If $A_{\overline{k}}\cong \frac{\overline{k}[x]}{(x^p)}$, show that $A\cong \frac{k[x]}{(x^p-c)}$ where $k$ is a field of characteristic $p$ Let $p$ be a prime number, $k$ be a field of characteristic $p$ and $A$ be a $k$-algebra.
Assume $A_\overline{k}=A\otimes_k \overline{k}$ is isomorphic to $\frac{\overline{k}[x]}{(x^p)}$, we want to show that $A\cong\frac{k[y]}{(y^p-c)}$ for some $c\in k$.
Facts that I can deduce:

*

*The $R$-algebra $A$ must be finite dimensional of rank $p$.

*We have that $\Omega^1_{A/k}\otimes_A A_\overline{k}\cong \Omega^1_{A_\overline{k}/\overline{k}}=A_\overline{k}\cdot dx$. By tag 05B2, we know $\Omega^1_{A/k}$ is a free $A$-module of rank 1.

*Let $a\in A\subset A\otimes_k \overline{k}$, then $a^p\in A\cap\overline{k}=k$. So $A^p\subset k$.

*If there exists $y\in A$ s.t. $\{dy\}$ is an $A$-basis of $\Omega^1_{A/k}$, then there exists a $k$-derivation $D$ of $A$ into itself s.t. $D(y)=1$. Then we can show $y^{\{0,...,p-1\}}$ is a linearly independent subset of $A$ over $k$ with a contraidiction arugment (if there exists a non-trivial relation with minimal degree, then apply $D$ to reduce the degree). By counting the rank we know it is a basis and we must have $A\cong \frac{k[Y]}{(Y^p-y^p)}$ as desired.

So the question reduces to showing $\Omega^1_{A/k}$ has a basis of the form $\{dy\}$ for some $y\in A$.
Possible generalization: Replace $k$ by a commutative ring $R$ and $k\to\overline{k}$ by a fppf/fpqc ring map $R\to S$. Show that
$$A\otimes_R S\cong \frac{S[x]}{(x^p)}\Rightarrow A\cong \frac{R[y]}{(y^p-c)}$$
for some $c\in R$.
 A: Thanks to Mohan's comment, here is the answer:
Claim: $A$ is local.
Proof: Let $Q$ be the maximal ideal of $A_{\overline{k}}$ and $P:=Q\cap A$. Then $A$ is prime and $\frac{A}{P}\to \frac{A_\overline{k}}{Q}$ is an integral ring extension. So $\frac{A}{P}$ is a field and $P$ is maximal by tag 00GR. To show $P$ is the only maximal ideal it suffices to show every element $a$ of $A\backslash P$ is a unit. Clearly $a\notin Q$ so $a$ is a unit in $A_\overline{k}$ with an inverse $b$, with an integral relation
$$b^m=-(a_{m-1}b^{m-1}+\cdots+a_0)$$where $a_i\in A$. Multiplying with $a^{m-1}$ on both sides we know $b\in A$. So $a$ is a unit in $A$.$\square$
Claim: There exists $y\in A$ s.t. $\{dy\}$ is a basis of $\Omega^1_{A/k}$.
Proof: Clearly $\{da|\ a\in A\}$ generates $\Omega^1_{A/k}$. So it generates $\frac{\Omega^1_{A/k}}{P\Omega^1_{A/k}}=\frac{A}{P}\otimes_A \Omega^1_{A/k}$ over $\frac{A}{P}$, which is a vector space of dimension 1. So $\exists y\in A$ s.t. $\{dy\}$ generates $\frac{\Omega^1_{A/k}}{P\Omega^1_{A/k}}$. By Nakayama, tag 00DV (8), $\{dy\}$ generates $\Omega^1_{A/k}$. Then it must be a basis by tag 05G8.$\square$
