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A implies B. True, as you said, because a finitely generated ring is Noetherian, and $X$ is glued from finitely many spectra of such.

A implies C. True (argument as above).

B implies A. False, e.g. $X = \mathrm{Spec }\ \mathbb{Q}$.

B implies C. False (I believe). There are rings $R$ whose spectrum is homeomorphic to the topological space $\{1, 2, \ldots \}$ with open sets $\{n, n+1, \ldots\}$, which is Noetherian but of infinite Krull dimension. I think something like $\mathrm{Spec }\ k[x_1, x_1 x_2, x_1 x_2 x_3, \ldots]$ should work, but I didn't check the details. EDIT. This doesn't seem to work (see comments), although I don't is nonsense - see easily whythe comments below and Fred Rohrer's answer.

C implies A. False, e.g. $X= \mathrm{Spec }\ \mathbb{Q}$.

C implies B. False, e.g. $X = \mathrm{Spec}\ k[x, x^{1/2}, x^{1/3}, \ldots]$.

2 added 85 characters in body

A implies B. True, as you said, because a finitely generated ring is Noetherian, and $X$ is glued from finitely many spectra of such.

A implies C. True (argument as above).

B implies A. False, e.g. $X = \mathrm{Spec }\ \mathbb{Q}$.

B implies C. False (I believe). There are rings $R$ whose spectrum is homeomorphic to the topological space $\{1, 2, \ldots \}$ with open sets $\{n, n+1, \ldots\}$, which is Noetherian but of infinite Krull dimension. I think something like $\mathrm{Spec }\ k[x_1, x_1 x_2, x_1 x_2 x_3, \ldots]$ should work, but I didn't check the details. EDIT. This doesn't seem to work (see comments), although I don't see easily why.

C implies A. False, e.g. $X= \mathrm{Spec }\ \mathbb{Q}$.

C implies B. False, e.g. $X = \mathrm{Spec}\ k[x, x^{1/2}, x^{1/3}, \ldots]$.

1

A implies B. True, as you said, because a finitely generated ring is Noetherian, and $X$ is glued from finitely many spectra of such.

A implies C. True (argument as above).

B implies A. False, e.g. $X = \mathrm{Spec }\ \mathbb{Q}$.

B implies C. False (I believe). There are rings $R$ whose spectrum is homeomorphic to the topological space $\{1, 2, \ldots \}$ with open sets $\{n, n+1, \ldots\}$, which is Noetherian but of infinite Krull dimension. I think something like $\mathrm{Spec }\ k[x_1, x_1 x_2, x_1 x_2 x_3, \ldots]$ should work, but I didn't check the details.

C implies A. False, e.g. $X= \mathrm{Spec }\ \mathbb{Q}$.

C implies B. False, e.g. $X = \mathrm{Spec}\ k[x, x^{1/2}, x^{1/3}, \ldots]$.