= Solution
Let $\mathcal F$ be any <quasi-coherent sheaf> on $X$, and choose $m$ from (i). Filter it by the powers of the <nilradical sheaf>:
$$
\mathcal F\supseteq\mathcal N\mathcal F\supseteq\cdots\supseteq\mathcal N^m\mathcal F=0.
$$
Each subquotient
$$
\mathcal Q_j=\mathcal N^j\mathcal F/\mathcal N^{j+1}\mathcal F,\qquad 0\leq j<m,
$$
is a <quasi-coherent sheaf> annihilated by $\mathcal N$. Apply (ii) to the <quasi-coherent sheaf> $\mathcal N^j\mathcal F$. Since $X_{\mathrm{red}}$ is an <affine scheme>, <vanishing of quasi-coherent cohomology on an affine scheme> gives
$$
H^1(X,\mathcal Q_j)\cong H^1(X_{\mathrm{red}},\mathcal Q_j)=0.
$$
The <short exact sequences>
$$
0\longrightarrow\mathcal N^{j+1}\mathcal F\longrightarrow\mathcal N^j\mathcal F\longrightarrow\mathcal Q_j\longrightarrow0
$$
yield the segments of the <long exact sequence in sheaf cohomology>
$$
H^1(X,\mathcal N^{j+1}\mathcal F)\longrightarrow H^1(X,\mathcal N^j\mathcal F)\longrightarrow H^1(X,\mathcal Q_j)=0.
$$
Starting from $\mathcal N^m\mathcal F=0$, descending <mathematical induction> makes every middle group zero, finally giving $H^1(X,\mathcal F)=0$.
This holds for every <quasi-coherent sheaf> $\mathcal F$. Since a <Noetherian scheme> is a <quasi-compact scheme>, the <Serre criterion for affineness> in (a) applies and proves
$$
\boxed{X_{\mathrm{red}}\text{ affine}\Longrightarrow X\text{ affine}.}
$$
This is an instance of the general fact that <affineness is invariant under nilpotent thickening>.
Back to article page
