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A nonempty topological space is irreducible when it is not the union of two proper closed subsets, equivalently when any two nonempty open subsets intersect.

The Zariski topology on ${\mathbb{A}}^n$ is Noetherian because $k[x_1,\dots ,x_n]$ satisfies the ascending chain condition on ideals. Suppose a closed set $X$ that is not a finite union of irreducible closed sets were minimal among such counterexamples. It is reducible, say $X=X_1\cup X_2$ with both $X_i$ proper closed subsets. Minimality expresses each $X_i$ as a finite union of irreducible closed sets, and combining those decompositions contradicts the choice of $X$. Thus every closed $X$ has a finite irreducible decomposition.

Using $x_3=x_2^2$, the second defining polynomial reduces toConsequentlyEach component is the image of $t\mapsto (\pm t,t,t^2)$ and is isomorphic to ${\mathbb{A}}^1$, hence irreducible. In characteristic two the two displayed components coincide.

Suppose a polynomial $P(x,y)={\sum }_{j=0}^m{p}_j(x)y^j$ vanishes on every $(x,e^x)$. Thenas an entire function. These exponential-polynomial functions are linearly independent: applying $(D-m{)}^{\mathrm{deg}{p}_m+1}$ kills the term with largest exponent, while it acts injectively on $p_j(x)e^{jx}$ for $j<m$; induction on the number of terms finishes the proof. Hence every $p_j$ is zero, so no nonzero polynomial vanishes on the graph. Its Zariski closure is all of ${\mathbb{A}}^2$.

Finally refine an open cover of an affine variety $X$ by distinguished opens $D(f_j)$, choosing one inside an original member around each point. Since these distinguished opens cover,The Hilbert Nullstellensatz implies that the $f_j$ generate the unit ideal. Thusfor finitely many of them, so $D(f_{j_1}),\dots ,D(f_{j_r})$ cover $X$. The corresponding original open sets form a finite subcover. This is quasi-compactness of an affine variety.

A covering map is locally a disjoint union of homeomorphisms onto the base. A space is simply connected when it is path-connected and every loop is null-homotopic. By the uniformization theorem, the simply connected Riemann surfaces areup to analytic isomorphism.

A lattice is a discrete subgroupwith ${\omega }_1/{\omega }_2\notin \mathbb{R}$. The Weierstrass elliptic functionconverges normally on compact subsets of $\mathbb{C}\setminus L$, defines a meromorphic map $\mathbb{C}\to {\mathbb{C}}_{\infty }$, and is nonconstant because it has a double pole at every lattice point. Reindexing the normally convergent derivative series shows that ${\wp }_L^{\prime }$ is $L$-periodic; evenness fixes the integration constants, so ${\wp }_L(z+\omega )={\wp }_L(z)$.

The invariance makeswell-defined and unique on $\mathbb{C}/L$. It is analytic because the quotient projection is locally biholomorphic.

Neither map is a covering. For any nonzero $\omega \in L$, periodicity and oddness of ${\wp }_L^{\prime }$ giveso ${\wp }_L^{\prime }(\omega /2)=0$. The map is not a local homeomorphism there; since the quotient projection is locally biholomorphic, the descended map fails for the same reason.

There is no covering $\mathbb{C}/L\to {\mathbb{C}}_{\infty }$. The torus is connected, while every connected covering of the simply connected sphere is a homeomorphism. A torus is not homeomorphic to a sphere, for example because their fundamental groups are ${\mathbb{Z}}^2$ and $0$.