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For $x$ in the derived algebra of a complex matrix Lie algebra $L$, define $y$ by multiplication by $\overline{\lambda }$ on each generalized eigenspace $V_{\lambda }$ of $x$. Polynomial interpolation and adjoint compatibility of additive Jordan decomposition make $\mathrm{ad}y$ a polynomial in $\mathrm{ad}x$ with zero constant term, so $[y,L]\subseteq [L,L]$. Trace orthogonality of $[L,L]$ and $L$ forces the displayed sum to vanish. Thus $x$ is a nilpotent endomorphism, and the Engel theorem implies solvability. The auxiliary $y$ and the Jordan components are not required to lie in $L$.

A finite-dimensional complex Lie algebra is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form is the symmetric bilinear formThe cyclic trace identity and ${\mathrm{ad}}_{[z,x]}=[{\mathrm{ad}}_z,{\mathrm{ad}}_x]$ giveThus the Killing form is an invariant bilinear form on a Lie algebra. In particular its radical $R=\{x:B_L(x,L)=0\}$ is an ideal.

We supply the trace argument needed for nondegeneracy rather than assuming the Cartan criterion for semisimplicity. The matrix form of the Cartan solvability criterion says: if $A\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ and $\mathrm{tr}(uv)=0$ for every $u\in [A,A]$, $v\in A$, then $A$ is solvable. To prove this direction, fix $u\in [A,A]$. Let $V={⨁}_{\lambda }{V}_{\lambda }$ be its generalized eigenspace decomposition, and define $s$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, the semisimple part of ${\mathrm{ad}}_u$ has eigenvalue $\lambda -\mu$, whereas ${\mathrm{ad}}_s$ acts by $\overline{\lambda }-\overline{\mu }=\overline{\lambda -\mu }$. Polynomial interpolation on the finitely many eigenvalues, with all derivatives through the sizes of the nilpotent blocks set to zero, therefore givesHere the same difference always has the same conjugate, so the interpolation is consistent; the prescribed zero derivatives remove every nilpotent block. Since ${\mathrm{ad}}_u(A)\subseteq [A,A]$, it follows that $[s,A]\subseteq [A,A]$.

Write $u={\sum }_j[a_j,b_j]$. By cyclicity and the trace hypothesis,On the other hand, $\mathrm{tr}(us)={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$. Thus all eigenvalues of $u$ vanish, and every element of $[A,A]$ is nilpotent. The Engel theorem makes $[A,A]$ nilpotent as a Lie algebra, hence solvable; $A/[A,A]$ is abelian, so $A$ is solvable. This is the Conjugate-spectrum proof of Cartan solvability.

Apply this to the Killing radical $R$. For $x,y\in R$, the adjoint actions preserve $R$ and act as zero on $L/R$, because $R$ is an ideal. Computing traces in a basis adapted to $R$ givesThe matrix Lie algebra ${\mathrm{ad}}_R(R)$ consequently satisfies the trace criterion and is solvable. Its kernel is $Z(R)$, an abelian ideal; a central extension of a solvable algebra is solvable. Thus $R$ is solvable. This proves the reusable assertion that the Killing radical is a solvable ideal. Semisimplicity forces $R=0$, and hence $B_L$ is nondegenerate.

A Cartan subalgebra is a nilpotent subalgebra $H$ which is self-normalizing:For a complex semisimple algebra this is equivalently a maximal toral subalgebra. The nilpotent, self-normalizing definition permits a proof of the restricted nondegeneracy without first assuming the toral characterization.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the adjoint action of $H$:For completeness, the stability underlying this decomposition follows directly from nilpotence of $H$. For fixed $h$, every ${\mathrm{ad}}_h$ on $H$ is nilpotent. If $T=\rho (h)$ and $S=\rho (y)$ in a finite-dimensional representation, then $({\mathrm{ad}}_T{)}^{r}S=0$ for sufficiently large $r$. The identityshows that $S$ preserves each generalized eigenspace of $T$. Starting with a basis of $H$, refine these primary decompositions successively; all summands remain $H$-invariant. Each resulting summand has only one eigenvalue for each basis element. The Lie theorem triangularizes the action on that summand, so those eigenvalues extend to a single linear character $\lambda$ on all of $H$. This proves the displayed decomposition and nilpotence of all shifted operators there.

Since $H$ is nilpotent, $H\subseteq L^0$. The space $L^0$ is a subalgebra: repeated use of the derivation rule for ${\mathrm{ad}}_h$ shows that the bracket of two generalized zero-eigenvectors is another such vector. If $L^0/H\ne 0$, the adjoint action of $H$ on this quotient consists entirely of nilpotent maps. The Engel theorem supplies a nonzero coset $x+H$ with $[H,x]\subseteq H$, contradicting self-normalization. Therefore the zero generalized weight space of a Cartan subalgebra is exactly $H$.

Finally, $B_L(H,L^{\lambda })=0$ when $\lambda \ne 0$. Choose $h\in H$ with $\lambda (h)\ne 0$ and put $T={\mathrm{ad}}_h$. On $H$, a power $T^m$ vanishes. On $L^{\lambda }$, $T$ is invertible. For $u\in H$ and $v\in L^{\lambda }$, write $v=T^{m}w$; invariance givesIf $u\in H$ is orthogonal to $H$, it is now orthogonal to every summand of $L$, so nondegeneracy of $B_L$ implies $u=0$. The restriction $B_L{|}_{H\times H}$ is nondegenerate. This establishes the nondegeneracy of the Killing form on a Cartan subalgebra for every Cartan subalgebra, without requiring a chosen root basis.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincare-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.

Let $N$ be the sum of all proper submodules. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper submodule and is therefore the unique maximal proper submodule. Its quotientis irreducible, and the nonzero image of $v_{\omega }$ is primitive of weight $\omega$. This proves existence for every $\omega \in H^{\ast }$ and gives the irreducible quotient of a Verma module. It also proves uniqueness up to isomorphism among irreducible modules generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega )$, whose proper kernel must be $N$.

These modules are not generally finite-dimensional. For example, in an ${\mathfrak{sl}}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfiesIf its irreducible highest-weight module is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^{m}v\ne 0$, the identity forces $\lambda =m\in {\mathbb{Z}}_{\ge 0}$. In general a finite-dimensional highest-weight module therefore requires nonnegative integral values on all simple coroots; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.

For simple coroots $h_1,\dots ,h_r$ forming a basis of $H$, the fundamental weights are the dual basis ${\omega }_1,\dots ,{\omega }_r\in H^{\ast }$:For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$, take the diagonal trace-zero Cartan subalgebra and positive roots corresponding to upper triangular matrix units. Write ${\epsilon }_i(\mathrm{diag}(t_1,t_2,t_3))=t_i$. The simple roots are ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$, and the simple coroots areThe fundamental weights of sl3 are thereforeChecking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean trace-zero plane they can also be writtenThe Fundamental representations of sl3 make these weights concrete: ${\mathbb{C}}^3$ has primitive vector $e_1$ of weight ${\omega }_1$, and ${\Lambda }^2{\mathbb{C}}^3$ has primitive vector $e_1\wedge e_2$ of weight ${\omega }_2$. Every finite-dimensional irreducible highest weight is $a{\omega }_1+b{\omega }_2$ with nonnegative integers $a,b$.

On $T=V^{\otimes n}$, define the permutation and diagonal actions byThe inverse in the first formula gives a left action. Applying $g$ to every tensor factor and then reordering produces the same tensor as reordering and then applying $g$. Thus the two actions commute. The Schur algebra isIt is the commutant of an operator algebra of the permutation action.

To identify this commutant, use the multilinear isomorphismConjugation by a permutation reorders the factors on the right. Therefore the invariant subspace is the space of symmetric tensors of degree $n$ in $\mathrm{End}(V)$. It is spanned by $a^{\otimes n}$. Indeed, the polarization identityexpresses every symmetrized elementary tensor as a linear combination of pure powers. Those symmetrized elementary tensors span the invariant subspace in characteristic zero.

One may restrict $a$ to invertible endomorphisms without changing the span. For fixed $a$, the vector-valued polynomial $(a+tI{)}^{\otimes n}$ has degree at most $n$. Choose $n+1$ distinct values of $t$ away from the finitely many roots of $\det (a+tI)$. Polynomial interpolation expresses its constant term $a^{\otimes n}$ as a linear combination of those invertible tensor powers. ConsequentlyThe span is an algebra, since products are $(gh{)}^{\otimes n}$.

The image $A_T$ of $\mathbb{C}{S}_n$ is semisimple, as a quotient of a semisimple algebra. The double-centralizer theorem for semisimple operator algebras gives $A_T^{\prime \prime }=A_T$. Since the preceding computation says that $A_T^{\prime }$ is the span of the general linear action, the two actions are mutual commutants. More explicitly, semisimple module decomposition givesHere $D_{\lambda }(V)={\mathrm{Hom}}_{S_n}(S^{\lambda },T)$ is the multiplicity space, with its natural general linear action. The commutant algebra is the product of the full endomorphism algebras of these multiplicity spaces. Hence each nonzero $D_{\lambda }$ is irreducible, and distinct spaces have nonisomorphic general linear representations, because the operators $g^{\otimes n}$ span that commutant. A primitive $e_t$ picks a one-dimensional factor from $S^{\lambda }$, so $e_{t}T$ is an equivalent realization of $D_{\lambda }$ as a Schur module.

The range of shapes is exact. If $\lambda$ has more than $m$ rows, a column antisymmetrizes more than $m$ vectors and its action vanishes by ${\Lambda }^{m+1}V=0$. Conversely, for at most $m$ rows place the $i$th basis vector in every tensor position belonging to row $i$ of $t$. Row symmetrization multiplies this tensor by ${\prod }_i{\lambda }_i!$. Column antisymmetrization is nonzero, because all vectors within each column are distinct, and its different permutations give distinct basis tensors. Thus $h_{t}T\ne 0$. This proves the length bound for a Schur module and the stated Schur–Weyl duality. For $n=0$, the empty partition and $V^{\otimes 0}=\mathbb{C}$ give the trivial version.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

Let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be a finite-dimensional Lie algebra representation. The form denoted $B_V$ isIt is the Trace form of a Lie algebra representation; the Killing form $B$ without a subscript is specifically the case $V=\mathfrak{g}$, $\rho =\mathrm{ad}$. The distinction matters: the form of the trivial representation cannot detect whether the algebra is semisimple.

Linearity of $\rho$ and trace proves bilinearity, and cyclicity of the trace gives symmetry. With $A=\rho (X)$, $D=\rho (Y)$ and $C=\rho (Z)$, the representation identity givesThus it is an invariant bilinear form on a Lie algebra, and in particular the Adjoint representation preserves the Killing form.

The Cartan solvability criterion has two useful formulations. For a finite-dimensional complex Lie algebra $\mathfrak{g}$,Its matrix version says that a Lie subalgebra $L\subseteq \mathfrak{gl}(V)$ is solvable precisely when $\mathrm{tr}(xy)=0$ for all $x\in [L,L]$ and $y\in L$. We prove the matrix version first, with ordinary trace in $V$.

If $L$ is solvable, the Lie theorem makes all its matrices upper triangular. Their commutators are strictly upper triangular, so multiplying such a matrix by an upper triangular one still has zero diagonal and hence zero trace. This proves the easy direction.

Conversely assume the trace-orthogonality condition and fix $x\in [L,L]$. We show that all eigenvalues of $x$ vanish. Use its Additive Jordan decomposition $x=x_s+x_n$ with $[x_s,x_n]=0$. On the generalized eigenspace $V_{\lambda }$ of $x$, define an auxiliary endomorphism $y$ to be $\overline{\lambda }I$. This $y$ need not be in $L$; we only need control of its commutator with $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}{x}_s$ acts by $\lambda -\mu$ and $\mathrm{ad}y$ by $\overline{\lambda }-\overline{\mu }$. Choose a polynomial $q$ with $q(0)=0$ and $q(\lambda -\mu )=\overline{\lambda -\mu }$ at the finitely many distinct differences. Polynomial interpolation supplies it because equal differences have equal conjugates. Therefore $\mathrm{ad}y=q(\mathrm{ad}{x}_s)$.

The adjoint compatibility of additive Jordan decomposition identifies $\mathrm{ad}{x}_s$ as the semisimple part of $\mathrm{ad}x$. Elementary Jordan–Chevalley decomposition gives $\mathrm{ad}{x}_s=p(\mathrm{ad}x)$ for a polynomial $p$ with $p(0)=0$. Hence $\mathrm{ad}y$ is a polynomial in $\mathrm{ad}x$ with zero constant term. Since $\mathrm{ad}x$ maps $L$ into $[L,L]$ and preserves that Lie algebra ideal, we obtainNo assumption that $x_s$, $x_n$ or $y$ belongs to $L$ was made.

Write $x={\sum }_i[a_i,b_i]$ with $a_i,b_i\in L$. Cyclicity and the assumed orthogonality now giveOn the other hand the nilpotent parts have zero trace on each generalized eigenspace, soThus all $\lambda$ are zero and $x$ is a nilpotent endomorphism. This is the Conjugate-spectrum proof of Cartan solvability.

The permitted Engel theorem, in the form needed here, states: a finite-dimensional Lie subalgebra of endomorphisms in which every element is nilpotent has a nonzero vector annihilated by all its elements, and iteration on quotients makes every element simultaneously strictly upper triangular. Apply it to $[L,L]$. The strictly upper triangular algebra is nilpotent, so $[L,L]$ is a nilpotent Lie algebra and therefore solvable. Since $L/[L,L]$ is abelian, the derived series of a Lie algebra of $L$ terminates too. This proves the matrix criterion.

Apply it to $L=\mathrm{ad}\mathfrak{g}$. The condition on $B$ is exactly the matrix condition on $L$. Thus $\mathrm{ad}\mathfrak{g}$ is solvable. The kernel of $\mathrm{ad}$ is the center of a Lie algebra, which is abelian, so $\mathfrak{g}$ is solvable as well: once the derived series maps to zero it is central, and its next term vanishes. Conversely a solvable $\mathfrak{g}$ has solvable adjoint image, proving the abstract Cartan solvability criterion in both directions.

A finite-dimensional Lie algebra is a semisimple Lie algebra when it has no nonzero solvable Lie algebra ideal, equivalently its solvable radical is zero. Let $R=\mathrm{rad}B=\{x:B(x,\mathfrak{g})=0\}$. Invariance makes $R$ an ideal. For $x,y\in R$, $\mathrm{ad}x$ induces zero on $\mathfrak{g}/R$, and the block trace gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$, where $B_R$ is the adjoint Killing form of $R$ itself. The Cartan solvability criterion therefore makes $R$ solvable. If $\mathfrak{g}$ is semisimple, $R=0$.

Conversely suppose $B$ is a nondegenerate bilinear form. If a nonzero solvable ideal existed, its last nonzero derived term $A$ would be an abelian ideal of $\mathfrak{g}$. For $a\in A$ and $x\in \mathfrak{g}$, the operator $\mathrm{ad}a\mathrm{ad}x$ has image in $A$ and is zero on $A$, so its square and its trace are zero. Thus $B(A,\mathfrak{g})=0$, contradicting nondegeneracy. This is Abelian ideals lie in the radical of the Killing form. We conclude the Cartan criterion for semisimplicity:

Write $g_t(y)=(y-t{)}_{+}^{k-1}$. Begin with distinct increasing knots so ordinary polynomial interpolation is unambiguous. The two interpolants agree at the $k-1$ common knots $t_{i+1},\dots ,t_{i+k-1}$. Their difference is consequently a scalar multiple of the monic polynomial ${\omega }_i$ of degree $k-1$.

The leading coefficient of an interpolant is its highest divided difference. Thus this scalar isThe equality is precisely the divided-difference recurrence. This proves the Lee interpolation identity:For $k=1$ the common-knot product is empty; the same argument is a difference of constants. Interpret $(y-t{)}_{+}^0$ as $1_{y>t}$ to obtain the usual half-open order-one B-splines. Repeated knots use the standard confluent or limiting interpretation, where the interpolation data are defined.

Sum over $i=1,\dots ,n$. The right side telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. When $t_k<t<t_{n+1}$, the first interpolation nodes all lie below $t$, so ${\ell }_1=0$. The last interpolation nodes all lie above $t$, so ${\ell }_{n+1}$ interpolates the polynomial $(x-t{)}^{k-1}$ of degree $k-1$ and equals it identically. Hence the Marsden identity follows:

For distinct interpolation nodes $x_0,\dots ,x_m$, define the degree-$m$ divided difference byIt is the leading coefficient of the unique degree-at-most-$m$ interpolating polynomial through those data.

DefineThe partial polynomial $p_m$ agrees with $p_{m-1}$ at $x_0,\dots ,x_{m-1}$. Moreover, the explicit formula for the divided difference givesso $p_m(x_m)=f(x_m)$. Induction shows that $p_n(x_j)=f(x_j)$ for every $j$. By uniqueness of polynomial interpolation, this is the required Newton interpolation polynomial.

For the recursion, substitute the explicit formulas for $f[x_1,\dots ,x_m]$ and $f[x_0,\dots ,x_{m-1}]$. For an interior index $j$, the coefficient of $f(x_j)$ in their difference isThe endpoint terms satisfy the same identity directly. Dividing by $x_m-x_0$ therefore gives

Define the cardinal polynomialsThey satisfy ${\ell }_k(x_i)={\delta }_{ik}$. Hence evaluating the proposed polynomial interpolation formula at $x_i$ gives $p_n(x_i)=a_i$, so the required coefficients areThe polynomial ${\sum }_k{f}_k{\ell }_k$ has degree at most $n$ and takes every prescribed value. Its uniqueness follows because the difference of two such polynomials would have $n+1$ distinct roots while having degree at most $n$.

Use Shamir's secret sharing over the finite field ${\mathbb{F}}_p$. The Chair independently chooses $a_1,\dots ,a_{r-1}$ from the uniform distribution on a finite set ${\mathbb{F}}_p$ and forms the random polynomialChoose distinct nonzero $x_1,\dots ,x_s\in {\mathbb{F}}_p$, with $s$ equal to the number of discs to be issued, and inscribe $(x_i,f(x_i))$ on disc $i$.

Any $r$ discs give $r$ distinct values of a polynomial of degree at most $r-1$. Their Vandermonde determinant is nonzero, so polynomial interpolation recovers $f$ uniquely and hence recovers the secret as $N=f(0)$.

Conversely, fix any $r-1$ observed discs and any candidate secret $n\in {\mathbb{F}}_p$. Interpolating through those $r-1$ points and $(0,n)$ gives exactly one polynomial of degree at most $r-1$. Thus every candidate $n$ is compatible with the observed discs in exactly the same number of ways. Because the coefficients were chosen uniformly, the conditional distribution of the secret is unchanged after seeing any $r-1$ discs. This proves the Perfect secrecy of Shamir's threshold scheme.

Use the nodes in the order $0,1,2,3$. Their first divided differences are $4,-2,6$, their second divided differences are $-3,4$, and their third divided difference is $7/3$. The Newton interpolation polynomial is thereforeThis is the unique interpolating polynomial of degree at most three through the four data points.

Shamir's secret sharing places a secret $N$ at the constant term of a random polynomial $f\in {\mathbb{F}}_p[X]$ of degree at most $r-1$ and issues distinct nonzero evaluation pairs $(x_i,f(x_i))$. Any $r$ shares recover $f$ by polynomial interpolation. Given fewer than $r$ shares, every candidate constant term has the same number of compatible coefficient tuples, which gives perfect secrecy.

Let $U\subseteq W\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ be vector subspaces and $M=\{T:[T,W]\subseteq U\}$. If $\alpha \in M$ and $\mathrm{tr}(\alpha \beta )=0$ for all $\beta \in M$, then $\alpha$ is a nilpotent endomorphism.

To prove it, let $\beta$ act by $\overline{\lambda }$ on the generalized eigenspace of $\alpha$ of eigenvalue $\lambda$. Polynomial interpolation on eigenvalue differences, together with adjoint compatibility of additive Jordan decomposition, expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Since $\mathrm{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta \in M$. Taking the matrix trace on each generalized eigenspace yields $0={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$, so every eigenvalue is zero. This is the linear-algebra step behind the Cartan solvability criterion; $U,W$ need not be Lie subalgebras.

For a fixed degree bound, two best approximating polynomials have a best approximating average. At each active point of that average their errors agree, hence the two polynomials agree. There must be at least two more active points than the degree: otherwise polynomial interpolation constructs a strictly improving sign-matching perturbation. The difference has too many zeros and must vanish. Finite-dimensional coefficient compactness also gives existence.