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Fix fundamental units ${\epsilon }_1,\dots ,{\epsilon }_r$ and write $x=\zeta \prod {\epsilon }_i^{b_i}$. Rank zero already gives a finite unit group. Otherwise put $B=\max (1,|b_i|)$. The logarithmic embedding in Dirichlet's theorem has a weighted sum-zero image and is injective on the free exponent lattice. Norm equivalence and that sum-zero condition give an archimedean embedding $\sigma$ withThis is the small archimedean value of a large unit. From the equation, $\sigma (\beta y)=1-\sigma (\alpha x)$ is exponentially close to one. Also $h(y)\le h(x)+O(1)$ by the arithmetic height inequalities for $y=(1-\alpha x)/\beta$, so the fundamental-unit exponents of $y$ have size $O(B)$.

For large $B$, take the small-branch logarithm of $\sigma (\beta y)$. With $y=\eta \prod {\epsilon }_i^{c_i}$ it is the nonzero formusing fixed logarithms and $\log (-1)=\pi i$. The integer $k$ corrects the branch, and nonzero follows since $\beta y=1$ would force $\alpha x=0$. Its upper bound is $|\Lambda |\le C^{\prime }{e}^{-cB}$. The assumed effective logarithmic-form estimate for the fixed algebraic numbers gives $\log |\Lambda |\ge -C^{\prime \prime }\log (2B)$. Thus $cB\le C^{\prime \prime }\log (2B)+O(1)$, effectively bounding $B$. The finitely many embeddings and torsion choices give uniform constants. Therefore there are only finitely many unit solutions, and their exponents are effectively bounded.

For the Thue application, take an irreducible integral binary form $F$ of degree $d\ge 3$ and $m\ne 0$. In a splitting field, write $F(X,Y)=a{\prod }_i(X-{\rho }_{i}Y)$ and put ${\theta }_i=a{\rho }_i$, which are algebraic integers. The integral factors $L_i=aX-{\theta }_{i}Y$ satisfy ${\prod }_i{L}_i=a^{d-1}m$. Each ideal $(L_i)$ therefore divides one fixed ideal, giving finitely many possibilities. Choose a generator ${\delta }_i$ for each principal possibility, so $L_i={\delta }_i{u}_i$ with units $u_i$.

For three distinct roots,Division by the third term gives a fixed-coefficient unit equation in $u_1/u_3,u_2/u_3$. Its solutions are finite, hence there are finitely many values of $L_1/L_2$. This ratio determines the rational direction $X:Y$, since the roots are distinct. In a fixed direction, homogeneity and $F(X,Y)=m\ne 0$ leave only finitely many integer scales. This proves Thue theorem, and the argument also works for a form with at least three distinct projective roots even if reducible.

The hyperelliptic alternative uses a related factorization for the usual nondegenerate case with squarefree $f$ of degree at least three. For $y^2=f(x)$, ideals of two different root factors can share primes only above the fixed root differences and the leading coefficient. At other primes their valuations are even. Finite exceptional-prime and ideal-class choices, together with units modulo squares, reduce the factors to finitely many expressions ${\delta }_i{z}_i^2$. Relations between distinct root factors then give simultaneous norm or Pell-type equations. Writing their solutions in fundamental units and applying the same small-value/logarithmic-form estimates bounds the unit exponents, hence $x,y$. The common-factor and square-class reductions are essential; a single quadratic norm equation alone can have infinitely many Pell solutions. This outlines the effective hyperelliptic treatment as well as the explicit Thue reduction.

Use the following fixed-number form of a logarithmic estimate: for chosen logarithms of fixed nonzero algebraic numbers $a_1,\dots ,a_s$, a nonzero form $\Lambda =\sum b_j\log a_j$ with integer coefficients $|b_j|\le B$ satisfies $|\Lambda |\ge c(2B{)}^{-C}$, for fixed positive constants $c,C$. This follows from the Baker lower bound for a homogeneous linear form in logarithms.

By the Dirichlet unit theorem, write units as $x=\zeta \prod {ϵ}_j^{m_j}$ and $y=\xi \prod {ϵ}_j^{n_j}$, with finitely many roots of unity and fixed fundamental units. If the unit rank is zero, there are already only finitely many choices. Otherwise put $B=\max (1,|m_j|,|n_j|)$. The real logarithmic embedding of the free unit group has full rank and lies in the product-formula hyperplane. Equivalence of norms in that finite-dimensional space shows that a unit with exponent size $B$ has some archimedean conjugate of modulus at most $e^{-c_{0}B}$: a large positive logarithmic coordinate forces a comparably large negative one, and a large negative coordinate itself suffices.

Interchanging the two terms if necessary, choose such a conjugate for $y$. At this embedding the equation makes $\sigma (\alpha x)=1-\sigma (\beta y)$ with $|\sigma (\beta y)|\le C_0{e}^{-c_{0}B}$. For large $B$ define the logarithm near one. It is nonzero because $\beta y\ne 0$, andOn fixed branches it isfor an integer $k=O(B)$. The bound on $k$ follows from summing the fixed imaginary parts and then reducing to the near-one branch. There are only finitely many possible embeddings and roots of unity, so the stated lower bound applies uniformly with coefficients $O(B)$. It contradicts the exponential upper bound when $B$ is large. Hence the exponents are bounded and $\boxed{\alpha x+\beta y=1\text{ has finitely many unit solutions}}$ in ${\mathcal{O}}_K^{\times }$. This is the unit equation over a number field proof; “unit” here means an algebraic-integer unit, not an arbitrary nonzero field element.

For the Thue theorem, let $F\in \mathbb{Z}[X,Y]$ be irreducible homogeneous of degree $d\ge 3$ and let $m\ne 0$ be fixed. In a splitting field $L$, write $F(X,Y)=a{\prod }_i(X-{\theta }_{i}Y)$ with distinct roots. Choose a fixed integer $c$ so that each $c{\theta }_i$ is integral and $c^{d}m/a$ is a nonzero integer. For a solution $(u,v)$ the nonzero algebraic integers $t_i=c(u-{\theta }_{i}v)$ satisfy ${\prod }_i{t}_i=c^{d}m/a$. Their principal ideals therefore divide a fixed ideal and have only finitely many possibilities. For each occurring principal ideal choose a generator ${\gamma }_i$, so $t_i={\gamma }_i{\epsilon }_i$ with ${\epsilon }_i$ a unit.

Any three distinct roots give the identityDivide by the third term and absorb the finitely many generator ratios into the coefficients. This is a fixed-coefficient unit equation in ${\epsilon }_1/{\epsilon }_3$ and ${\epsilon }_2/{\epsilon }_3$, so it has finitely many solutions for each generator choice. Thus $\rho =t_1/t_3$ has finitely many values. The relation $(1-\rho )u=({\theta }_1-\rho {\theta }_3)v$ fixes a rational direction whenever an integer solution exists; $\rho =1$ simply gives $v=0$. On each such direction, $(u,v)=k(u_0,v_0)$ for a primitive integer pair, and $k^{d}F(u_0,v_0)=m$ allows finitely many $k$. This proves the desired finiteness. When $m=0$, irreducibility excludes a nonzero rational root direction, leaving only $(0,0)$.

The argument also proves finiteness for any binary form with at least three distinct projective roots: irreducibility was used only to guarantee distinct roots and to dispose of the zero-value case. A rational invertible change of variables can move every projective root away from infinity; after clearing its fixed denominators, the resulting pairs lie in an integer lattice and the same proof applies. This stronger nonzero-value version will apply to the binomial forms in question 6.

Here representations mean finite-dimensional rational representations of $S{L}_2(\mathbb{C})$. Put $W={\mathbb{C}}^2$. The irreducibles areThe invariant alternating form on $W$ identifies $W$ with $W^{\ast }$, so these can equally be realized as binary forms of degree $d$. With basis $e_j=x^{d-j}{y}^j$, the standard Lie algebra generators act asThe distinct torus weights and these nonzero lowering and raising coefficients show irreducibility: any nonzero invariant subspace contains a weight vector, and repeated applications of $E,F$ obtain the entire basis.

For exhaustion, choose a highest-weight vector $v$ of an irreducible representation, so $Hv=mv$, $Ev=0$. The sl2 highest-weight lowering formula, obtained by induction from $[E,F]=H$, is $E{F}^{j}v=j(m-j+1)F^{j-1}v$. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, applying $E$ to the latter equality gives $(r+1)(m-r)F^{r}v=0$, hence $m=r$. The vectors $v,Fv,\dots ,F^{r}v$ have distinct weights and span an invariant submodule, so irreducibility makes it the whole representation. After rescaling basis vectors it is $V_r$. Complete reducibility from Question 2 now describes every representation as a finite direct sum of these symmetric powers. Their central element $-I$ acts by $(-1{)}^d$.

The symbolic method for binary-form invariants works as follows. Write a degree-$d$ form as $f(x,y)={\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{d}{j}\right)f_j{x}^{d-j}{y}^j$ and introduce a formal letter $a=(a_0,a_1)$. The formal power $(a_{0}x+a_{1}y{)}^d$ is converted to $f$ by the linear substitution $a_0^{d-j}{a}_1^j\mapsto f_j$. This is an umbral linear rule, not a claim that every form is an actual power of a linear form.

For a coefficient polynomial of degree $r$, polarize it into $r$ copies of the form, represented by letters $a,b,\dots$. Each letter must occur with degree $d$. In characteristic zero, polarization and evaluation on equal forms recover the original coefficient polynomial. The basic invariant expressions in the letters are bracketsThey are invariant because $\det g=1$. Invariant tensors on copies of a two-dimensional space are spanned by pairwise alternating contractions; equivalently, the invariant polynomials in the letters are generated by these brackets. One can see the tensor statement from the two-row Young-symmetrizer description of the determinant powers in tensor space. Their basic relations include $(ab)=-(ba)$ and $(ab)(cd)-(ac)(bd)+(ad)(bc)=0$. Multiply brackets, ensure the required degree in each letter, symmetrize letter copies as needed, and apply the umbral substitutions. This gives all coefficient invariants, and also works for joint invariants of several forms.

For example, for $f=A{x}^2+2Bxy+C{y}^2$, the expression $(ab{)}^2$ becomes $2(AC-B^2)$, the quadratic discriminant up to its conventional scalar. Contractions leaving variable indices unpaired give covariants. A useful systematic construction is the transvectant of binary forms:For forms of degrees $m,n$, this is an equivariant form of degree $m+n-2r$. The corresponding tensor decomposition isIts character identity follows by multiplying the weight sums; complete reducibility then gives the decomposition, and alternating contractions supply the displayed maps onto its summands.

A vector in a rational representation is an unstable vector in a rational representation when $0\in \overline{Gv}$, with Zariski closure. For $G=S{L}_n(\mathbb{C})$, the Hilbert-Mumford criterion for the affine null cone isEvery algebraic one-parameter subgroup is conjugate to $\mathrm{diag}(t^{a_1},\dots ,t^{a_n})$ with integer $a_j$ summing to zero. This follows by decomposing the standard representation into characters of the multiplicative group, then imposing determinant one. In its weight decomposition, a limit of zero means that only strictly positive weights occur.

The reverse implication in the criterion is immediate: the subgroup lies in $G$ and supplies points of the orbit tending to zero. For the harder implication, the complex orbit is locally closed and Zariski open in its closure. A nonempty Zariski-open subset of a complex irreducible variety is dense also in the usual topology. Thus choose $g_{j}v\to 0$ in the ordinary vector-space topology. Singular value decomposition, with determinant phases adjusted, givesUse an $S{U}_n$-invariant Hermitian inner product on $V$. By compactness, after a subsequence ${\ell }_j\to \ell$. Put $w=\ell v$, and decompose it into the weights of the diagonal torus, $w={\sum }_{\chi }{w}_{\chi }$. These weight spaces are orthogonal because the unitary diagonal torus preserves the inner product. HenceFor every $\chi$ with $w_{\chi }\ne 0$, its final norm is bounded away from zero for large $j$, forcing $\chi (b_j)\to -\infty$. There are only finitely many such weights, so choose a single real vector $b$ with sum zero and $\chi (b)<0$ for all of them. The inequalities are strict with integer coefficients. A nearby rational vector still satisfies them; multiply its negative by a common denominator to obtain integers $a_r$ of sum zero with $\chi (a)>0$ throughout this support. Then $\mu (t)=\mathrm{diag}(t^{a_r})$ takes $w$ to zero. Conjugating gives $\lambda (t)={\ell }^{-1}\mu (t)\ell$ taking $v$ to zero, completing the proof. The zero vector itself satisfies the criterion trivially.

For degree-six binary forms, a nontrivial algebraic one-parameter subgroup of $S{L}_2$ is, after conjugation and choosing its direction, $\mathrm{diag}(t^a,t^{-a})$ with $a>0$. On $x^{6-j}{y}^j$ the weight is $a(6-2j)$. Strictly positive weights permit only $j=0,1,2$, so the form is divisible by $x^4$. Conversely every such form tends to zero under that subgroup. Changing basis replaces $x$ by any linear factor. The root-multiplicity criterion for unstable binary forms therefore givesA triple root alone is insufficient: $x^3{y}^3$ has weight zero and is not unstable. Nor does merely having a repeated root suffice.

For $\mathrm{diag}(t,t^{-1})$, the monomial $x^{d-i}{y}^i$ has weight $d-2i$. Positive weights force divisibility by $x^{\lfloor d/2\rfloor +1}$. Conjugating the subgroup replaces $x$ by any nonzero linear factor. Thus a nonzero binary form is unstable exactly when a projective root has multiplicity greater than half its degree. Equality gives zero weight and does not imply instability.

A point is stable when a positive-degree homogeneous invariant is nonzero there and its orbit is closed in that invariant's nonvanishing affine open set, with finite stabilizer. This is stronger than merely having a closed orbit or a nonzero invariant. For binary forms of degree four, the stable points have four distinct projective roots.

Replace a degree-$d$ binary form by the formal power $(a_{0}x+a_{1}y{)}^d$ and recover its coefficients with the linear rule $a_0^{d-i}{a}_1^i\mapsto f_i$. For a coefficient invariant of degree $r$, use $r$ symbolic letters, each of degree $d$, polarize and then apply this rule to each letter. Products of brackets $(ab)=a_0{b}_1-a_1{b}_0$ generate the invariant expressions before the final coefficient substitution.