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Let $I$ denote the identity endomorphism. The two-root minimal polynomial gives $(\alpha -{\lambda }_{1}I)(\alpha -{\lambda }_{2}I)=0$. Every vector has the decompositionThe first summand lies in the eigenspace $E_1=\ker (\alpha -{\lambda }_{1}I)$, and the second in $E_2$. If a vector belongs to both, subtracting its two eigenvalue equations gives $({\lambda }_1-{\lambda }_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.

For distinct real roots ${\lambda }_1,\dots ,{\lambda }_m$, define polynomial projectorsThe Lagrange interpolation polynomials in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence ${\sum }_i{P}_i=I$. Also $(\alpha -{\lambda }_{i}I)P_i=0$, because its numerator is the minimal polynomial evaluated at $\alpha$. Thus every vector is a sum of eigenvectors, $v={\sum }_i{P}_{i}v$. On $E_j$, $P_i$ acts as ${\delta }_{ij}I$, so applying $P_i$ to any zero sum of vectors from the eigenspaces shows that every summand is zero. ThereforeChoosing a basis in each eigenspace proves diagonalizability over the reals. Conversely, if a real matrix is similar over the reals to a diagonal matrix, a polynomial annihilates it exactly when it vanishes at each distinct real diagonal entry. Its minimal polynomial is therefore the product of those distinct linear factors. We have proved real diagonalizability is equivalent to a minimal polynomial with distinct real roots.

The monic orthogonal polynomial $p_n$ is orthogonal to every polynomial of degree less than $n$, since $p_0,\dots ,p_{n-1}$ form a basis of that polynomial space. Let $r_1,\dots ,r_k$ be its distinct interior roots at which its sign changes, and put $q(x)={\prod }_{j=1}^k(x-r_j)$, with $q=1$ if $k=0$. The real polynomial $p_{n}q$ has constant sign throughout $(a,b)$: multiplication by each linear factor removes its corresponding sign change, while even-multiplicity roots do not change sign. It is not identically zero. Since $w>0$, its weighted integral is nonzero.

If $k<n$, orthogonality instead gives ${\int }_a^b{p}_{n}qwdx=0$, a contradiction. Thus $k\ge n$. Degree $n$ also gives $k\le n$, so $k=n$ and all roots have multiplicity one, with no roots left outside the interval. Therefore there are exactly $n$ distinct simple zeros in $(a,b)$. The weighted inner products are understood to exist, as required by the stipulated orthogonality.

Choose a flex as $O=[0:1:0]$, with tangent line $Z=0$. The flex coordinates for a smooth plane cubic giveThe coefficient $a$ is nonzero because $Z=0$ is not a component, and $b\ne 0$ expresses smoothness at $O$. Since the field characteristic is not two, the projective linear changecompletes the square. The equation becomes $Y^{\prime 2}Z=P_3(X,Z)$ after dividing by a nonzero constant, where $P_3$ is a degree-three homogeneous polynomial with nonzero $X^3$ coefficient.

On $Z=1$, the polynomial $P_3(x,1)$ has three distinct roots $r_1,r_2,r_3$ in the algebraically closed field. A repeated root $r$ would make $(r,0)$ a singular point of $y^{\prime 2}=P_3(x,1)$, since both first partial derivatives vanish there. PutA further nonzero scaling of $y^{\prime }$ absorbs the leading coefficient and $(r_2-r_1{)}^3$; the required square root exists in the algebraically closed field. These are projective linear changes of the original coordinates. Renaming the new coordinates gives the Legendre form of an elliptic curveIts projective completion is $Y^{2}Z=X(X-Z)(X-\lambda Z)$, with the chosen flex at infinity.

Every projective line on the smooth algebraic surface through $P$ lies in the projective tangent plane $T_{P}S$, since its tangent direction is contained in the surface's Zariski tangent space. The plane section $S\cap T_{P}S$ is a nonzero plane cubic. It cannot be the entire plane, since a smooth cubic surface is irreducible and has no plane component.

Each distinct projective line through $P$ is therefore a distinct degree-one factor of the cubic restricted to $T_{P}S$. A degree-three polynomial has at most three such factors. Hence there are at most three lines through any point. This is the lines through a point of a smooth cubic surface bound; equality is allowed when the tangent section consists of three concurrent lines.

Choose one projective line from each of two different pairs. They are skew lines by the preceding disjointness result. After a projective linear coordinate change call themFor $A\in L$, $B\in M$, their joining line meets the cubic surface at $A,B$ and a third point. This defines a rational map $L\times M⇢S$ on the Zariski-open set where that third intersection is distinct and the joining line is not contained in $S$.

More explicitly, because $F$ vanishes on both projective lines,The third point is $[VA-UB]$. Both polynomials $U,V$ are nonzero: if either vanished identically, one of the two projective lines would lie in the singular locus. Thus $UV\ne 0$ supplies a nonempty Zariski-open set.

Conversely, a point $P=[x_0:x_1:x_2:x_3]$ outside $L\cup M$ lies on the unique joining line whose endpoints have coordinates $[x_0:x_1]$ on $L$ and $[x_2:x_3]$ on $M$. This is a rational inverse on the same generic locus. The rational parametrization of a cubic surface from two skew lines givesThe latter contains the affine plane as a dense open subset, as does ${\mathbb{P}}^2$. Hence $S$ is birational to ${\mathbb{P}}^2$.

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.

We first give an auxiliary-integral proof, rather than deducing both requested results from a theorem only stated later. The following symmetric Hermite integral obstruction will do both jobs. Supposewhere $k\ne 0$ and $b_j$ are integers, the distinct nonzero algebraic exponents form a Galois-stable set, and their weights are invariant under conjugation. Choose $P\in \mathbb{Z}[X]$ vanishing at those exponents, with $P(0)\ne 0$, and a positive integer $c$ for which every $c{\beta }_j$ is an algebraic integer. For a large prime number $p$, letAt zero, derivatives below $p-1$ vanish; the derivative of order $p-1$ equals $c^{D}P(0{)}^p$. All higher derivatives at zero are integers divisible by $p$. At each ${\beta }_j$, derivatives below $p$ vanish and all higher derivatives belong to $p{\mathcal{O}}_L$ in a Galois splitting field $L$. Indeed their coefficients contain $r!/(p-1)!$, divisible by $p$ for $r\ge p$, and every $c^D{\beta }_j^u$ is integral for $u\le D$. ConsequentlyGalois invariance makes the sum rational, and integrality makes it an integer. For prime numbers not dividing $kcP(0)$ it is nonzero.

But $F_p-F_p^{\prime }=f_p$, so integration along the straight segment to $\beta$ givesMultiply by the weights and use the assumed exponential relation. This expresses $-I_p$ as the sum of these integrals. On the finitely many fixed segments, their absolute values are at most $C^p/(p-1)!$ for a fixed $C$: all polynomial and exponential factors have fixed bounds, and $D$ is linear in $p$. Thus $|I_p|\to 0$, contradicting its being a nonzero integer. The obstruction is proved.

If $e$ were algebraic, an integral polynomial relation would give $a_0+{\sum }_{j=1}^s{a}_j{e}^j=0$ with $a_0\ne 0$. Integer exponents and weights satisfy the obstruction, so $e$ is transcendental.

If $\pi$ were algebraic, put $\alpha =i\pi$ and let ${\alpha }_1,\dots ,{\alpha }_d$ be its algebraic conjugates. Since $1+e^{\alpha }=0$, the product ${\prod }_j(1+e^{{\alpha }_j})$ vanishes. Expand it and collect equal subset sums. The zero sums contribute a positive integer $k\ge 1$; the nonzero sums give a Galois-stable list ${\beta }_j$ with positive integral multiplicities. This is precisely the prohibited relation. Hence $\pi$ is transcendental.

The general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Suppose the sine quotient were algebraic, say $\gamma$. The denominator is nonzero: $\sin \beta =0$ would give a forbidden relation between the exponentials of the distinct algebraic numbers $i\beta ,-i\beta$. The quotient identity would giveThe four exponents are distinct under the stated nonzero and non-opposite assumptions, so this also contradicts linear independence. Therefore

We prove the Mahler alternative in full. Let $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. Its series converges locally uniformly in the unit disc and satisfiesIt is transcendental as a function. At a root of unity $\zeta$ with ${\zeta }^{l^r}=1$, the radial tail ${\sum }_{n\ge r}{t}^{l^n}$ in $f(t\zeta )$ is real and tends to infinity as $t\uparrow 1$, while the initial terms remain bounded. These roots are dense on the unit circle. An algebraic function can have singularities only at the finitely many zeros of its leading coefficient and polynomial discriminant: elsewhere its defining polynomial and the implicit-function theorem give local analytic branches. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Assume now that $\alpha$ and $f(\alpha )$ are algebraic, with $0<|\alpha |<1$, and put both in a number field $K$ of degree $d$. For each $N$, the $(N+1{)}^2$ coefficients of a polynomial $P_N(X,Y)$ of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of $P_N(z,f(z))$. The equations have integer coefficients, so a nonzero integer solution exists after clearing denominators. Functional transcendence makes $R_N(z)=P_N(z,f(z))$ nonzero. Its order $T_N$ at zero satisfies $T_N\ge (N+1{)}^2-1$.

At ${\alpha }_r={\alpha }^{l^r}$ the functional equation gives $f({\alpha }_r)=f(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}\in K$. Thus ${\eta }_r=R_N({\alpha }_r)\in K$ is nonzero for all sufficiently large $r$ andUse the absolute logarithmic height $h$. The elementary inequalities for sums and polynomial evaluation giveHere $h(f({\alpha }_r))\le h(f(\alpha ))+(l^r-1)h(\alpha )/(l-1)+r\log 2$. The Liouville height inequality gives $\log |{\eta }_r|\ge -dh({\eta }_r)$. Divide by $l^r$ and let $r\to \infty$ to obtainThe left grows quadratically in $N$, the right only linearly. Choose $N$ large to contradict this. Hence every such algebraic $\alpha$ has transcendental $f(\alpha )$. This is the auxiliary-value height argument for the Fredholm series.

For the rational continuation, let $a_n={\sum }_{j=0}^n(p/q{)}^{l^j}$ and $Q_n=q^{l^n}$. This is a rational number with reduced denominator $B_n\le Q_n$, and its positive tail satisfies, for large $n$,A rational value $a/b$ would instead have distance at least $1/(b{Q}_n)$ from each distinct $a_n$, already impossible. If the value were algebraic irrational, choose $ϵ>0$ with $2+ϵ<\kappa$. The distinct truncations have unbounded reduced denominators, and the displayed errors are eventually smaller than $B_n^{-2-ϵ}$. This contradicts Roth theorem. Thus the requested rational values are transcendental by the Roth criterion for rational Fredholm values.

For completeness, the elliptic alternative's conclusion also has a short auxiliary-divisor certificate. If $\Phi$ is Frobenius on $E/{\mathbb{F}}_q$, then $\mathrm{deg}\Phi =q$ and $\mathrm{deg}(1-\Phi )=N=\#E({\mathbb{F}}_q)$: the latter map is separable and its kernel is exactly the rational points. The divisor identity for $x(P)-x(Q)$ on $E\times E$ gives the parallelogram rule $\mathrm{deg}(u+v)+\mathrm{deg}(u-v)=2\mathrm{deg}u+2\mathrm{deg}v$. Polarizing it yieldsAll these degrees are nonnegative. In particular $\mathrm{deg}(-t+2\Phi )=4q-t^2\ge 0$, so $|t|\le 2\sqrt{q}$. The Frobenius characteristic polynomial is $X^2-tX+q$; its complex roots have modulus $\sqrt{q}$, including the double-root case. The zeta function has numerator $1-tT+q{T}^2$, whose zeros consequently have modulus $q^{-1/2}$, the elliptic Riemann hypothesis. This supplements the fully proved Mahler branch with the degree-form proof of the Hasse bound; it is an auxiliary-divisor proof, not a claim that an unspecified transcendence theorem proves the bound.

A standard form of the Schneider-Lang theorem is this. Let $f_1,\dots ,f_s$ be meromorphic functions with $f_j^{\prime }\in K[f_1,\dots ,f_s]$ for a fixed number field $K$. Suppose $f_1,f_2$ have algebraic independence over $\mathbb{C}$ and have finite growth orders ${\rho }_1,{\rho }_2$. There are at most $[K:\mathbb{Q}]({\rho }_1+{\rho }_2)$ points at which all the functions are finite and all their values lie in $K$.

Take $f_1(z)=z$ and $f_2(z)=e^z$, of orders zero and one. Their derivatives belong to $K[z,e^z]$. They have algebraic independence: a polynomial relation ${\sum }_j{p}_j(z)e^{jz}=0$ is impossible by exponential-polynomial independence, or by letting real $z$ tend to infinity and considering the largest $j$.

If nonzero algebraic $\alpha$ had algebraic $e^{\alpha }$, put $K=\mathbb{Q}(\alpha ,e^{\alpha })$. At each of the distinct points $z=n\alpha$, both $z$ and $e^z=(e^{\alpha }{)}^n$ belong to $K$. Infinitely many points contradict the theorem's finite bound. Therefore

The Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $\beta$ is algebraic irrational, then every value $\exp (\beta \lambda )$ with $e^{\lambda }=a$ is transcendental. Fix such a nonzero logarithm $\lambda$, and suppose $\gamma =e^{\beta \lambda }$ is algebraic. The field $K=\mathbb{Q}(a,\beta ,\gamma )$ then contains all values needed in Schneider lattice proof of the Gelfond–Schneider theorem.

Here is an outline including the parameter balance. For a large $L$, choose polynomial degrees $M\asymp L\log L$, $N\asymp L/\log L$, with constants large enough that $(M+1)(N+1)>[K:\mathbb{Q}]L^2$. ImposeExpand these $K$-linear equations in a rational basis and clear denominators. Siegel lemma gives a nonzero integer polynomial $P$ with $\log H(P)=O(L^2/\log L)$. This follows because each coefficient in the grid equations has arithmetic height $O(M\log L+NL)$ and the ratio of unknowns to constraints stays bounded away from one. The exponential polynomial $F(z)=P(z,e^{\lambda z})$ is not identically zero: distinct exponentials are independent over the polynomial ring, as one sees by successively applying powers of $D-j\lambda$ to kill all but one term.

The irrationality of $\beta$ makes the $L^2$ grid points distinct. Their zeros make subsequent grid values extremely small. More precisely, if a whole $S\times S$ grid has been obtained, take an outer radius $R\asymp S\log S$. Growth gives $\log {\max }_{|z|=R}|F(z)|\le \log H(P)+O(M\log R+NR)$. The product form of Schwarz's lemma contributes $(C/\log S{)}^{S^2}$ at the next grid points. Thereforefor large initial $L$, unless the value is zero. But every such value is algebraic in the same $K$, since $e^{\lambda (s+t\beta )}=a^s{\gamma }^t$. Its arithmetic height is $O(\log H(P)+M\log S+NS)$, so the Liouville lower bound is incompatible with the displayed upper bound for a nonzero value. Induction extends the zero grid indefinitely.

A fixed nonzero exponential polynomial has only $O(R)$ zeros in a disc of radius $R$, by its exponential-type growth and Jensen's formula centered at a nonzero value. The grids give $S^2$ distinct zeros in a disc of radius $O(S)$, a contradiction. Thus $a^{\beta }$ is transcendental for every choice of logarithm, completing the Schneider lattice proof of the Gelfond–Schneider theorem.

In Siegel's original convention, an E-function is an entire series $F(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$ with coefficients in a fixed number field, satisfying a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, together with these arithmetic bounds: for every $ϵ>0$, all conjugates of $a_0,\dots ,a_n$ have size $O_{ϵ}(n^{ϵn})$, and some positive integer $d_n=O_{ϵ}(n^{ϵn})$ makes every $d_n{a}_j$, $j\le n$, an algebraic integer. A frequently used stricter convention replaces both bounds by $C^{n+1}$. The original and strict growth definitions should not simply be declared equivalent; the exponential examples below satisfy the strict one.

For sums, work in the compositum of the coefficient fields. Coefficient conjugate bounds add. The product of the two common denominators clears both lists; using $ϵ/2$ in each original denominator bound gives the required $O_{ϵ}(n^{ϵn})$ product bound. Finally all derivatives of $F+G$ lie in the sum of the two finite-dimensional derivative spaces over that rational function field, so they satisfy a nontrivial linear dependence, giving a differential equation for the sum. Thus sums are again E-functions, the addition closure of Siegel E-functions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $F^{\prime }=A(z)F$ with rational-function matrix over $\overline{\mathbb{Q}}$, then at every nonzero algebraic $\xi$ which is not a pole of $A$,In particular algebraically independent functions give algebraically independent values.

Take $F_j(z)=e^{{\gamma }_{j}z}$ for algebraic numbers ${\gamma }_j$ linearly independent over $\mathbb{Q}$. Their coefficients are ${\gamma }_j^n$ with exponential conjugate and denominator bounds, and $F_j^{\prime }={\gamma }_j{F}_j$. A polynomial relation among the functions expands into a relation among exponentials with distinct exponents $\sum m_j{\gamma }_j$ and polynomial coefficients after clearing rational-function denominators; exponential-polynomial independence excludes it. At $\xi =1$ the theorem gives algebraic independence of the $e^{{\gamma }_j}$. For arbitrary distinct algebraic ${\alpha }_i$, choose a rational basis of their span and divide that basis by a common denominator. The $e^{{\alpha }_i}$ are distinct Laurent monomials in the corresponding algebraically independent exponential values, so are linearly independent over $\overline{\mathbb{Q}}$. This recovers the general Lindemann–Weierstrass theorem. Its simplest instance uses just $F(z)=e^z$ at nonzero algebraic $\xi$.

The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne 0$, positive integer bases, and exponents $n\ge 3$. Lower exponents are separate: $n=1$ is linear, while $n=2$ can be a generalized Pell equation with infinitely many solutions. For example $x^2-2y^2=1$ has the infinite sequence generated by $(3+2\sqrt{2}{)}^j$. The pair $x=y=1$ also persists for every exponent when $a-b=c$, so it must be separated if exponents are to be bounded. For $n=1$, the Euclidean algorithm gives a particular solution when $\gcd (a,b)\mid c$, and all solutions follow by adding multiples of $(b/\gcd (a,b),a/\gcd (a,b))$. For $n=2$, put $X=ax$ to obtain $X^2-ab{y}^2=ac$ with the congruence $a\mid X$. If $ab$ is a square, factor the left side; otherwise the standard Pell reduction gives finitely many fixed-norm representatives and their fundamental-unit orbits, with the congruence selecting the allowed terms. This is an effective description even when the set is infinite.

For $Z=\max (x,y)>1$ and sufficiently large $Z^n$, the equation makes $a{x}^n$ and $b{y}^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio givesIf $x\ne y$, a two-logarithm estimate has one fixed algebraic argument $a/b$ and one variable rational argument $x/y$, of arithmetic height at most $\log Z$. It gives $\log |\Lambda |\ge -C(a,b)\log Z\log (2n)$. A zero fixed logarithm is omitted, leaving the valid one-logarithm case. Therefore $n\le C\log (2n)+O(1)$, effectively bounding $n$. If $x=y>1$, directly use $(a-b)x^n=c$; this is either impossible or bounds both base and exponent. Cases where $Z^n$ stays small are finite and directly enumerable.

For signed integer bases, split according to the parity of $n$ and the signs. Cases reducing to a sum of positive powers are elementary because their sum is fixed; zero or absolute-value-one bases give separately detectable constant-power families.

For each bounded $n\ge 3$, the form $a{X}^n-b{Y}^n$ has $n$ distinct projective roots. The three-factor unit-equation argument of Question 5 gives an effective finite list of pairs, even if the binomial is reducible. Consequently all nondegenerate solutions can be effectively found, with the linear/Pell families and constant-base exception treated separately. This is the exponent bound for a binomial power equation.

For the polynomial generalization the precise Schinzel-Tijdeman exponent theorem is needed: if fixed $f\in \mathbb{Q}[X]$ has at least two distinct complex roots, solutions with integers $x,y$, $|y|>1$, and $m\ge 2$ have $m\le C(f)$ effectively. The hypotheses cannot be dropped. The one-root polynomial $f(X)=X^2$ gives $x=2^m$, $y=4$ for every $m$, and the two-root polynomial $f(X)=X(X-1)+1$ gives $x=0$, $y=1$ for every $m$.

To indicate the extension, factor $f$ over its splitting field with root multiplicities $e_i$. Distinct factors have common ideal divisors only over a fixed finite collection of primes. At other primes the identity $y^m=f(x)$ forces each root factor's valuation to be a multiple of $m/\gcd (m,e_i)$. Ideal classes and fundamental units supply the remaining finitely specified factors. Generalized logarithmic-form estimates for these power equations control the unit terms and exceptional primes and bound those reduced exponents; since $\gcd (m,e_i)\le e_i\le \mathrm{deg}f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.