The cusp conductor and its two semigroup-ring generators

$t^{2}k[t]\subseteq A$: for $n\ge 0$ the monomial $t^{n+2}$ has exponent at least $2$, and every exponent $\ge 2$ lies in the numerical semigroup generated by $2$ and $3$, so $t^{n+2}\in A$; the same argument gives $t^{3}k[t]\subseteq A$. Hence $t^2,t^3\in \mathfrak{c}$, and since $\mathfrak{c}$ is an ideal, $(t^2,t^3)\subseteq \mathfrak{c}$.

The semigroup $⟨2,3⟩=\{0,2,3,4,\dots \}$ contains every integer $n\ge 2$ and misses $1$, so the least exponent $m$ with $t^n\in A$ for all $n\ge m$ is $m=2$. The element $t^3$ is not a multiple of $t^2$ inside $A$: an equation $t^3=t^{2}a$ with $a\in A\subseteq k[t]$ would force $a=t\mathrm{\notin }A$ since $k[t]$ is a domain. Hence $t^2$ alone does not generate the conductor, and the two monomials $t^2,t^3$ are both needed.

Conversely let $a\in A$ satisfy $ak[t]\subseteq A$. If $a\mathrm{\notin }(t^2,t^3)$, it has a nonzero constant coefficient $c$ and all remaining terms have exponent at least two. Thus $at$ has nonzero coefficient $c$ at exponent one, so $at\mathrm{\notin }A$, a contradiction; $a=0$ lies in $(t^2,t^3)$. Hence $a\in (t^2,t^3)$ and $\mathfrak{c}\subseteq (t^2,t^3)$; with step 1.1, $\mathfrak{c}=(t^2,t^3)$.

Therefore the conductor of the cusp extension $k[t^2,t^3]\subseteq k[t]$ is the maximal ideal $(t^2,t^3)$ of the semigroup ring, whose generators are the first two positive elements of the semigroup: the least exponent from which all monomials lie in $k[t^2,t^3]$ is $2$, and neither generator alone suffices.