The totally ramified cases

Proof. The transformations $$([u:v],[x:y])\longmapsto([\lambda^{-5}u:v],[\lambda x:y])$$ and $([u:v],[x:y])\mapsto([v:u],[y:x])$ preserve the curve and induce an action of the reductive group $G=\mathbb G_m\rtimes\boldsymbol\mu_2$ on $X_0$. The induced action on the base of $\phi$ has no point fixed by all of $G$: the torus fixes the two coordinate points, and the involution interchanges them.

Let $F$ be a $G$-invariant prime divisor over $X_0$, and let $Z=c_{X_0}(F)$. Its image under $\phi$ is a closed irreducible $G$-invariant subset of $\mathbb P^1$, and consequently is the whole base. Thus $Z$ meets a smooth fibre at a point $p$. By Lemma 90.4, $A_{X_0}(F)/S_{X_0}(F)\ge\delta_p(X_0)>1$. The equivariant valuative criterion for K-polystability [3, Theorem 1.22] proves the assertion. This is the Abban–Zhuang proof underlying [3, Corollary 5.71]. ◻

Proof. Choose the totally ramified source point to be $[0:1]$, and its image to be $[1:0]$. After a target coordinate change, the equation has the form $$u x^5+v\left(y^5+\sum_{i=1}^4 b_i x^i y^{5-i}\right)=0.$$ Here the coefficient of $y^5$ is nonzero because $f,g$ are coprime; the $x^5$ term in the second quintic has been absorbed into $u$. For $\lambda\in\mathbb A^1$ define $$C_\lambda=\left\{u x^5+v\left(y^5+ \sum_{i=1}^4\lambda^i b_i x^i y^{5-i}\right)=0\right\}.$$ The two quintics defining this graph have no common zero for any $\lambda$. Thus the total curve is a smooth relative graph over $\mathbb A^1$, and its blow-up in $M\times\mathbb A^1$ is a smooth projective family. Its fibres are Fano members of Family 3.5. For $\lambda\ne0$, the change $$x'=\lambda x,\quad y'=y,\quad u'=\lambda^{-5}u,\quad v'=v$$ identifies $C_\lambda$ with $C_1$, and hence identifies $X_\lambda$ with $X$. The action $$(\lambda,x,y,u,v)\longmapsto (s\lambda,s^{-1}x,y,s^5u,v)$$ makes the relative anticanonically polarized family a test configuration. Its central fibre is the variety in Lemma 91.1. Openness of K-semistability, applied at the central fibre and then using the isomorphisms over $\mathbb A^1\setminus\{0\}$, gives K-semistability of $X$. ◻

Proof of Theorem 89.2. The assertion without total ramification is proved in Theorem 101.1 below. The preceding lemma proves the semistability assertion. If there are two index-five points, their contributions exhaust $\deg R=8$. The map is then equivalent to $[x:y]\mapsto[x^5:y^5]$, and the threefold is $X_0$.

If there is exactly one, the ramification locus has at least three points: one further point could not contribute the remaining four without also having index five. The group of automorphisms of $\mathbb P^1$ preserving this finite set is finite. Using $\operatorname{Aut}(X)=\operatorname{Aut}(S,C)$ [3, Section 7.5], we conclude that $\operatorname{Aut}(X)$ is finite. Thus $X$ is not isomorphic to $X_0$, which has a torus of automorphisms. The test configuration above has Donaldson–Futaki invariant zero: this is the Futaki invariant of the induced product action on its K-polystable central fibre. It is not a product test configuration, since its central fibre is not isomorphic to $X$. Therefore $X$ is not K-polystable, and is strictly K-semistable. Equivalently, one may apply [3, Corollary 1.13]. ◻