A noninvertible ideal in a singular one-dimensional domain

The inclusion $R\subseteq k[t]$ is integral because $t$ satisfies the monic equation $X^2-t^2=0$ with coefficient $t^2\in R$. The ring $k[t]$ is a one-variable polynomial ring over a field, hence a nonfield principal ideal domain and therefore one-dimensional. Thus [L2] gives $\dim R=1$. The same monic equation shows that $t$ is integral over $R$, but $t\mathrm{\notin }R$, so $R$ is not integrally closed and therefore is not Dedekind.

In the local ring $R_{\mathfrak{m}}$ one has ${\mathfrak{m}}_{\mathfrak{m}}^2=(t^4,t^5,t^6)R_{\mathfrak{m}}$. If $t^2$ lay in ${\mathfrak{m}}_{\mathfrak{m}}^2$, we could write $t^2=(a{t}^4+b{t}^5+c{t}^6)\mathrm{/}s$ with $a,b,c\in R$ and $s\mathrm{\notin }\mathfrak{m}$, so $s=t^2(a+bt+c{t}^2)\in \mathfrak{m}$, contradiction. Hence $t^2\in {\mathfrak{m}}_{\mathfrak{m}}\setminus {\mathfrak{m}}_{\mathfrak{m}}^2$. If ${\mathfrak{m}}_{\mathfrak{m}}$ were principal, say ${\mathfrak{m}}_{\mathfrak{m}}=a{R}_{\mathfrak{m}}$, then $t^2=au$ for some $u\in R_{\mathfrak{m}}$. Because $t^2\mathrm{\notin }{\mathfrak{m}}_{\mathfrak{m}}^2=a{\mathfrak{m}}_{\mathfrak{m}}$, the element $u$ is a unit, so ${\mathfrak{m}}_{\mathfrak{m}}=t^2{R}_{\mathfrak{m}}$. But then $t^3\in {\mathfrak{m}}_{\mathfrak{m}}$ implies $t=t^3\mathrm{/}{t}^2\in R_{\mathfrak{m}}$, impossible: if $t=r\mathrm{/}s$ with $r\in R$ and $s\mathrm{\notin }\mathfrak{m}$, then $r=ts$ would have a linear $t$-term while elements of $R=k[t^2,t^3]$ have no such term. Therefore ${\mathfrak{m}}_{\mathfrak{m}}$ is not principal.

Since ${\mathfrak{m}}_{\mathfrak{m}}$ is not principal, [L1] shows that $\mathfrak{m}$ is not invertible.