A system of parameters need not minimally generate the maximal ideal

Put $T=k[x,y]\mathrm{/}(y^2-x^3)$. Division by the monic polynomial in $y$ writes every element uniquely as $a(x)+yb(x)$; substituting $x=t^2$, $y=t^3$ maps this to $a(t^2)+t^{3}b(t^2)$, whose even and odd powers cannot cancel. Thus $T$ embeds in $k[t]$ and is a domain. Its fraction field is algebraic over $k(x)$ because $y^2=x^3$, so [L2] gives $\dim T=1$; localizing at $(x,y)$ gives $\dim R\le 1$. Also $R\mathrm{/}(\bar{x})\cong k[y{]}_{(y)}\mathrm{/}(y^2)$ is Artinian local, so $(\bar{x})$ has radical $\mathfrak{m}$.

If $\mathfrak{m}$ were principal, then $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ would be one-dimensional over the residue field. But the classes of $\bar{x}$ and $\bar{y}$ are linearly independent modulo ${\mathfrak{m}}^2$, so $\mathfrak{m}$ is not principal. In particular $\mathfrak{m}\mathrm{\ne }(0)$, so $R$ is not a field.

The ring $R$ is a domain by step 1.1, so $(0)$ is prime. Its nonzero maximal ideal $\mathfrak{m}$ from step 1.2 gives a strict prime chain $(0)⊊\mathfrak{m}$, hence $\dim R\ge 1$ directly. Together with $\dim R\le 1$ from step 1.1, this forces $\dim R=1$. Then [L1] makes $(\bar{x})$ a parameter ideal.

Therefore $(\bar{x})$ is a parameter ideal while $\mathfrak{m}$ is not principal, so a system of parameters need not minimally generate the maximal ideal.