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If $X$ is an affine scheme, each reduced irreducible component is a closed subscheme, hence an affine scheme. We prove the converse by explicit affine gluing along closed subschemes.

First suppose $X$ is the scheme-theoretic union of two closed subschemes which are affine schemes $Y$ and $Z$: their defining ideal sheaves have zero intersection. Their scheme-theoretic intersection $W$ is a closed subscheme of both, hence an affine scheme. WriteThe two maps $A\to C$ and $B\to C$ are surjective. Form the fiber product of ringsBoth projections are surjective. Let $I=\ker (R\to A)$ and $J=\ker (R\to B)$. ThenFor the last equality, the common image map $R\to C$ is surjective, and its kernel of a ring homomorphism consists exactly of pairs whose two coordinates map to zero. Therefore $\mathrm{Spec}R$ has closed pieces isomorphic to $Y$ and $Z$, with intersection $W$. They cover it because each prime ideal contains $I$ or $J$, as $IJ=0$.

A finite closed cover determines the topology: a subset is closed if and only if its intersections with both closed pieces are closed. The identifications of $Y,Z,W$ therefore give a homeomorphism between $X$ and $\mathrm{Spec}R$. To check the structure sheaf, on $X$ the elementary ideal-intersection sequence givesIt is exact on stalks, by ${\mathcal{I}}_Y\cap {\mathcal{I}}_Z=0$. The same sequence holds on $\mathrm{Spec}R$, using $I\cap J=0$. Under the homeomorphism, the two middle sheaves and their restriction maps identify, so their kernels identify as sheaves of rings. This identifies the structure sheaves, giving an isomorphism of schemes $X\simeq \mathrm{Spec}R$. The intersection $W$ need not be reduced.

Now a Noetherian scheme has finitely many irreducible components $Y_1,\dots ,Y_r$. Induct on $r$. For $r=1$, the reduced scheme is its single reduced irreducible component, so is an affine scheme. For $r>1$, let $Y=Y_1$ and let $Z$ be the reduced union of the remaining irreducible components. Its components are the corresponding $Y_i$, so $Z$ is an affine scheme by induction. The defining ideal sheaves of $Y$ and $Z$ have zero intersection: their intersection is the ideal sheaf of the reduced union of all components, which is zero because $X$ is reduced. The two-piece argument proves that $X$ is an affine scheme. The empty scheme is $\mathrm{Spec}0$ and causes no exception. Hence