Over a nonperfect field, smooth and regular are not equivalent. For example, let \(k_0\) be a field of characteristic \(p > 0\), let \(k = k_0(t)\), and let \(X \subseteq \AA_k^2\) be the curve defined by \(y^2 = x^p - t\).
Show that every local ring of \(X\) is a regular local ring, but that \(X\) is not smooth over \(k\).
Solution
Let \[ A=k[x,y]/(y^2-x^p+t), \qquad X=\Spec A. \]
1. The element\[ u=x^p-y^2\in k_0[x,y] \]is transcendental over \(k_0\).
The polynomial ring \(k_0[x,y]\) is a domain and \(u\) is nonconstant. If a nonzero polynomial \[ F(T)=c_mT^m+\cdots+c_0\in k_0[T] \] satisfied \(F(u)=0\), then regarding \(F(x^p-y^2)\) as a polynomial in \(x\), its highest \(x\)-degree term would be \[ c_mx^{pm}, \] which cannot cancel with any lower power of \(u\). Thus \(F(u)\ne0\).
2. The coordinate ring \(A\) is a localization of the polynomial ring \(k_0[x,y]\).
Write \[ S=k_0[t]\setminus\{0\}. \] Since \(k=k_0(t)=S^{-1}k_0[t]\), we have \[ \begin{aligned} A &=k_0(t)[x,y]/(y^2-x^p+t)\\ &\cong S^{-1}\!\left(k_0[t,x,y]/(t-(x^p-y^2))\right). \end{aligned} \] Eliminating \(t\) identifies the ring inside parentheses with \(k_0[x,y]\), and under this identification an element \(q(t)\in S\) becomes \[ q(x^p-y^2). \] Step 1 shows that none of these elements is zero. Hence \[ \boxed{ A\cong \bigl\{q(x^p-y^2):0\ne q\in k_0[t]\bigr\}^{-1}k_0[x,y]. } \]
3. Every local ring of \(X\) is regular.
The polynomial ring \(k_0[x,y]\) is regular. Every localization of a regular ring is regular, and step 2 expresses \(A\) as such a localization. Therefore \(A\) is a regular ring.
For every point \(P\in X\), the local ring \[ \OO_{X,P}=A_{\mathfrak p} \] is a further localization of \(A\), hence is a regular local ring.
4. After a purely inseparable field extension, \(X\) acquires a nonregular point.
Let \[ k'=k(\alpha), \qquad \alpha^p=t. \] Then \[ X_{k'} =\Spec k'[x,y]/(y^2-x^p+\alpha^p). \] Put \(v=x-\alpha\). In characteristic \(p\), \[ x^p-\alpha^p=(x-\alpha)^p=v^p, \] so \[ X_{k'}\cong\Spec k'[v,y]/(y^2-v^p). \]
Consider the point \(P=(v,y)=(0,0)\). Its local ring is \[ B=\bigl(k'[v,y]/(y^2-v^p)\bigr)_{(v,y)}. \] This is a hypersurface local ring of dimension one. The defining equation \(y^2-v^p\) lies in \((v,y)^2\), so it imposes no linear relation in the cotangent space. Hence \[ \dim_{k'}\mathfrak m_B/\mathfrak m_B^2=2. \] A noetherian local ring is regular exactly when its embedding dimension equals its Krull dimension. Here \[ 2\ne1, \] so \(B\) is not regular.
This argument includes \(p=2\): then \(y^2-v^2=(y-v)^2\), making the failure of geometric regularity especially visible through nonreducedness.
5. The curve \(X\) is not smooth over \(k\).
Smoothness is preserved by arbitrary base change. If \[ X\longrightarrow\Spec k \] were smooth, then \[ X_{k'}\longrightarrow\Spec k' \] would be smooth as well. A scheme smooth over a field is regular by the smoothness--regularity theorem. But step 4 exhibits a nonregular local ring on \(X_{k'}\). This contradiction proves that \(X\) is not smooth over \(k\).
6. Q.E.D.
Steps <1>1–<1>3 prove that every local ring of \(X\) is regular, while steps <1>4–<1>5 prove that \(X\) fails to be smooth over the imperfect field \(k\).