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These are exactly the zero-dimensional reduced commutative rings (a.k.a. "absolutely flat rings"). Clearly semiprimitive rings are reduced. Conversely, assume $R$ is zero-dimensional and reduced. For every $\mathfrak{p}\in\mathrm{Spec}(R)$, $R_\mathfrak{p}$ is local, zero-dimensional and reduced, hence $\mathfrak{p}R_\mathfrak{p}=0$ (in other words, $R_\mathfrak{p}$

[EDIT: what follows is a field)correct but much too complicated. So if $x$ is inSee the Jacobson radical of $R$, then $x$ is zero in every $R_\mathfrak{p}$, hence $x=0$comment by Luc Guyot.]

Conversely, assume $R$ is zero-dimensional and reduced. For every $\mathfrak{p}\in\mathrm{Spec}(R)$, $R_\mathfrak{p}$ is local, zero-dimensional and reduced, hence $\mathfrak{p}R_\mathfrak{p}=0$ (in other words, $R_\mathfrak{p}$ is a field). So if $x$ is in the Jacobson radical of $R$, then $x$ is zero in every $R_\mathfrak{p}$, hence $x=0$.

These are exactly the zero-dimensional reduced commutative rings (a.k.a. "absolutely flat rings"). Clearly semiprimitive rings are reduced. Conversely, assume $R$ is zero-dimensional and reduced. For every $\mathfrak{p}\in\mathrm{Spec}(R)$, $R_\mathfrak{p}$ is local, zero-dimensional and reduced, hence $\mathfrak{p}R_\mathfrak{p}=0$ (in other words, $R_\mathfrak{p}$ is a field). So if $x$ is in the Jacobson radical of $R$, then $x$ is zero in every $R_\mathfrak{p}$, hence $x=0$.

These are exactly the zero-dimensional reduced commutative rings (a.k.a. "absolutely flat rings"). Clearly semiprimitive rings are reduced.

[EDIT: what follows is correct but much too complicated. See the comment by Luc Guyot.]

Conversely, assume $R$ is zero-dimensional and reduced. For every $\mathfrak{p}\in\mathrm{Spec}(R)$, $R_\mathfrak{p}$ is local, zero-dimensional and reduced, hence $\mathfrak{p}R_\mathfrak{p}=0$ (in other words, $R_\mathfrak{p}$ is a field). So if $x$ is in the Jacobson radical of $R$, then $x$ is zero in every $R_\mathfrak{p}$, hence $x=0$.

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These are exactly the zero-dimensional reduced commutative rings (a.k.a. "absolutely flat rings"). Clearly semiprimitive rings are reduced. Conversely, assume $R$ is zero-dimensional and reduced. For every $\mathfrak{p}\in\mathrm{Spec}(R)$, $R_\mathfrak{p}$ is local, zero-dimensional and reduced, hence $\mathfrak{p}R_\mathfrak{p}=0$ (in other words, $R_\mathfrak{p}$ is a field). So if $x$ is in the Jacobson radical of $R$, then $x$ is zero in every $R_\mathfrak{p}$, hence $x=0$.