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A singular two-dimensional affine toric variety in the normal fan construction must come from a full-dimensional pointed cone; cones of smaller dimension give smooth charts with a torus factor. Let its primitive boundary generators be $v_0,v_{t+1}$. Choose an oriented lattice basis with $v_0=(1,0)$, and write the other generator as $(a,r)$, $r>0$. Primitivity gives $\gcd (a,r)=1$. Changing the second basis vector by a multiple of the first lets us write the cone asHere $r>1$ because determinant one would give a smooth chart. The sublattice $N^{\prime }=\mathbb{Z}(1,0)+\mathbb{Z}(-q,r)$ has index $r$, and its affine chart is ${\mathbb{C}}^2$. The quotient $N/N^{\prime }$ is cyclic, generated by $(0,1)$, whose coordinates in the two boundary generators are $(q/r,1/r)$. Thus the kernel of the corresponding finite map of tori acts on the two affine coordinates as $(x,y)\mapsto ({\zeta }^{q}x,\zeta y)$. More explicitly, invariant monomials are exactly those $x^i{y}^j$ with $qi+j\equiv 0(\mathrm{mod}r)$, which are the elements of the original dual lattice in the first quadrant. Their ring is $\mathbb{C}[{\sigma }^{\vee }\cap M]$. Hence the singularity is the cyclic quotientThe choice of the two boundary axes is unordered; exchanging them replaces $q$ by its inverse modulo $r$.

The Hirzebruch–Jung resolution is computed by the negative Euclidean algorithm. Put $r_0=r$, $r_1=q$ and recursively setThe nonzero remainders strictly decrease, $0\le r_{i+1}<r_i$, and their greatest common divisor stays one; the process terminates with $r_t=1$, $r_{t+1}=0$. All $b_i\ge 2$, and unwinding the recursion gives the negative continued fractionStart with $v_0=(1,0)$ and $v_1=(0,1)$ and define $v_{i+1}=b_i{v}_i-v_{i-1}$. The determinant of every consecutive pair is one, by induction. Moreoveragain by the same recursion and the initial values $r,q$. Each intermediate ray is therefore inside $\sigma$. To check the ordering explicitly, the first coordinates after $v_1$ are negative, the second coordinates positive, and successive determinants are positive. At the last step the determinant with the endpoint is zero. The vector $v_{t+1}$ is primitive, points in the same direction as $(-q,r)$, and hence equals it. Subdivide by all the intermediate rays $v_1,\dots ,v_t$. Every new cone is regular, and the subdivision has the same support, so it gives a proper smooth resolution.

Each internal ray has a compact invariant curve $E_i\cong {\mathbb{P}}^1$, because the quotient fan at it has the two opposite primitive rays. The relation $v_{i-1}+v_{i+1}=b_i{v}_i$ gives its self-intersection $E_i^2=-b_i$. This can also be checked directly: on the chart for $(v_{i-1},v_i)$ use coordinates $(x,y)$ with $E_i=\{y=0\}$; on the next chart the coordinates are $(u,v)=(x^{b_i}y,x^{-1})$, with $E_i=\{u=0\}$. The normal-coordinate transition is that of ${\mathcal{O}}_{{\mathbb{P}}^1}(-b_i)$, proving the self-intersection. Neighboring exceptional curves meet transversely once and nonneighbors do not meet.

Since $b_i\ge 2$, there is no exceptional minus-one curve. Combinatorially, deleting $v_i$ gives a cone of determinant $b_i>1$. Deleting a longer consecutive block cannot give a regular cone either: determinants of successively more distant rays satisfy the same recursion with coefficients at least two and grow from $0,1$ to integers at least two. Thus no proper smooth coarsening of this subdivision exists. This proves the minimality of the minimal resolution of a cyclic quotient surface singularity; equivalently its exceptional chain has no curve that can be blown down to a smooth point. The number of exceptional curves is the length $t$ of the negative continued fraction.

If $t=1$, then $r/q=b_1$ is an integer, so coprimality gives $q=1$. Conversely $q=1$ gives the single coefficient $[r{]}^{-}$ and one curve. If $t=2$, thenThe numerator and denominator on the right are coprime, so $r=b_1{b}_2-1$ and $q=b_2\ge 2$. Therefore $q>1$ and $q\mid r+1$. Conversely, if these conditions hold, then $b_1=(r+1)/q\ge 2$, $b_2=q\ge 2$ give exactly this two-term expansion. Hence the requested equivalences are

Now consider a complete fan in dimension two with three rays and at most one singularity. Its three two-dimensional cones are the consecutive ray pairs. At least two cones are regular, and they share a ray. After an oriented lattice change, take their consecutive generators to be $v_1=(1,0)$, $v_2=(0,1)$ and $v_3=(-1,b)$: the first cone is a basis, and $\det (v_2,v_3)=1$ forces the first coordinate $-1$. Completeness forces $b<0$, because the third pair must span the missing sector with positive determinant $\det (v_3,v_1)=-b$. Write $b=-r$, $r\ge 1$. The unique positive primitive relation is $v_1+r{v}_2+v_3=0$, so the fan is that of the weighted projective plane $\mathbb{P}(1,r,1)$, or $\mathbb{P}(1,1,r)$ after permuting weights. For $r=1$ this is ${\mathbb{P}}^2$; for $r>1$ its remaining cone has type $X_{r,1}$. This proves the full three-ray classificationThese are the complete three-ray toric surfaces with at most one singular point, including the smooth case.

For four rays and exactly the specified singularity, place the three regular cones consecutively. Normalize the middle one to have generators $(1,0),(0,1)$. Its neighboring regular cones then force the cyclic ray listThe remaining cone has determinant $\det (v_4,v_1)=1-ab=r$, so $ab=1-r$. Thus $a,b$ have opposite signs, and $s=|a|$ is a positive divisor of $r-1$. In the oriented lattice basis $v_4,(0,-1)$,so the singular cone has quotient parameter $a\mathrm{mod}r$. It follows that its parameter is congruent to $\pm s$. If identification with the specified singularity exchanges the two boundary rays, the parameter is inverted; this does not change the criterion, since for $st=r-1$ we have $s^{-1}\equiv -t(\mathrm{mod}r)$, and $t$ is the complementary positive factor. The same identity with a minus sign handles $(-s{)}^{-1}$.

Conversely suppose $s\mid r-1$ and $s\equiv \epsilon q(\mathrm{mod}r)$, $\epsilon \in \{1,-1\}$. Choose $a=\epsilon s$ and $b=-\epsilon (r-1)/s$ in the displayed four rays. They are primitive and cyclically ordered, with all four gaps less than a half-plane; the opposite signs of $a,b$ ensure completeness. Three consecutive determinants are one, and the fourth is $r$. Its quotient parameter is $a\equiv q(\mathrm{mod}r)$, so the only singular point has the required type. We have proved the four-ray completion criterion for a cyclic quotient surface singularity:

Take the weighted projective plane $X=\mathbb{P}(1,1,2)$. Its fan in $N={\mathbb{Z}}^2$ has rays generated byand all three two-dimensional cones spanned by adjacent rays. Their union is $N_{\mathbb{R}}$, so the fan is complete and the toric variety $X$ is proper.

The determinants of the cones $⟨v_1,v_2⟩$ and $⟨v_2,v_0⟩$ have absolute value one, whileThe smoothness criterion for a toric variety therefore shows that the affine chart for $⟨v_0,v_1⟩$ is singular; it is the cyclic quotient singularity of type $\frac{1}{2}(1,1)$. Thus $X$ is a singular proper toric surface.

Let $q:N_{\mathbb{R}}\to N(\tau {)}_{\mathbb{R}}=N_{\mathbb{R}}/\mathrm{span}(\tau )$ be the quotient map. The images $q(\sigma )$ for $\tau ⪯\sigma$ satisfy the face and intersection axioms, so $\mathrm{Star}(\tau )$ is a fan.

To prove completeness, take $\bar{v}\in N(\tau {)}_{\mathbb{R}}$, choose a lift $v$, and choose $w$ in the relative interior of $\tau$. Since $\Sigma$ is complete, each $v+mw$ lies in some cone of $\Sigma$. There are finitely many cones, so one cone $\sigma$ contains $v+m_{j}w$ for an unbounded sequence $m_j$. Closedness gives $w\in \sigma$. Because $w$ lies in the relative interior of the cone $\tau$ and $\sigma \cap \tau$ is a face of $\tau$, this forces $\tau ⪯\sigma$. Finally,Thus every point of the quotient lies in the support of $\mathrm{Star}(\tau )$, proving that it is a complete fan.

A toric variety is proper exactly when its fan is complete.