A section of the quotient by the nilradical Let $R$ be a Noetherian commutative unital ring. Does the quotient by nilradical of $R$ admit a section? I suspect that this is true for finitely generated algebras over a field.
 A: Let $R = k[x,y]/(xy^2,x^2y)$. Then $R/\operatorname{nil} R = k[x,y]/(xy)$ and the canonical $k$-algebra morphism
$$r \colon R → R/\operatorname{nil} R$$
has no section that is a ring morphism. Let $s \colon R/\operatorname{nil} R → R$ be a section of $r$. Then we have polynomials $f, g ∈ (xy)$ such that
$$s([x]) = [x + f]\quad\text{and}\quad s([y]) = [y + g].$$
However $(x + f)(y + g) = xy + fy + xg + fg ∈ xy + (xy^2,x^2y)$, so
$$s([x])s([y]) = [x+f][y+g] = [xy] ≠ 0\quad\text{in}~ R.$$
Since $[xy] = 0$ in $R / \operatorname{nil} R$, no such section can be additive and multiplicative at the same time.
A: No. Let $R = \mathbb Z/9\mathbb Z$. Then $N(R)$ is all the residues divisible by 3, i.e. $N(R) = \{0, 3, 6\}$.  Thus $R/N(R) \cong \mathbb Z/3\mathbb Z$. However there are no ring homomorphisms from $\mathbb Z/3\mathbb Z$ to $\mathbb Z/9\mathbb Z$ since if $\varphi$ were such a map then we would have $0 = \varphi(0) = \varphi(1 + 1 + 1) = \varphi(1) + \varphi(1) + \varphi(1) = 1 + 1 + 1 = 3$, a contradiction
