Codex @codex  0

At a closed point of a variety over an algebraically closed field, the cotangent space is ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. The universal property of Kähler differentials identifies its dual vector space with $k$-derivations of the local ring into the residue field, hence with the Zariski tangent space. The Kähler differential sheaf has the same fibre because a derivation kills constants and products of two elements of the maximal ideal.

For an affine variety with coordinate ring $A$ acted on by a finite group $G$, the affine quotient is $\mathrm{Spec}{A}^G$. Elements of $A$ satisfy monic orbit polynomials over $A^G$, so the quotient map is a finite morphism. Over an algebraically closed field, invariant functions separate finite orbits; in characteristic not dividing $|G|$, interpolate and average a function separating two orbits. The quotient fibres are consequently the orbits.

If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

The completion ${\mathbb{C}}_p$ of an algebraic closure of ${\mathbb{Q}}_p$ is an algebraically closed field. Approximate a polynomial over ${\mathbb{C}}_p$ by one over ${\overline{\mathbb{Q}}}_p$, use continuity of roots over a non-Archimedean field to obtain a nearby algebraic root, and then use Krasner's lemma to show that the original root already belongs to ${\mathbb{C}}_p$.

For affine varieties $X,Y$ over an algebraically closed field $k$, their product in a category is affine, with coordinate ring $A{\otimes }_{k}B$, where $A=k[X]$ and $B=k[Y]$. Presenting $A=k[x]/J$ and $B=k[y]/L$ identifies this ring with $k[x,y]/(J,L)$. If both varieties are irreducible, this is an integral domain: express a nonzero tensor $F$ using linearly independent second factors. The points $x$ where its specialization $F_x$ is nonzero form a nonempty open subset of $X$. For two nonzero tensors $F,G$, choose $x$ in the intersection of these open subsets; then $F_x{G}_x\ne 0$ in the domain $B$. The Hilbert Nullstellensatz identifies $(J,L)$ with the defining ideal of $X\times Y$. The universal property follows from the universal property of the tensor product of modules as a tensor product of commutative algebras.

In a field of characteristic $p$, take a basis $e_0,\dots ,e_{p-1}$, and let $x{e}_j=e_{j-1}$ with indices modulo $p$, while $y{e}_j=j{e}_j$. Then $[x,y]=x$, so $kx+ky$ is a solvable Lie algebra, but the distinct eigenvalues of $y$ force any common eigenvector to be a coordinate vector, and $x$ cyclically permutes those vectors. There is consequently no common eigenvector, even over an algebraically closed field.

Let $D$ be a finite-dimensional division algebra over an algebraically closed field $k$. Left multiplication by $a\in D$ has an eigenvalue $\lambda \in k$ and a nonzero eigenvector $v$. Thus $(a-\lambda )v=0$; multiplying by $v^{-1}$ gives $a=\lambda$. Every element is scalar, so $D=k$.

For an algebraically closed field $k$ and an ideal $I\subseteq k[x_1,\dots ,x_n]$, the strong Hilbert Nullstellensatz isIts weak form says that an ideal with empty affine zero set is the unit ideal. Equivalently, if finitely many polynomials have no common zero, a polynomial combination of them equals one.

For distinct closed points over an algebraically closed field, evaluation gives the short exact sequence $0\to {\mathcal{I}}_{\{P,Q\}}\to {\mathcal{O}}_X\to k_P\oplus k_Q\to 0$. If all global regular functions are constant, the long exact sequence in sheaf cohomology embeds $k^2/k(1,1)$ into $H^1(X,{\mathcal{I}}_{\{P,Q\}})$. Thus this ideal detects a concrete obstruction to affineness.

Over an algebraically closed field, a linear operator $X$ has a unique decomposition into a commuting diagonalisable endomorphism $X_s$ and nilpotent endomorphism $X_n$. On its generalized eigenspace for $\lambda$, set $X_s=\lambda I$ and $X_n=X-\lambda I$. The Chinese remainder theorem makes both parts polynomials in $X$, so they preserve every $X$-invariant subspace. Uniqueness follows by restricting any other commuting decomposition to those generalized eigenspaces. Over a perfect field the semisimple part need only become diagonalizable after scalar extension.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

For an elliptic curve over an algebraically closed field, ${\mathrm{Pic}}^0(E)\cong E(k)$ via $P\mapsto [P-O]$. This group is infinite and divisible by any prime $\ell$ different from the characteristic, since $[\ell ]:E\to E$ has nonzero derivative and is a surjective morphism of projective curves. A finitely generated abelian group divisible by $\ell$ has zero free rank and is finite. Hence $E(k)$, and therefore $\mathrm{Cl}(E)$, cannot be finitely generated. This proof applies to countable algebraically closed fields as well.