Corollary

— For a commutative ring to be reduced (V, p. 34, Def. 2) it is necessary and sufficient for it to be isomorphic to a subring of a product of fields.

The condition is clearly sufficient.

Let $\mathbf{A}$ be reduced. By Prop. 2 the intersection $n$ of the set of prime ideals of $\mathbf{A}$ is reduced to 0. For every prime ideal $\mathfrak{p}$ of $\mathbf{A}$, let $k(\mathfrak{p})$ be the field of fractions of $\mathbf{A}/\mathfrak{p}$ and $\varphi_n$ the canonical homomorphism of $\mathbf{A}$ into $k(\mathfrak{p})$. Let $\varphi$ be the homomorphism of $\mathbf{A}$ into $\prod_{\mathfrak{p}} k(\mathfrak{p})$ whose component of index $\mathfrak{p}$ is $\varphi_{\mathfrak{p}}$. The kernel of $\varphi_{\mathfrak{p}}$ is $\mathfrak{p}$, hence that of $\varphi$ is $n = 0$; therefore $\varphi$ is an isomorphism of $\mathbf{A}$ onto a subring of $\prod_{\mathfrak{p}} k(\mathfrak{p})$.