$\dim(R/x) = \dim(R)-1$ for Noetherian integral domains?

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Let $R$ be a Noetherian integral domain of finite Krull-dimension and $0 \neq x \in R$ a non-unit. Do we have $\dim(R/x) = \dim(R) -1$ in general?

If this is wrong, does it change something if we further assume that $R$ is positively graded, finitely generated by homogeneous elements of degree one over $R_0$, which is Artinian local, and $x$ is homogeneous of degree one?

Context: This dimension formula is used here.

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In general $\dim R/(x)$ can be $< \dim R - 1$ - see here. However it does hold if $R$ is (in addition) local. Thus in the specific setting, where $R$ is a standard graded Noetherian domain (so that $R_0$ is a field, not just Artinian), then if $\DeclareMathOperator{\m}{\mathfrak{m}}$$\m$ is the homogeneous maximal ideal (which is maximal), we have $\dim R - 1 = \dim R_m - 1 = \dim R_m/xR_m = \dim (R/x)_m = \dim R/(x)$ (since $\dim R = {}^*\!\dim R$ if $R$ is positively graded with $R_0$ local).