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Suppose first that the affine variety $V$ is irreducible. If $fg\in I(V)$, thenBoth pieces are Zariski closed in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a prime ideal.

Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the Hilbert Nullstellensatz would giveChoose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of irreducibility and prime vanishing ideal:

For the semicubical parabolaconsiderCertainly $(y^2-x^3)\subseteq \ker \theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker \theta$, thenThe first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. ThereforeThe latter is a subring of the integral domain $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and

An affine variety of dimension $d$ is smooth at $p$ when its Zariski tangent space has dimension $d$; equivalently, the Jacobian criterion requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient isAt $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the coordinate ring and singularity of the semicubical parabola gives

Let $F:X\to Y$ be a nonconstant holomorphic map of degree $d$ between compact connected Riemann surfaces. At $p\in X$, let $e_p$ be the ramification index of a holomorphic map. Choose a triangulation of $Y$ containing every branch value among its vertices, and lift it to $X$.

If the target triangulation has $V$ vertices, $E$ edges, and $T$ triangles, then the lifted triangulation has $dE$ edges and $dT$ triangles, because no branch value lies in an edge or face interior. For a target vertex $v$, local degree counting givesConsequently the number of vertices above $v$ isWriting $R={\sum }_{p\in X}(e_p-1)$, the Euler characteristic is thereforeSince a compact orientable surface of genus $g$ has Euler characteristic $2-2g$, this Triangulation proof of the Riemann-Hurwitz formula givesor equivalently

Now extend a cubic polynomial $f:\mathbb{C}\to \mathbb{C}$ to a degree-three holomorphic mapof the Riemann sphere. Both genera are zero, so Riemann--Hurwitz givesThe point at infinity is totally ramified with $e_{\infty }=3$, contributing two. Thus the finite points contribute exactly two. There are consequently two cases.

If there is one finite ramification point $\alpha$, it has index three. Choose an affine source map $h$ sending zero to $\alpha$. Thenfor some $c\ne 0$, and integration givesAn affine target map subtracts $d$ and rescales by $3/c$, yielding $z^3$.

Otherwise there are two distinct finite ramification points, each of index two. Choose an affine source map $h$ sending $-1$ and $1$ to them. The derivative of $f\circ h$ is then a nonzero multiple of $z^2-1$:HenceThe affine target map $g(w)=(w-d)/c$ givesTherefore the affine normal forms of a complex cubic polynomial are precisely

The Minkowski convex body theorem states that if $\Lambda \subset {\mathbb{R}}^n$ is a full lattice and $C$ is a measurable, convex, centrally symmetric set withthen $C$ contains a nonzero point of $\Lambda$.

Here $d\not\equiv 3(\mathrm{mod}4)$, so the ring of integers of a quadratic field isUnder its complex embedding it has real-lattice covolume $\sqrt{d}$. An integral ideal $I$ has index $N(I)$ in ${\mathcal{O}}_K$, so its covolume is $\sqrt{d}N(I)$. Apply Minkowski's theorem to the disk $|z|\le R$, whose area is $\pi R^2$. Taking $R^2$ down to $4\sqrt{d}N(I)/\pi$ gives a nonzero $\alpha \in I$ withSince the field norm is $N_{K/\mathbb{Q}}(\alpha )=|\alpha {|}^2$ in this imaginary quadratic field, this proves the imaginary quadratic Minkowski disk bound

Because $(\alpha )\subseteq I$, the idealis integral, represents the inverse class of $I$, and hasThus every ideal class has a bounded-norm representative. Only finitely many integral ideals have bounded norm, proving the finiteness of the ideal class group.

Now take $K=\mathbb{Q}(\sqrt{-22})$. Its ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, andEvery class therefore has an integral representative of norm at most five. The polynomial $X^2+22$ is irreducible modulo three and modulo five, so both primes are inert and there are no ideals of norm three or five. At two it reduces to $X^2$, and the splitting of rational primes in a quadratic field givesAn ideal of norm four is ${\mathfrak{p}}_2^2=(2)$ and is principal. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm two, whereashas no integer solution. Consequently the Ideal class group of Q of square root of minus twenty-two is

Suppose finally that $y^3=x^2+22$. If $x$ were even, the right-hand side would be $2$ modulo four, which is not a cube modulo four. Hence $x$ and $y$ are odd. In ${\mathcal{O}}_K$,The two ideals on the left are coprime. Indeed, a common prime ideal would divide both $2x$ and $2\sqrt{-22}$. It cannot lie above two because their norms are odd. It would therefore lie above eleven and force $11\mid x$; but then $x^2+22$ has $11$-adic valuation exactly one, impossible for the cube $y^3$.

By unique factorization of ideals in a number field, coprimality impliesfor some ideal $\mathfrak{a}$. The class group has order two, so it has no element of order three; $[\mathfrak{a}{]}^3=1$ therefore forces $\mathfrak{a}$ to be principal. The only units are $\pm 1$, both of which can be absorbed into a cube, and hencefor integers $a,b$. Comparing the coefficient of $\sqrt{-22}$ givesThus $b=\pm 1$. For $b=1$ this requires $3a^2=23$, and for $b=-1$ it requires $3a^2=21$; neither is possible. Therefore the result on no integral solutions to y cubed equals x squared plus twenty-two is