Diagonal F-threshold

Question

Let $(R, \mathfrak{m},K)$ be a standard graded $K$-algebra of characteristic prime $p$. We denote the diagonal F-threshold by $c^{ \mathfrak{m}}( \mathfrak{m})$. It is known that $\dim(R) \geq c^{ \mathfrak{m}}( \mathfrak{m})$.

If $R$ is also a integral domain. Is it true that $\dim(R) = c^{ \mathfrak{m}}( \mathfrak{m})$?

Answer 1

The answer is "no". We take $R=K[x,y,z]/(xy-z^2)$ of characteristic $p>2$. In this ring is a integral domain, $\dim(R)=2$. and $c^{\mathfrak{m}}(\mathfrak{m})=\frac{3}{2}$.