:p_1:p_2]$ is a nonsingular point of an algebraic curve, its projective tangent line isThe coefficients are not all zero, and the Euler theorem for homogeneous functions gives $p_0{P}_x(p)+p_1{P}_y(p)+p_2{P}_z(p)=dP(p)=0$, so the line passes through $p$. Changing the representative of the projective point only rescales the equation.

The strong form of Bézout's theorem says that two projective plane curves $C,D$ of degrees $d,e$, with no common irreducible component, have finitely many intersections andEach intersection multiplicity is a positive integer. At $p\in C\cap D$,This is the criterion for a transverse intersection of plane curves. In particular, an intersection at a singular point of either curve contributes at least two.

If $d=1$, $C$ is a line and has no singular points. Suppose $d>1$ and take $P$ to be an irreducible polynomial. First, $P_x$ is not the zero polynomial: otherwise $P\in \mathbb{C}[y,z]$, and a homogeneous polynomial in two variables of degree greater than one factors into linear factors over $\mathbb{C}$, contrary to irreducibility.

Let $D=V(P_x)$ be the polar curve in this coordinate direction. Since $\mathrm{deg}{P}_x=d-1<d$, the irreducible polynomial $P$ cannot divide $P_x$. Thus $C,D$ have no common irreducible component, and Bézout's theorem applies. One may count repeated factors of $P_x$ with their multiplicities; alternatively, passing to its reduced equation only decreases $\mathrm{deg}D$.

Every singular point $p$ of $C$ satisfies $P_x(p)=0$ and has $I_p(C,D)\ge 2$ by part (a). Consequently, if $s$ is the number of distinct singular points,This argument also proves that the singular locus is finite.

There is a nonsingular point of an algebraic curve on $C$, since part (b) makes its singular locus finite whereas a projective curve has infinitely many complex points. Apply a projective transformation taking one such point to $p=[1:0:0]$, and use $P$ for the transformed homogeneous polynomial. As in part (b), $P_x$ is nonzero and $D=V(P_x)$ has no common irreducible component with $C$.

The Euler theorem for homogeneous functions and its differentiated identities giveandSince $p$ is nonsingular on $C$, at least one of $P_y(p),P_z(p)$ is nonzero. Because $d>1$, the polar curve $D$ is also nonsingular at $p$, and these identities show that $T_{p}D=T_{p}C$. Hence $I_p(C,D)\ge 2$.

This additional intersection is separate from all the singular points of $C$. Thus Bézout's theorem givesThe improvement comes from making the polar curve tangent at a chosen nonsingular point, thereby reserving at least two of the available intersections.

A nonsingular point of an algebraic curve $p$ is an inflection point of an algebraic plane curve, or flex, if its projective tangent line has contact of order at least three:Here the projective tangent line must not be a component of $C$. Ordinary tangency has contact order two.

Let $C=V(F)$, where $F$ is a homogeneous polynomial of degree $d\ge 2$. Its Hessian curve is defined byEach entry has degree $d-2$, so a nonzero $H_F$ has degree $3(d-2)$. By the Hessian criterion for a flex, the flexes are exactly the points of $C\cap V(H_F)$. The stated no-common-component assumption and Bézout's theorem therefore giveThis counts distinct points; higher contact can use more than one unit of intersection multiplicity at a single point.

For completeness, the Hessian criterion for a flex follows by putting $p=[0:0:1]$ and $T_{p}C=\{y=0\}$, with $F_y(p)\ne 0$. The Euler theorem for homogeneous functions gives $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, whenceThe restriction $F(x,0,1)$ already has vanishing constant and linear terms. Its quadratic term vanishes exactly when the intersection multiplicity with the tangent is at least three, proving the criterion.

For $d\ge 3$ the bound is the actual maximum number of flexes of a smooth plane curve: a general nonsingular curve has only contact of order three, and the corresponding intersections with its Hessian curve are simple. One way to see genericity is to vary a point and a line through it, a three-dimensional family. Contact of order at least four imposes four independent linear conditions on the coefficients of a degree-$d$ equation for $d\ge 3$. Thus the locus of equations with such contact is a proper exceptional locus. Locally, if the curve is $y=h(x)$, its Hessian curve restricted to $C$ is a nonvanishing factor times $h^{\prime \prime }(x)$; contact exactly three gives a simple zero.

A nonsingular plane conic has no flexes, agreeing with the formula at $d=2$. A line is excluded from this formulation: its tangent is itself and its Hessian matrix is zero, so the no-common-component hypothesis fails. The formula is therefore intended for $d\ge 2$, rather than giving the meaningless negative value at $d=1$.

A projective variety is a Zariski-closed subset of projective space ${\mathbb{P}}_k^n$, equivalently the common zero set of homogeneous polynomials. Homogeneity makes their vanishing independent of the nonzero scalar used to represent a projective point.

A morphism is a map locally given by regular coordinate functions in affine charts. Equivalently, near each point of a projective source embedded in ${\mathbb{P}}^n$, a morphism to ${\mathbb{P}}^m$ can be writtenusing homogeneous polynomials of a common degree that are not all zero on that neighbourhood. The expressions must agree on overlaps, and for a specified target variety their values must lie in it. A single global tuple is a sufficient description when available, but is not part of the general definition.

At an equilibrium point, $x=-y^2$ and $x^2=a^{2}y$, so $y(y^3-a^2)=0$. With $d=|a{|}^{2/3}>0$, the two equilibrium points and their Jacobian matrix areAt $(-d^2,d)$, $\det J=-3a^2<0$. The trace-determinant stability criterion therefore gives an unstable saddle equilibrium. At the origin, the eigenvalues are $\lambda =\pm i|a|$: its linearization is a center equilibrium. The linearization stability theorem does not classify this nonhyperbolic equilibrium of the nonlinear system.

Nonlinear classification. In fact the answer depends on $|a|$:Here is an explicit derivation, rather than an inference from the imaginary eigenvalues. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the first integralhasIts Hessian matrix at the origin is the identity. Small level curves are closed ovals on which the vector field is nonzero, proving the center equilibrium classification and Lyapunov stability.

For $a^2\ne 1$, a higher-order Lyapunov function calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau =kt$, $p=k^{-1}$ and $q=k^{-2}$. ThenWrite $R^2=U^2+V^2$ and introduce the differential operatorsSuccessively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these homogeneous polynomials:Direct differentiation verifies $\mathcal{L}{H}_3=-\mathcal{N}{R}^2$, $\mathcal{L}{H}_4=-\mathcal{N}{H}_3$ and $\mathcal{L}{H}_5=-\mathcal{N}{H}_4$. The remaining degree-six term has circular averageFor example, the only nonzero averages of degree-six monomials needed here are $⟨U^6⟩=⟨V^6⟩=5/16$ and $⟨U^4{V}^2⟩=⟨U^2{V}^4⟩=1/16$. Since $\mathcal{L}={\partial }_{\theta }$ in polar coordinates, the zero-average part can be cancelled by a degree-six homogeneous polynomial $H_6$ satisfyingIndeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive Lyapunov function $W=R^2+H_3+H_4+H_5+H_6$ satisfiesFor $k>1$ this is strictly negative near the origin, giving asymptotic stability. For $k<1$ it is strictly positive, and $W_{\tau }\ge C{W}^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally ${\theta }_{\tau }=1+O(R)>0$, so the nearby trajectories rotate: these are weak foci, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne 0$ would miss this distinction.

On a monomial $X^a{Y}^b$, direct differentiation givesFor example $(EF-FE)(X^a{Y}^b)=[a(b+1)-b(a+1)]X^a{Y}^b=(a-b)X^a{Y}^b$. The same calculation with $H$ givesThese are exactly the brackets of $h,e,f$, so the linear map $\rho$ respects every bracket and is a Lie algebra representation.

For $m\ge 0$, let $W_m$ be the space of homogeneous polynomials of total degree $m$. Each operator preserves total degree. Every polynomial is a finite sum of homogeneous polynomials, soIn $W_m$, the monomials $v_j=X^{m-j}{Y}^j$ have distinct weights $m-2j$ and exactly the actions described in part (i). To see simplicity directly, a nonzero invariant subspace is invariant under $H$ and hence contains one of its weight components: polynomial interpolation in $H$ projects onto each of its distinct eigenspaces. Starting with a nonzero multiple of $v_j$, repeated application of $E$ reaches $v_0$, and repeated application of $F$ then reaches all $v_j$. All coefficients used are nonzero in characteristic zero. Therefore $W_m$ is simple and isomorphic to $L_m$.

ThusThis is an algebraic direct sum, not an infinite product or completion. Its simple submodules are precisely the $W_m$: they are pairwise nonisomorphic and each occurs once, so projection onto homogeneous degrees and Schur's lemma exclude diagonal copies between distinct degrees. Complete reducibility here follows from the explicit decomposition, rather than applying a finite-dimensional theorem directly to the infinite-dimensional $W$.

Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.