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3 correct hypothesis

I realized that I completely missed the second part of the question (the example). Note that ZMT implies that $f$ is a quasi-affine morphism. Then $X\to \mathit{Spec}(f_*\mathcal O_X)$ is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if $f_*\mathcal O_X$ is finite over $\mathcal O_Y$.

Some comments: one should note that in general, the quasi-coherent algebra $f_*\mathcal O_X$ is not finite over $\mathcal O_Y$ and even worse, the morphism $\mathit{Spec}(f_*\mathcal O_X)\to Y$ may not be of finite type (take $Y$ an algebraic variety and $f$ an open immersion. Then $\mathcal O(X)$ is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation $X\to Z\to Y$. If the complementary of $X$ in $Z$ only consists in points of depth at least 2 (see discussions here), then $f_*\mathcal O_X=h_*\mathcal O_Z$ is finite and we are happy. This happens when $X$ is normal (or with non-normal locus finite over $Y$) and surjective to $Y$. Y$with complementary in$Z$of codimension at least 2. But I don't have a general criterion. 2 typo in URL I realized that I completely missed the second part of the question (the example). Note that ZMT implies that$f$is a quasi-affine morphism. Then$X\to \mathit{Spec}(f_*\mathcal O_X)$is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if$f_*\mathcal O_X$is finite over$\mathcal O_Y$. Some comments: one should note that in general, the quasi-coherent algebra$f_*\mathcal O_X$is not finite over$\mathcal O_Y$and even worse, the morphism$\mathit{Spec}(f_*\mathcal O_X)\to Y$may not be of finite type (take$Y$an algebraic variety and$f$an open immersion. Then$\mathcal O(X)$is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation$X\to Z\to Y$. If the complementary of$X$in$Z$only consists in points of depth at least 2 (see discussions here), then$f_*\mathcal O_X=h_*\mathcal O_Z$is finite and we are happy. This happens when$X$is normal (or with non-normal locus finite over$Y$) and surjective to$Y$. But I don't have a general criterion. 1 I realized that I completely missed the second part of the question (the example). Note that ZMT implies that$f$is a quasi-affine morphism. Then$X\to \mathit{Spec}(f_*\mathcal O_X)$is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if$f_*\mathcal O_X$is finite over$\mathcal O_Y$. Some comments: one should note that in general, the quasi-coherent algebra$f_*\mathcal O_X$is not finite over$\mathcal O_Y$and even worse, the morphism$\mathit{Spec}(f_*\mathcal O_X)\to Y$may not be of finite type (take$Y$an algebraic variety and$f$an open immersion. Then$\mathcal O(X)$is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation$X\to Z\to Y$. If the complementary of$X$in$Z$only consists in points of depth at least 2 (see discussions here), then$f_*\mathcal O_X=h_*\mathcal O_Z$is finite and we are happy. This happens when$X$is normal (or with non-normal locus finite over$Y$) and surjective to$Y\$. But I don't have a general criterion.