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All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore

Write the cubic in the Weierstrass equation of an elliptic curve as $f(x)=x^3+a_2{x}^2+a_{4}x+a_6$. A suitable invariant differential on an elliptic curve isIt is regular and nonzero wherever $y\ne 0$. At a point with $y=0$, nonsingularity gives $f^{\prime }(x)\ne 0$, and differentiating the equation shows $\omega =dy/f^{\prime }(x)$, again regular and nonzero. At the identity $O$, use the parameter $t=-x/y$: the expansions start $x=t^{-2}+\cdots$, $y=-t^{-3}+\cdots$, so $\omega =(1+\cdots )dt$. Thus it is a nowhere-vanishing regular differential on the whole smooth projective curve.

Here is a direct proof of translation invariance of a Weierstrass differential. Fix $Q=(u,v)$, let $P=(x,y)$ vary, and write $P+Q=(X,Y)$. On the open set where the chord-and-tangent group law uses an ordinary chord, put $h=(y-v)/(x-u)$. The addition formulas are $X=h^2-a_2-x-u$ and $Y=h(x-X)-y$. The line-intersection identity isDifferentiate this polynomial in $T$ at $T=x$ to obtain $f^{\prime }(x)-2yh=(x-u)(x-X)$. Differentiating $h$ along the curve now gives $dh/dx=(x-X)/(2y)$, and henceThis proves translation invariance on a dense open set. Since the translation on an elliptic curve is an automorphism and both sides are regular differentials, it proves invariance everywhere, including the exceptional addition cases. Translation by $O$ is the identity.

Translation invariance also gives ${\mu }^{\ast }\omega ={\mathrm{pr}}_1^{\ast }\omega +{\mathrm{pr}}_2^{\ast }\omega$ for the addition map: its differential on a tangent pair is the sum of the two translated tangent vectors. Therefore $[j{]}^{\ast }\omega =j\omega$ for every integer $j$.

Let $d_j=\mathrm{deg}[j]$, with the zero endomorphism assigned degree zero. The assumed degree identity, applied to $[j]$ and $[1]$, givesInduction gives $\mathrm{deg}[j]=j^2$ for positive $j$; negation is an automorphism, so the result holds for negative $j$ as well. This is quadratic degree recursion for elliptic multiplication.

For a prime $\ell \ne p$, the invariant differential on an elliptic curve has nonzero pullback under $[\ell ]$, so that map is a separable isogeny. Translation identifies all its fibres and their local multiplicities, so its geometric kernel has exactly its degree ${\ell }^2$ distinct points. It is killed by $\ell$, and thereforeThe algebraic closure is essential here; the assertion is not generally true for the rational-point subgroup over ${\mathbb{F}}_p$ itself. This is prime-to-characteristic geometric torsion.

For $[p]$, the pullback differential vanishes. Its total degree is $p^2$, and its inseparable degree is at least $p$, so its kernel has at most $p$ geometric points. Thus $E({\mathbb{F}}_p)[p]$ has dimension at most one over ${\mathbb{F}}_p$, while every other prime-torsion subgroup has dimension at most two. The rational-point group is finite and abelian. The structure theorem for finite abelian groups says that the number of cyclic factors of an $\ell$-primary component equals the dimension of its subgroup killed by $\ell$. Each component therefore has at most two factors. Combining the smaller primary factors into one cyclic group and the larger ones into another yieldsOne may also choose $p\nmid m$, since the $p$-primary component is cyclic. This explains two generators for elliptic curves over finite fields.

Let $C$ be the projective cubic of a Weierstrass equation of an elliptic curve, allowing its elliptic-curve discriminant to vanish, and let $O=[0:1:0]$. Use only the smooth locus of a variety $C_{\mathrm{sm}}(K)$: the singular point, if present, is excluded. For $P,Q$, intersect their chord with $C$, using the tangent if $P=Q$ and counting intersection multiplicity. If the third intersection is $R$, define $P+Q=-R$, wherein general Weierstrass equation of an elliptic curve coordinates. The line through $R$ and $O$ gives this reflection. A vertical chord gives $P+(-P)=O$, and the tangent at $O$ meets $C$ three times at $O$, so $O$ is the identity. The construction is symmetric in $P,Q$. It stays in the nonsingular locus: a line through a singular point has intersection multiplicity at least two there, and therefore cannot also contain two smooth intersections counted with multiplicity.

For the nonsingular case, prove associativity by transporting a known abelian group law. The Abel-Jacobi map of a genus-one curveis bijective over an algebraic closure. Indeed, the Riemann-Roch theorem in genus one says that every divisor class of degree one has a unique effective representative consisting of one point: existence follows from $\ell (D)=1$, and uniqueness follows because two distinct representatives would produce a degree-one map to the projective line, impossible for a genus one curve. Subtracting $O$ gives the claimed bijection.

Any line section represents the same divisor class as $3O$, including tangencies. Thus $P+Q+R\sim 3O$ as divisors on an algebraic curve, while the vertical line gives $R+(-R)+O\sim 3O$. HenceAddition in the Picard group is associative, soThe chord-and-tangent group law is defined over $K$, so the group operation restricts to the $K$-rational points.

For completeness, a singular Weierstrass equation of an elliptic curve produces the smooth-locus group of a singular Weierstrass cubic, not an elliptic curve. Over an algebraic closure, its normalization of an algebraic curve is the projective line; deleting the two preimages of a nodal crossing gives the multiplicative algebraic group, while deleting the single preimage of a cusp gives the additive group. For example, on $y^2=x^3$, the coordinate $u=x/y$, with $u(O)=0$, makes the smooth-locus law addition. On $y^2=x^2(x+1)$ in characteristic different from two, put $t=y/x$ and $z=(t+1)/(t-1)$, with $z(O)=1$; the chord relation gives $z(P)z(Q)z(R)=1$, so the group law is multiplication of $z$. A nonsplit node gives the corresponding form of the multiplicative algebraic group over $K$. These descriptions also establish the singular-case group laws.

Classical Kummer theory relates extraction of $n$th roots to Galois cohomology. In characteristic zero the exact sequenceand Hilbert theorem 90 identify $H^1(K,{\mu }_n)$ with $K^{\times }/K^{\times n}$. For an elliptic curve, the corresponding Kummer exact sequence of an elliptic curve isMultiplication by $n$ is surjective over the algebraic closure. If $nQ=P\in E(K)$, the cocycle $\sigma \mapsto \sigma Q-Q$ takes values in $E[n]$. Changing the choice of $Q$ changes it by a coboundary, and changing $P$ by an element of $nE(K)$ does not change its class. Conversely, a trivial cocycle class allows $Q$ to be adjusted by an $n$-torsion point to become $K$-rational. Thus the Kummer map of an elliptic curve is an injectionThis is the elliptic form of Kummer theory. The full cohomology group need not be finite; the arithmetic restriction on these classes is essential.

Choose a finite set $S$ of places containing the archimedean places, the primes over $n$, and all primes of bad reduction of an elliptic curve. At a finite place outside $S$, $E$ has good reduction and $n$ is a unit. A division point of the reduction of $P$ exists over the algebraic closure of the residue field. Smooth lifting gives a point over a finite unramified extension whose multiple differs from $P$ by an element of the kernel of reduction of an elliptic curve. In the formal group of an elliptic curve, $[n](T)=nT+O(T^2)$ is an isomorphism by the invertible morphism criterion for formal group laws, and its integral inverse converges on the maximal ideal. Correcting that difference produces a division point in the maximal unramified extension. Hence the Kummer map of an elliptic curve class is unramified outside $S$.

To prove finiteness explicitly, choose a finite Galois extension $L/K$ containing all $E[n]$ and all $n$th roots of unity, and enlarge $S$ to include its ramified places. A basis of $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$ identifies it over $L$ with ${\mu }_n^2$. Classical Kummer theory then identifiesThe restriction of every class in the image of $\delta$ belongs to $L(S_L,n{)}^2$, where the S-unramified power class group isIndeed an unramified local Kummer extension at residue characteristic prime to $n$ has valuation divisible by $n$: in an unramified field containing a root, $nv(a^{1/n})=v(a)$ with integral valuations.

The finiteness of S-unramified Kummer classes follows from the exact sequenceTo see the final map, write the ideal of $a$ away from $S_L$ as ${\mathfrak{a}}^n$ and take the ideal class of $\mathfrak{a}$. Its kernel is represented by an S-unit, after division by an $n$th power; conversely an $n$-torsion ideal class yields such an $a$. The S-unit group is finitely generated by the Dirichlet unit theorem together with the finitely many inverted primes. The ideal class group of the localized ring is a quotient of the finite ordinary ideal class group. Both outer groups are therefore finite.

Finally restriction has finite kernel: inflation-restriction puts it in the finite group $H^1(\mathrm{Gal}(L/K),E[n])$. Thus the image of $\delta$ has finite restriction image and finite kernel, andThis proves the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the full Mordell-Weil theorem. Local restrictions at every place refine the finite group used here to the Selmer group of an elliptic curve, which is useful for explicit descent calculations.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the Going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The Going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the Going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

Over the complex numbers, every finite-dimensional representation of a solvable Lie algebra has a basis in which every representing matrix is upper triangular. The Lie theorem is often stated first as the existence of a common eigenvector in every nonzero finite-dimensional Lie algebra representation of a solvable Lie algebra. Applying that assertion successively to quotient representations gives an invariant complete flag, and hence the upper triangular form. The same proof works over any algebraically closed field of characteristic zero.

We prove the common eigenvector assertion by induction on $\dim \mathfrak{g}$, writing the action as $av$. The zero Lie algebra is immediate. If $\mathfrak{g}\ne 0$ is solvable, its derived series of a Lie algebra shows that $[\mathfrak{g},\mathfrak{g}]\ne \mathfrak{g}$. Choose a codimension-one ideal of a Lie algebra $\mathfrak{a}$ containing $[\mathfrak{g},\mathfrak{g}]$, and choose $x\notin \mathfrak{a}$. By induction there are $v\ne 0$ and a linear functional $\lambda :\mathfrak{a}\to \mathbb{C}$ such that $av=\lambda (a)v$ for all $a\in \mathfrak{a}$.

Let $W$ be the span of $v,xv,x^{2}v,\dots$. If $m=\dim W$, the first $m$ of these vectors form a basis, and $W$ is $x$-invariant. We claim that for each $a\in \mathfrak{a}$,For $j=0$ this is the definition of $v$. For the induction step, use $ax=xa+[a,x]$ and $[a,x]\in \mathfrak{a}$. Applying the induction hypothesis to both $a$ and $[a,x]$ proves the claim. Consequently $W$ is $\mathfrak{a}$-invariant, and every $a\in \mathfrak{a}$ acts on $W$ by an upper triangular matrix with all diagonal entries $\lambda (a)$.

Since both $x$ and $a$ preserve $W$, the matrix trace of their commutator on $W$ is zero. The claim applied to $[x,a]\in \mathfrak{a}$ givesHere characteristic zero is essential: $m\ne 0$ in the field, so $\lambda ([x,a])=0$. Now the common weight spaceis nonzero and $x$-invariant. Indeed, for $w\in E_{\lambda }$,An endomorphism of a nonzero finite-dimensional complex vector space has an eigenvector, so choose an eigenvector of $x$ in $E_{\lambda }$. It is a common eigenvector for $\mathfrak{g}=\mathfrak{a}+\mathbb{C}x$. This proves the Lie theorem.

For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector. Index the standard basis by $e_0,\dots ,e_{p-1}$. The cyclic entry in the PDF givesThe $p$ diagonal eigenvalues of $y$ are distinct in $k$. Thus every eigenvector of $y$ is a scalar multiple of a single $e_j$. Since $p\ge 2$, $x{e}_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an algebraically closed field.

For $1\le j\le p-1$,At the cyclic boundary,Hence $[x,y]=x$. The two-dimensional Lie subalgebra $kx+ky$ has derived algebra $kx$, whose own derived algebra is zero, so it is solvable. It nevertheless has no common eigenvector, including after extending $k$ to its algebraic closure. This is a failure of Lie theorem in positive characteristic. In the proof above, the obstruction is precisely that $\dim W=p$ can vanish as a scalar in $k$.

The derived algebra of a complex solvable Lie algebra is nilpotent. First suppose $\mathfrak{g}\subseteq \mathfrak{gl}(V)$. By the Lie theorem, put every element of $\mathfrak{g}$ in upper triangular form. The diagonal of a commutator of upper triangular matrices is zero, so $\mathfrak{d}=[\mathfrak{g},\mathfrak{g}]$ consists of strictly upper triangular matrices. The Lie algebra of all such matrices is a nilpotent Lie algebra: if $F_r$ consists of matrices whose entries vanish whenever $j-i<r$, thenTherefore the lower central series of a Lie algebra of $\mathfrak{d}$ reaches zero. Alternatively, every element of $\mathfrak{d}$ is a nilpotent linear map, and the Engel theorem states that a finite-dimensional Lie subalgebra consisting of nilpotent linear maps is a nilpotent Lie algebra.

For an abstract complex solvable Lie algebra $\mathfrak{g}$, apply the preceding result to its Adjoint representation. The Lie algebra $\mathrm{ad}\mathfrak{d}=[\mathrm{ad}\mathfrak{g},\mathrm{ad}\mathfrak{g}]$ is nilpotent. Since $\ker (\mathrm{ad}{|}_{\mathfrak{d}})=\mathfrak{d}\cap Z(\mathfrak{g})$ is central in $\mathfrak{d}$, some term of the lower central series of a Lie algebra of $\mathfrak{d}$ lies in that central ideal; the next term is zero. Thus $\mathfrak{d}$ itself is nilpotent.

Conversely, every nilpotent Lie algebra is solvable, since its derived series of a Lie algebra is contained term by term in its lower central series of a Lie algebra. If $\mathfrak{d}$ is nilpotent, it is therefore solvable, and $\mathfrak{g}/\mathfrak{d}$ is Abelian. More directly, the derived series of a Lie algebra of $\mathfrak{g}$, after its first term, is the derived series of a Lie algebra of $\mathfrak{d}$. Consequently the derived algebra nilpotence criterion is

Normalize $v_K$ by $v_K({\pi }_K)=1$ and put $v_K(0)=+\infty$. Write $f(X)={\sum }_{j=0}^d{a}_j{X}^j$, with $a_d=1$ and $a_0\ne 0$. In the usual coefficient-exponent convention, the Newton polygon $N_K(f)$ is the lower boundary of the convex hull of the upward vertical rays starting atIt runs from $(0,v_K(a_0))$ to $(d,0)$, with nondecreasing slopes from left to right. Zero coefficients contribute no finite point. Equivalently it is the largest convex piecewise-linear function lying below all these coefficient points.

The Newton polygon root valuation theorem says that a segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity, using the extended valuation on an algebraic closure. The nonzero constant coefficient excludes a zero root.

For a short justification, use the weighted Gauss valuation $w_t(P)={\min }_j(v(c_j)+jt)$ for $P=\sum c_j{X}^j$. It is multiplicative: after scaling $X$ by an element of valuation $t$ in a valued extension, the initial nonzero residue polynomials multiply without vanishing. Rational $t$ suffice. Factoring a monic polynomial into linear factors gives $w_t(f)={\sum }_r\min (t,v(r))$. This supporting-line function changes derivative at precisely the root valuations; its derivative drops count their multiplicities. Supporting lines to the lower polygon have slope $-t$, proving the sign and horizontal-length assertion.

Slope convention for part (b). Reflecting the horizontal axis, by plotting $(d-j,v_K(a_j))$, produces the reflected Newton polygon convention; its slopes are the root valuations themselves. Part (b)'s wording with $m$ instead of $-m$ is correct under this positive-root-valuation convention. The proof below gives both forms explicitly, so the result does not depend on silently changing the sign.

A finite field extension $L/K$ is a separable field extension if every element of $L$ has a minimal polynomial over $K$ with distinct roots in an algebraic closure. The primitive element theorem says that a finite separable field extension $L/K$ is $K(\alpha )$ for some $\alpha \in L$.

If $K$ is finite, then so is $L$. The multiplicative group of a finite field is cyclic, so a generator of $L^{\times }$ generates $L$ as a field. To justify the group fact, a finite subgroup of a field's multiplicative group is cyclic: in a finite abelian group its exponent is attained as an element order, while the polynomial root bound forces its size to be at most that exponent.

If $K$ is infinite, first consider $L=K(a,b)$. The minimal polynomials of $a$ and $b$ have distinct roots $a_i,b_j$ in an algebraic closure. Choose $c\in K$, $c\ne 0$, avoiding the finitely many values for which $a_i+c{b}_j=a+cb$ for a pair other than $(a,b)$. Set $\alpha =a+cb$. Over $K(\alpha )$, the polynomials $f(T)$ and $g((\alpha -T)/c)$ have precisely one common root, $a$. Their monic greatest common divisor is therefore $T-a$, so $a\in K(\alpha )$ and $b=(\alpha -a)/c\in K(\alpha )$. Thus $K(a,b)=K(\alpha )$. Induct over finitely many generators of the finite separable field extension. This proves every finite separable extension is simple.

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

An algebraic closure of a field $K$ is an algebraically closed field that is an algebraic extension of $K$.

By definition, every element ofis algebraic over $\mathbb{Q}$, and the question allows us to use that this set is a field. It remains to prove that it is algebraically closed.

Let $f\in \overline{\mathbb{Q}}[X]$ be nonconstant. Its finitely many coefficients generate a finite extension $E/\mathbb{Q}$. By the fundamental theorem of algebra, $f$ has a root $z\in \mathbb{C}$. Since $z$ is a root of a polynomial over $E$, it is algebraic over $E$; since $E/\mathbb{Q}$ is algebraic, transitivity of algebraic extensions makes $z$ algebraic over $\mathbb{Q}$. Thus $z\in \overline{\mathbb{Q}}$. Repeating after division by $X-z$ shows that $f$ splits there, so $\overline{\mathbb{Q}}$ is algebraically closed and is an algebraic closure of $\mathbb{Q}$.

Choose finite generating sets $I=(g_1,\dots ,g_s)$ and $J=(f_1,\dots ,f_t)$, using the Hilbert basis theorem. The assumed inclusion says that each $f_j$ vanishes on $V_{\mathbb{C}}(I)$. By the Strong Hilbert Nullstellensatz, for every $j$ there is an exponent $N_j$ such that

The coefficients in an expression $f_j^{N_j}={\sum }_i{q}_{ij}{g}_i$ solve a finite system of linear equations with rational coefficients. Since it has a complex solution, Gaussian elimination gives a rational solution. Clearing the finitely many denominators produces a nonzero integer $D$ such thatfor every $j$.

For any prime $p\nmid D$, reduce these identities modulo $p$. At a common zero of ${\pi }_p(I)$ in the algebraic closure ${\overline{\mathbb{F}}}_p$, they give ${\pi }_p(f_j{)}^{N_j}=0$, hence ${\pi }_p(f_j)=0$, for all $j$. Thereforefor every prime except the finitely many divisors of $D$. This is the spreading out of an affine zero-set inclusion.

For a singular projective Weierstrass cubic, its smooth locus of a variety contains the point at infinity and carries the chord-and-tangent group law. The singular point is excluded. Over an algebraic closure, a node gives the multiplicative algebraic group and a cusp gives the additive group. A node whose branches are not defined over the base field gives a nonsplit one-dimensional algebraic torus. This group is distinct from an elliptic curve, whose projective model is smooth.