Purely inseparable field algebras separate regularity from smoothness

The class $\alpha$ of $t$ satisfies ${\alpha }^p=a$, and $L=k[\alpha ]$ is a quotient of $k[t]$, hence of finite type over $k$ by [F3]. The polynomial $t^p-a$ is nonconstant, and [F1] with $F=k$, $c=a$ and $n=1$ makes it irreducible in $k[t]$; by [F2] the quotient $L$ is therefore a field. The structural map $\phi :k\to L$ is a field homomorphism by [F19] and $1_L\mathrm{\ne }{0}_L$ by [F18]; for $x\mathrm{\ne }0$ in $k$ the inverse relation $x{x}^{-1}=1$ of [F18] is preserved by [F19], giving $\phi (x)\phi (x^{-1})=\phi (1)=1\mathrm{\ne }0$, so $\phi (x)\mathrm{\ne }0$. Hence $\phi$ is injective and exhibits $k$ as a subfield of $L$.

For any field $F$, the quotient ring $R_F=F[u]\mathrm{/}(u^p)$ has the classes of $1,u,\dots ,u^{p-1}$ as an $F$-basis, because division by the monic polynomial $u^p$ leaves unique remainders of degree less than $p$. Hence $R_F$ is a commutative $F$-algebra of dimension $p$ over $F$, and it is a Noetherian ring by [F10]. Since $p\ge 2$, the ring $R_F$ is nonzero and $u\mathrm{\ne }0$ in $R_F$.

Let $K\mathrm{/}k$ be a field extension and let $\beta \in K$ satisfy ${\beta }^p=a$. Then $K$ has characteristic $p$, and [F16] with $x=z$, $y=-\beta$ and $n=p$ gives $(z-\beta {)}^p={\sum }_{k=0}^p\left(\genfrac{}{}{0px}{}{p}{k}\right)z^{p-k}(-\beta {)}^k$ in the commutative ring $K[z]$; the intermediate coefficients $\left(\genfrac{}{}{0px}{}{p}{k}\right)$ with $0<k<p$ vanish by [F17], while $\left(\genfrac{}{}{0px}{}{p}{0}\right)=\left(\genfrac{}{}{0px}{}{p}{p}\right)=1$, so $(z-\beta {)}^p=z^p+(-\beta {)}^p=z^p-{\beta }^p=z^p-a$, where $(-\beta {)}^p=-{\beta }^p$ holds in every characteristic.

$X=\mathrm{Spec}L$ is of finite type over $k$: over the affine base $\mathrm{Spec}k$ the source is covered by the single affine chart $\mathrm{Spec}L$, and the ring map $k\to L$ is of finite type by step 1.1, so [F4] applies.

$X$ is locally Noetherian and regular. Since $L$ is a field, [F11] makes $L$ a Noetherian ring, and the one-chart cover $\{\mathrm{Spec}L\}$ witnesses local Noetherianity by [F12]. The field $L$ has the single prime ideal $(0)$, so $\dim L=0$ by [F15]; its maximal ideal is $(0)$, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2=0$ and $\mathrm{edim}L=0=\dim L$, which makes $L$ a regular local ring by [F9]. The single point $(0)$ of $X$ has local ring ${\mathcal{O}}_{X,(0)}\cong L_{(0)}=L$ by [F14], so it is regular by [F13]; being the only point, it makes $X$ regular.

For a field extension $K\mathrm{/}k$, write $X_K$ for the base change of $X$ along $\mathrm{Spec}K\to \mathrm{Spec}k$. The inverse image of the affine open $U=X$ under $X_K\to X$ is $\mathrm{Spec}(L{\otimes }_{k}K)$ by [F6], and it is all of $X_K$; applying [F7] with $A=k$, the variable $t$, the ideal $I=(t^p-a)\subseteq k[t]$ and $C=K$ gives a ring isomorphism $L{\otimes }_{k}K\cong K[z]\mathrm{/}(z^p-a)$.

For any field $F$, the ring $R_F=F[u]\mathrm{/}(u^p)$ is local with unique prime $(u)$. By the basis of step 1.2 every element of $R_F$ has a unique expression $c_0+c_{1}u+\cdots +c_{p-1}{u}^{p-1}$. If $c_0\mathrm{\ne }0$, write the element as $c_0(1+uh)$; then $(uh{)}^p=u^p{h}^p=0$, so $1+uh$ has inverse $1-uh+(uh{)}^2-\cdots +(-uh{)}^{p-1}$ and the element is a unit. If $c_0=0$, the element lies in $(u)$ and is not a unit, because $u\mathrm{\ne }0$ is nilpotent and a nilpotent element of a nonzero commutative ring cannot be a unit. Hence $(u)$ is the unique maximal ideal. Since $u^p=0$, every prime ideal contains $u$ and hence contains $(u)$; and $(u)$ is prime because $R_F\mathrm{/}(u)\cong F$ is a field. So $(u)$ is the only prime ideal.

Fix a field extension $K\mathrm{/}k$ and $\beta \in K$ with ${\beta }^p=a$. Combining steps 2.3 and 1.3 and substituting $u=z-\beta$ gives $K$-algebra isomorphisms $L{\otimes }_{k}K\cong K[z]\mathrm{/}(z^p-a)=K[z]\mathrm{/}((z-\beta {)}^p)\cong K[u]\mathrm{/}(u^p)=R_K$, with $R_K$ as in steps 1.2 and 2.4; in particular, taking $K=L$ and $\beta =\alpha$, which is legitimate by step 1.1, the base change $X_L=\mathrm{Spec}(L{\otimes }_{k}L)$ is isomorphic to $\mathrm{Spec}{R}_L$ by [F8].

For any field $F$, the ring $R_F$ is not a regular local ring. It is nonzero, Noetherian by step 1.2, and local with maximal ideal $(u)$ by step 2.4. As $(u)$ is the only prime ideal, [F15] gives $\dim R_F=0$. Every element of $(u)$ is congruent modulo $(u{)}^2$ to $cu$ for some $c\in F$ by the basis of step 1.2, and $u\mathrm{\notin }(u{)}^2$ because $u$ has basis coefficient $1$ in degree $1$ while every element of $(u{)}^2$ has basis coefficients only in degrees at least $2$; hence $(u)\mathrm{/}(u{)}^2$ is one-dimensional over $F$ with basis the class of $u$, and $\mathrm{edim}{R}_F=1$. Thus $\mathrm{edim}{R}_F=1\mathrm{\ne }0=\dim R_F$, and [F9] shows that $R_F$ is not regular local.

The scheme $X_L$ has a nonregular local ring. By step 3.1, $X_L\cong \mathrm{Spec}{R}_L$, and by step 2.4 the ring $R_L$ is local with unique maximal ideal $(u)$, so $\mathrm{Spec}{R}_L$ consists of the single point $(u)$. Its local ring is ${\mathcal{O}}_{\mathrm{Spec}{R}_L,(u)}\cong (R_L{)}_{(u)}=R_L$ by [F14], the last equality because every element outside $(u)$ is a unit by step 2.4. Step 3.2 says that $R_L$ is not a regular local ring, so this local ring of $X_L$ is not regular.

$X$ is not smooth over $k$. By [F5], the AC-carrying geometric-regularity characterization, $X\to \mathrm{Spec}k$ is smooth only if every local ring of every base change $X_K$, with $K\mathrm{/}k$ a field, is regular. The field extension $K=L$ of step 1.1 and the nonregular local ring of $X_L$ exhibited in step 4.1 contradict that condition, so $X\to \mathrm{Spec}k$ is not smooth. Meanwhile $X$ is of finite type over $k$ by step 2.1 and regular by step 2.2, so an imperfect base field separates regularity from smoothness.

Boundary and hypothesis checks. (i) The hypothesis $a\mathrm{\notin }{k}^p$ is exactly what step 1.1 needs, and by [F20] the existence of some such $a$ is equivalent to imperfection of $k$ in characteristic $p$; a perfect field of characteristic $p$ has no such $a$, so the conclusion of step 5.1 cannot arise there. (ii) If instead $a=b^p$ lies in $k^p$, then $t^p-a=(t-b{)}^p$ and $L\cong k[u]\mathrm{/}(u^p)$ is the nonreduced ring of steps 1.2 and 3.2, which is not even regular; so the hypothesis is used, not decorative. (iii) The nilpotent class $u$ survives: steps 2.3 and 3.1 are isomorphisms of the actual tensor product, and no reduction or radical is taken, so the nonreduced base change is retained. (iv) The extension is genuinely needed: for $K=k$ the base change is $X$ itself, which is regular by step 2.2, and the witness $K=L$ is the field generated over $k$ by one $p$th root of $a$. (v) At $p=2$ the ring $R_F=F[u]\mathrm{/}(u^2)$ is the classical dual-number ring of dimension $0$ and embedding dimension $1$; steps 1.2 through 3.2 divide by nothing except the monic polynomial $u^p$, so characteristic $2$ is included. (vi) AC is declared in [F21] and used only through [F5]; the proof exhibits the single extension $K=L$ and one point, so it makes no simultaneous choice and invokes no dependent choice.