Write $m=ae_1^*+be_2^*\in M$. The inequalities defining the <dual cone> are
$$
\langle m,3e_1-2e_2\rangle=3a-2b\geq0,
\qquad
\langle m,e_2\rangle=b\geq0.
$$
Thus
$$
\sigma^\vee\cap M
=\{(a,b)\in\mathbb Z^2:b\geq0, 3a\geq2b\}.
$$
Its <Hilbert basis of a rational cone> is
$$
(1,0),\qquad(1,1),\qquad(2,3).
$$
Indeed, after subtracting copies of $(2,3)$ one reduces to $b=0,1,$ or $2$, and the remaining point is generated by $(1,0)$ and $(1,1)$. The <coordinate ring of an affine toric variety> is therefore
$$
\mathbb C[\sigma^\vee\cap M]
=\mathbb C[x,xy,x^2y^3]
\subseteq\mathbb C[x^{\pm1},y^{\pm1}].
$$
Putting $A=x$, $B=xy$, and $C=x^2y^3$ gives the alternative presentation
$$
\mathbb C[U_\sigma]
\cong\mathbb C[A,B,C]/(B^3-AC).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 152 / 1 / d / i / Solution

{title2=$\sigma^\vee$}
{wiki}

For a cone $\sigma\subseteq N_{\mathbb R}$, its dual cone is
$$
\sigma^\vee=\{m\in M_{\mathbb R}:\langle m,v\rangle\geq0\text{ for all }v\in\sigma\}.
$$


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