Lemma 29.46.10. Let $A$ be a ring. Let $x, y \in A$.
If $x^3 = y^2$ in $A$, then $A \to B = A[t]/(t^2 - x, t^3 - y)$ induces bijections on residue fields and a universal homeomorphism on spectra.
If there is a prime number $p$ such that $p^ px = y^ p$ in $A$, then $A \to B = A[t]/(t^ p - x, pt - y)$ induces a universal homeomorphism on spectra.
Proof. We will use the criterion of Lemma 29.45.5 to check this. In both cases the ring map is integral. Thus it suffices to show that given a field $k$ and a ring map $\varphi : A \to k$ the $k$-algebra $B \otimes _ A k$ has a unique prime ideal whose residue field is equal to $k$ in case (1) and purely inseparable over $k$ in case (2). See Lemma 29.10.2.
In case (1) set $\lambda = 0$ if $\varphi (x) = 0$ and set $\lambda = \varphi (y)/\varphi (x)$ if not. Then $B = k[t]/(t^2 - \lambda ^2, t^3 - \lambda ^2)$. Thus the result is clear.
In case (2) if the characteristic of $k$ is $p$, then we obtain $\varphi (y) = 0$ and $B = k[t]/(t^ p - \varphi (x))$ which is a local Artinian $k$-algebra whose residue field is either $k$ or a degree $p$ purely inseparable extension of $k$. If the characteristic of $k$ is not $p$, then setting $\lambda = \varphi (y)/p$ we see $B = k[t]/(t - \lambda ) = k$ and we conclude as well. $\square$
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