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Over an algebraically closed field, the blowup of a smooth algebraic surface at a closed point replaces the point by an exceptional smooth rational curve $E\cong {\mathbb{P}}^1$, with $E^2=-1$. Locally it is the blowup of the affine plane at the origin. It is a proper birational morphism, an isomorphism off the exceptional curve, and the blown-up surface is again smooth and projective when the original is projective.

Two successive point blowups of a smooth algebraic surface suffice. Let $X_2\to X_1={\mathbb{A}}^2$ be the blowup of the affine plane at the origin. In its chart $x=uy$, with coordinates $(y,u)$, the total-transform equation isRemoving the exceptional factor gives the strict transformAbove the original origin it has just one point, $(y,u)=(0,0)$, which is still singular. The other chart is $y=vx$, where the strict transform has equation $1-v^5{x}^3=0$ and does not meet the exceptional divisor $x=0$. Thus there are no other points above the origin to resolve.

Blow up the remaining point to obtain $X_3\to X_2$. In the chart $u=vy$, with coordinates $(y,v)$, the total transform of $C_2$ isand hence its strict transform isThe derivative of $v^2-y$ with respect to $y$ is $-1$, so this is nonsingular, even in characteristics two or five. In the other chart $y=wu$, the strict transform has equation $1-w^{3}u=0$ and does not meet the exceptional divisor $u=0$. Therefore the only point of $C_3$ mapping to the original origin is the smooth point $(y,v)=(0,0)$.

The required sequence isThe local parameter $v$ there gives $y=v^2$ and $x=v^5$, also verifying the resolved branch directly. This is the resolution of the (2,5) cusp by two blowups. Its tangency to an exceptional divisor does not affect the requested nonsingularity of the strict transform; making the whole total transform have normal crossings is a stronger task.

Use the two charts of the blowup of the affine plane at the origin. In the chart $y=xv$, the total transform isso the proper transform has equation $1-x{v}^3=0$. It misses the exceptional divisor $x=0$ and is smooth, since its derivative with respect to $x$ is $-v^3\ne 0$ there.

In the chart $x=yu$, the total transform is $y^2(u^2-y)$, so the proper transform isIts equation $u^2-y=0$ has partial derivative $-1$ with respect to $y$, so it is smooth even at its unique intersection $(u,y)=(0,0)$ with the exceptional divisor. The two chart descriptions agree on their overlap. Thus one blowup makes the cusp nonsingular, although it is tangent to the exceptional divisor there.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

The toric morphism ${\mathbb{A}}_{x,y}^2\to {\mathbb{A}}_t^1$ with $t=xy$ is induced on cocharacter lattices byThe fan of ${\mathbb{A}}^2$ is the cone $⟨e_1,e_2⟩$ and its faces, while the fan of ${\mathbb{A}}^1$ is ${\mathbb{R}}_{\ge 0}$ and its zero face. The blowup of the affine plane at the origin is the star subdivision obtained by inserting the ray through $e_1+e_2$; its maximal cones areBecause $\phi (e_1)=\phi (e_2)=1$ and $\phi (e_1+e_2)=2$, every cone maps into ${\mathbb{R}}_{\ge 0}$, giving the composite toric morphism $f:X\to {\mathbb{A}}^1$.